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Page No 372:
Question 1:
Answer:
(i)
Applying Thales' theorem, we get:
AD = 3.6 cm , AB = 10 cm, AE = 4.5 cmâ
DB = 10 3.6 = 6.4 cm
(ii)
(iii)
(iv)
Page No 372:
Question 2:
(i)
Applying Thales' theorem, we get:
AD = 3.6 cm , AB = 10 cm, AE = 4.5 cmâ
DB = 10 3.6 = 6.4 cm
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
(iii)
Page No 372:
Question 3:
(i)
(ii)
(iii)
Answer:
(i) We have:
(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 4.2 = 7 cm
Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.
(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 4 = 5.6 cm
Now,
(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 6.4 = 3.6 cm
Now,
Page No 372:
Question 4:
(i) We have:
(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 4.2 = 7 cm
Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.
(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 4 = 5.6 cm
Now,
(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 6.4 = 3.6 cm
Now,
Answer:
(i)
(ii)
(iii)
(iv)
Hence, BC = 3 + 4.2 = 7.2 cm
Page No 373:
Question 5:
(i)
(ii)
(iii)
(iv)
Hence, BC = 3 + 4.2 = 7.2 cm
Answer:
Given: ABCD is a parallelogram
To prove:
Proof: In â³DMC and â³NMB
DMC =NMB (Vertically opposite angle)
DCM =NBM (Alternate angles)
By AAA- similarity
â³DMC ~ â³NMB
Adding 1 to both sides, we get
Page No 373:
Question 6:
Given: ABCD is a parallelogram
To prove:
Proof: In â³DMC and â³NMB
DMC =NMB (Vertically opposite angle)
DCM =NBM (Alternate angles)
By AAA- similarity
â³DMC ~ â³NMB
Adding 1 to both sides, we get
Answer:
Let the trapezium âbe ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.
In
Applying Thales' theorem, we get:
Thus. EF is parallel to both AB and DC.
This completes the proof.
Page No 373:
Question 7:
Let the trapezium âbe ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.
In
Applying Thales' theorem, we get:
Thus. EF is parallel to both AB and DC.
This completes the proof.
Answer:
In trapezium ABCD, and the diagonals AC and BD intersect at O.
Therefore,
Page No 373:
Question 8:
In trapezium ABCD, and the diagonals AC and BD intersect at O.
Therefore,
Answer:
In â³ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN (âµBM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In â³ABC,
∠A + ∠B + ∠C = 180o .....(1)
(Angle Sum Property of triangle)
Again In In â³AMN,
∠A +∠AMN + ∠ANM = 180o ......(2)
(Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B = 2∠AMN
⇒∠B = ∠AMN
Since, ∠B and ∠AMN are corresponding angles.
∴ MNâ¥BC
Page No 373:
Question 9:
In â³ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN (âµBM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In â³ABC,
∠A + ∠B + ∠C = 180o .....(1)
(Angle Sum Property of triangle)
Again In In â³AMN,
∠A +∠AMN + ∠ANM = 180o ......(2)
(Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B = 2∠AMN
⇒∠B = ∠AMN
Since, ∠B and ∠AMN are corresponding angles.
∴ MNâ¥BC
Answer:
Similarly, applying Thales' theorem in , we get:
Applying the converse of Thales' theorem, we conclude that .
This completes the proof.
Page No 373:
Question 10:
Similarly, applying Thales' theorem in , we get:
Applying the converse of Thales' theorem, we conclude that .
This completes the proof.
Answer:
Join BX and CX.
It is given that BC is bisected at D.
∴ BD = DC â
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO
and
Applying Thales' theorem in ABX, we get:
From (1) and (2), we have:
Applying the converse of Thales' theorem in .
This completes the proof.
Page No 373:
Question 11:
Join BX and CX.
It is given that BC is bisected at D.
∴ BD = DC â
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO
and
Applying Thales' theorem in ABX, we get:
From (1) and (2), we have:
Applying the converse of Thales' theorem in .
This completes the proof.
Answer:
Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = AC ...(i)
Also, it is given that CQ = AC ...(ii)
Dividing equation (ii) by (i), we get:
or, CQ =
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in
In .
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.
Page No 373:
Question 12:
Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = AC ...(i)
Also, it is given that CQ = AC ...(ii)
Dividing equation (ii) by (i), we get:
or, CQ =
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in
In .
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.
Answer:
Given:
AD = AE ...(i)
AB = AC ...(ii)
Subtracting AD from both sides, we get:
AB AD = AC AD
AB AD = AC AE (Since, AD = AE)
BD = EC ...(iii)
Dividing equation (i) by equation (iii), we get:
Applying the converse of Thales' theorem, DEBC
Hence, quadrilateral BCED is cyclic.
Therefore, B,C,E and D are concyclic points.
Page No 374:
Question 13:
Given:
AD = AE ...(i)
AB = AC ...(ii)
Subtracting AD from both sides, we get:
AB AD = AC AD
AB AD = AC AE (Since, AD = AE)
BD = EC ...(iii)
Dividing equation (i) by equation (iii), we get:
Applying the converse of Thales' theorem, DEBC
Hence, quadrilateral BCED is cyclic.
Therefore, B,C,E and D are concyclic points.
Answer:
In triangle âBQP, BR bisects angle B.
Applying angle bisector theorem, we get:
This completes the proof.
Page No 399:
Question 1:
In triangle âBQP, BR bisects angle B.
Applying angle bisector theorem, we get:
This completes the proof.
Answer:
(i)
We have:
Therefore, by AAA similarity theorem,
(ii)
We have:
But, (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.
(iii)
We have:
Also,
Therefore, by SAS similarity theorem, .
(iv)
We have
Therefore, by SSS similarity theorem,
(v)
Therefore, by AA similarity theorem,
Page No 400:
Question 2:
(i)
We have:
Therefore, by AAA similarity theorem,
(ii)
We have:
But, (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.
(iii)
We have:
Also,
Therefore, by SAS similarity theorem, .
(iv)
We have
Therefore, by SSS similarity theorem,
(v)
Therefore, by AA similarity theorem,
Answer:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
In , we have:
(iii)
It is given that
Therefore,
(iv)
Again,
Therefore,
Page No 400:
Question 3:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
In , we have:
(iii)
It is given that
Therefore,
(iv)
Again,
Therefore,
Answer:
(i) Let OA be x cm.
Hence, OA = 5.6 cm
(ii) Let OD be y cm
Hence, DO = 4 cm
Page No 401:
Question 4:
(i) Let OA be x cm.
Hence, OA = 5.6 cm
(ii) Let OD be y cm
Hence, DO = 4 cm
Answer:
Given:
ADE = ABC and
Let DE be x cm
Therefore, by AA similarity theorem,
Hence, DE = 2.8 cm
Page No 401:
Question 5:
Given:
ADE = ABC and
Let DE be x cm
Therefore, by AA similarity theorem,
Hence, DE = 2.8 cm
Answer:
It is given that triangles ABC and PQR are similar.
Therefore,
Page No 401:
Question 6:
It is given that triangles ABC and PQR are similar.
Therefore,
Answer:
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.
Let the perimeter of âABC be x cm.
Therefore,
Thus, the perimeter of âABC is 35 cm.
Page No 401:
Question 7:
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.
Let the perimeter of âABC be x cm.
Therefore,
Thus, the perimeter of âABC is 35 cm.
Answer:
Page No 401:
Question 8:
Answer:
It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.
Hence, BC = 8.1 cm
Page No 401:
Question 9:
It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.
Hence, BC = 8.1 cm
Answer:
It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.
CD = cm
Page No 401:
Question 10:
It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.
CD = cm
Answer:
We have:
This completes the proof.
Page No 402:
Question 11:
We have:
This completes the proof.
Answer:
We have:
DA BC
(Alternate angles)
(AA similarity theorem)
or,
This completes the proof.
Page No 402:
Question 12:
We have:
DA BC
(Alternate angles)
(AA similarity theorem)
or,
This completes the proof.
Answer:
In , we have:
This completes the proof.
Page No 402:
Question 13:
In , we have:
This completes the proof.
Answer:
Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m
Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.
In , we have:
Therefore, by AA similarity theorem, we get:
Therefore, PQ = 36 m
Hence, the height of the tower is 36 m.
Page No 402:
Question 14:
Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m
Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.
In , we have:
Therefore, by AA similarity theorem, we get:
Therefore, PQ = 36 m
Hence, the height of the tower is 36 m.
Answer:
Disclaimer: It should be instead of âACP ∼ âBCQ
It is given that ABC is an isosceles triangle.
Therefore,
CA = CB
Also,
Thus, by SAS similarity theorem, we get:
This completes the proof.
Page No 402:
Question 15:
Disclaimer: It should be instead of âACP ∼ âBCQ
It is given that ABC is an isosceles triangle.
Therefore,
CA = CB
Also,
Thus, by SAS similarity theorem, we get:
This completes the proof.
Answer:
We have:
Page No 402:
Question 16:
We have:
Answer:
So, PQ|| BC, and .....(1)
Similarly, In â³ADC, .....(2)
Now, In â³BCD, .....(3)
Similarly, In â³ABD, .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.
Page No 402:
Question 17:
So, PQ|| BC, and .....(1)
Similarly, In â³ADC, .....(2)
Now, In â³BCD, .....(3)
Similarly, In â³ABD, .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.
Answer:
Given: AB and CD are two chords
To Prove:
Proof: In
(Vertically Opposite angles)
(Angles in the same segment are equal)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
Page No 403:
Question 18:
Given: AB and CD are two chords
To Prove:
Proof: In
(Vertically Opposite angles)
(Angles in the same segment are equal)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
Answer:
Given: AB and CD are two chords
To Prove:
Proof:
.....(1)
(Opposite angles of a cyclic quadrilateral are supplementary)
....(2)
(Linear Pair Angles)
Using (1) and (2), we get
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Page No 403:
Question 19:
Given: AB and CD are two chords
To Prove:
Proof:
.....(1)
(Opposite angles of a cyclic quadrilateral are supplementary)
....(2)
(Linear Pair Angles)
Using (1) and (2), we get
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Answer:
We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In â³BDC, we get
â³CQD ∼ â³DQB
Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
(b)
Similarly, â³APD ∼ â³DPB
Page No 403:
Question 20:
We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In â³BDC, we get
â³CQD ∼ â³DQB
Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
(b)
Similarly, â³APD ∼ â³DPB
Answer:
Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DC = and QM = MR = ...(1)
Now,
ΔABC ~ ΔPQR
As we know, corresponding sides of similar triangles are proportional.
Thus, ...(2)
Also, ...(3)
From (1) and (2), we get
Now, in ΔABD and ΔPQM
By SAS similarity,
ΔABD ~ ΔPQM
Therefore, .
Hence, .
Page No 417:
Question 1:
Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DC = and QM = MR = ...(1)
Now,
ΔABC ~ ΔPQR
As we know, corresponding sides of similar triangles are proportional.
Thus, ...(2)
Also, ...(3)
From (1) and (2), we get
Now, in ΔABD and ΔPQM
By SAS similarity,
ΔABD ~ ΔPQM
Therefore, .
Hence, .
Answer:
It is given that .
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Hence, BC = 11.2 cm
Page No 417:
Question 2:
It is given that .
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Hence, BC = 11.2 cm
Answer:
It is given that .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Hence, QR = 6 cm
Page No 417:
Question 3:
It is given that .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Hence, QR = 6 cm
Answer:
Hence, QR = 6 cm
Page No 417:
Question 4:
Hence, QR = 6 cm
Answer:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.
Hence, the longest side of the smaller triangle is 22 cm.
Page No 417:
Question 5:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.
Hence, the longest side of the smaller triangle is 22 cm.
Answer:
It is given that âABC ∼ âDEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the altitude of âABC be AP, drawn from A to BC to meet BC at P and the altitude of âDEF be DQ, drawn from D to meet EF at Q.
Then,
Hence, the altitude of âDEF is 3.5 cm
Page No 417:
Question 6:
It is given that âABC ∼ âDEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the altitude of âABC be AP, drawn from A to BC to meet BC at P and the altitude of âDEF be DQ, drawn from D to meet EF at Q.
Then,
Hence, the altitude of âDEF is 3.5 cm
Answer:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectivelyâ.
It is given that .
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Hence, the ratio of their areas is 4 : 9
Page No 418:
Question 7:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectivelyâ.
It is given that .
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Hence, the ratio of their areas is 4 : 9
Answer:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
Hence, the altitude of the other triangle is 4.9 cm.
Page No 418:
Question 8:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
Hence, the altitude of the other triangle is 4.9 cm.
Answer:
Let the two triangles be ABC and PQR with medians AM and PN, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.
Hence, the median of the larger triangle is 7 cm.
Page No 418:
Question 9:
Let the two triangles be ABC and PQR with medians AM and PN, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.
Hence, the median of the larger triangle is 7 cm.
Answer:
We have:
Also,
By SAS similarity , we can conclude that âAPQâABC.
Hence proved.
Page No 418:
Question 10:
We have:
Also,
By SAS similarity , we can conclude that âAPQâABC.
Hence proved.
Answer:
It is given that DE ⥠BC
By AA similarity , we can conclude that .
Hence, area of triangle ABC is 60 cm2.â
Page No 418:
Question 11:
It is given that DE ⥠BC
By AA similarity , we can conclude that .
Hence, area of triangle ABC is 60 cm2.â
Answer:
In , we have:
By AA similarity, we can conclude that .
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.
Hence, the ratio of areas of both the triangles is 169 : 25
Page No 418:
Question 12:
In , we have:
By AA similarity, we can conclude that .
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.
Hence, the ratio of areas of both the triangles is 169 : 25
Answer:
It is given that DE BC.
Applying AA similarity theorem, we can conclude that .
Page No 418:
Question 13:
It is given that DE BC.
Applying AA similarity theorem, we can conclude that .
Answer:
It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE BC.
Hence, by B.P.T., we get:
Also, .
Applying SAS similarity theorem, we can conclude that .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Page No 441:
Question 1:
It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE BC.
Hence, by B.P.T., we get:
Also, .
Applying SAS similarity theorem, we can conclude that .
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Answer:
For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
Thus, the given triangle is not right-angled.
(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
Thus, the given triangle is a right-angled.
(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
Thus, the given triangle is right-angled.
(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
Thus, the given triangle is not right-angled.
(v)
p = (a 1) cm, q = 2 cm and r = (a + 1) cm
Then,
Thus, the given triangle is right-angled.
Page No 441:
Question 2:
For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
Thus, the given triangle is not right-angled.
(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
Thus, the given triangle is a right-angled.
(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
Thus, the given triangle is right-angled.
(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
Thus, the given triangle is not right-angled.
(v)
p = (a 1) cm, q = 2 cm and r = (a + 1) cm
Then,
Thus, the given triangle is right-angled.
Answer:
Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.
We need to find AC.
In right-angled triangle ABC, we have:
Hence, the man is 170 m away from the starting point.
Page No 441:
Question 3:
Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.
We need to find AC.
In right-angled triangle ABC, we have:
Hence, the man is 170 m away from the starting point.
Answer:
Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right DEF, we have:
DE = 10 m, EF = 24 m
Hence, the man is 26 m away from the starting point.
Page No 441:
Question 4:
Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right DEF, we have:
DE = 10 m, EF = 24 m
Hence, the man is 26 m away from the starting point.
Answer:
Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:
Hence, the distance of the foot of the ladder from the building is 5 m
Page No 441:
Question 5:
Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:
Hence, the distance of the foot of the ladder from the building is 5 m
Answer:
Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:
Hence, the length of the ladder is 25 m.
Page No 441:
Question 6:
Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:
Hence, the length of the ladder is 25 m.
Answer:
Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m and AC = 5 m
We need to find AD, the distance between their tops.
Applying Pythagoras theorem in right-angled triangle ACD, we have: â
Hence, the distance between the tops of the two poles is 13 m.
Page No 442:
Question 7:
Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m and AC = 5 m
We need to find AD, the distance between their tops.
Applying Pythagoras theorem in right-angled triangle ACD, we have: â
Hence, the distance between the tops of the two poles is 13 m.
Answer:
Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the stake should be driven m far from the base of the pole.
Page No 442:
Question 8:
Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the stake should be driven m far from the base of the pole.
Answer:
Applying Pythagoras theorem in right-angled triangle POR, we have:
In â PQR,
Therefore, by applying Pythagoras theorem, we can say that âPQR is right-angled at P.
Page No 442:
Question 9:
Applying Pythagoras theorem in right-angled triangle POR, we have:
In â PQR,
Therefore, by applying Pythagoras theorem, we can say that âPQR is right-angled at P.
Answer:
It is given that is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
are right-angled triangles.
Applying Pythagoras theorem, we have:
Hence,
BC = 2(BD) = 2 = 24 cm
Page No 442:
Question 10:
It is given that is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
are right-angled triangles.
Applying Pythagoras theorem, we have:
Hence,
BC = 2(BD) = 2 = 24 cm
Answer:
In isosceles â ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC =
Applying Pythagoras theorem in right-angled âABD, we have:
Page No 442:
Question 11:
In isosceles â ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC =
Applying Pythagoras theorem in right-angled âABD, we have:
Answer:
Let AD, BE and CF be the altitudes of âABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.
In right-angled âABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
Similarly,
BE = units and CF = units
Page No 442:
Question 12:
Let AD, BE and CF be the altitudes of âABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.
In right-angled âABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
Similarly,
BE = units and CF = units
Answer:
Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
Hence, the height of the given triangle is 6 cm.
Page No 442:
Question 13:
Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
Hence, the height of the given triangle is 6 cm.
Answer:
Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm
Applying Pythagoras theorem in right-angled triangle ABC, we get:
Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm
Page No 442:
Question 14:
Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm
Applying Pythagoras theorem in right-angled triangle ABC, we get:
Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm
Answer:
Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the rhombus is 13 cm.
Page No 442:
Question 15:
Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the rhombus is 13 cm.
Answer:
In right-angled triangle AEB, applying Pythagoras theorem, we have:
...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:
...(ii)
This completes the proof.
Page No 442:
Question 16:
In right-angled triangle AEB, applying Pythagoras theorem, we have:
...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:
...(ii)
This completes the proof.
Answer:
Given: ∠ACB = 900 and CD ⊥ AB
To Prove:
Proof:
In
(Given)
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
In
(Given)
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Dividing (2) by (1), we get
Page No 442:
Question 17:
Given: ∠ACB = 900 and CD ⊥ AB
To Prove:
Proof:
In
(Given)
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
In
(Given)
(Common)
By AA similarity-criterion
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Dividing (2) by (1), we get
Answer:
(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:
(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
(iii)
Adding (i) and (ii), we get:
(iv)
Subtracting (ii) from (i), we get:
Page No 443:
Question 18:
(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:
(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
(iii)
Adding (i) and (ii), we get:
(iv)
Subtracting (ii) from (i), we get:
Answer:
Draw AEBC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:
Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD
Page No 443:
Question 19:
Draw AEBC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:
Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD
Answer:
Applying Pythagoras theorem in right-angled triangle ABC, we get:
...(i)
Page No 443:
Question 20:
Applying Pythagoras theorem in right-angled triangle ABC, we get:
...(i)
Answer:
Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in hours =
Distance covered by plane B in hours =
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the distance between two planes after hours is m.
Page No 443:
Question 21:
Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in hours =
Distance covered by plane B in hours =
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the distance between two planes after hours is m.
Answer:
(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
(c) Adding (2) and (4), we get
Page No 443:
Question 22:
(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
(c) Adding (2) and (4), we get
Answer:
Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In â³BMC
By using Pythagoras theorem, we have
BC2 = CM2 + MB2
= (2.4)2 + (1.8)2
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In â³BC'M
By using Pythagoras theorem, we have
C'M2 = BC'2 − MB2
= (2.4)2 − (1.8)2
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m
Page No 446:
Question 1:
Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In â³BMC
By using Pythagoras theorem, we have
BC2 = CM2 + MB2
= (2.4)2 + (1.8)2
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In â³BC'M
By using Pythagoras theorem, we have
C'M2 = BC'2 − MB2
= (2.4)2 − (1.8)2
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m
Answer:
The two triangles are similar if and only if
1. The corresponding sides are in proportion.
2. The corresponding angles are equal.
Page No 446:
Question 2:
The two triangles are similar if and only if
1. The corresponding sides are in proportion.
2. The corresponding angles are equal.
Answer:
If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.
Page No 446:
Question 3:
If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.
Answer:
If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Page No 446:
Question 4:
If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Answer:
The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.
Page No 446:
Question 5:
The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.
Answer:
If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.
Page No 446:
Question 6:
If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.
Answer:
If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.
Page No 446:
Question 7:
If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.
Answer:
Page No 446:
Question 8:
Answer:
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
Page No 446:
Question 9:
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
Answer:
The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.
Page No 446:
Question 10:
The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.
Answer:
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle
Page No 446:
Question 11:
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle
Answer:
By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In â³ABC and â³EFD
∠ABC = ∠EFD (Opposite angles of a parallelogram)
∠BCA = ∠EDF (Opposite angles of a parallelogram)
By AA similarity criterion, â³ABC ∼ â³EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the ratio of the areas of â³DEF and â³ABC is 1 : 4.
Page No 447:
Question 12:
By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In â³ABC and â³EFD
∠ABC = ∠EFD (Opposite angles of a parallelogram)
∠BCA = ∠EDF (Opposite angles of a parallelogram)
By AA similarity criterion, â³ABC ∼ â³EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the ratio of the areas of â³DEF and â³ABC is 1 : 4.
Answer:
Now, In â³ABC and â³PQR
∠A = ∠P = 700 (Given)
By SAS similarity criterion, â³ABC ∼ â³PQR
Page No 447:
Question 13:
Now, In â³ABC and â³PQR
∠A = ∠P = 700 (Given)
By SAS similarity criterion, â³ABC ∼ â³PQR
Answer:
When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, â³ABC ∼ â³DEF
Page No 447:
Question 14:
When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, â³ABC ∼ â³DEF
Answer:
In â³ADE and â³ABC
∠ADE = ∠ABC (Corresponding angles in DEâ¥BC)
∠AED = ∠ACB (Corresponding angles in DEâ¥BC)
By AA similarity criterion, â³ADE ∼ â³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
Hence, the value of x is 2.
Page No 447:
Question 15:
In â³ADE and â³ABC
∠ADE = ∠ABC (Corresponding angles in DEâ¥BC)
∠AED = ∠ACB (Corresponding angles in DEâ¥BC)
By AA similarity criterion, â³ADE ∼ â³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
Hence, the value of x is 2.
Answer:
Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
Page No 447:
Question 16:
Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
Answer:
We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
Hence, the length of the altitude of an equilateral triangle of side 2a cm is cm.
Page No 447:
Question 17:
We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
Hence, the length of the altitude of an equilateral triangle of side 2a cm is cm.
Answer:
We have â³ABC ∼ â³DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Page No 447:
Question 18:
We have â³ABC ∼ â³DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Answer:
In â³AOB and COD
∠ABO = ∠CDO (Alternte angles in ABâ¥CD)
∠AOB = ∠COD (Vertically opposite angles)
By AA similarity criterion, â³AOB ∼ â³COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Page No 447:
Question 19:
In â³AOB and COD
∠ABO = ∠CDO (Alternte angles in ABâ¥CD)
∠AOB = ∠COD (Vertically opposite angles)
By AA similarity criterion, â³AOB ∼ â³COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Answer:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Page No 447:
Question 20:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Answer:
We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
Now, In right triangle ADC
By using Pythagoras theorem, we have
Page No 447:
Question 21:
We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
Now, In right triangle ADC
By using Pythagoras theorem, we have
Answer:
Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
∴AB2 = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm
Page No 447:
Question 22:
Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
∴AB2 = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm
Answer:
If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, â³DEF ∼ â³GHK
∴∠E = ∠H = 57â
Now, In â³DEF
∠D + ∠E + ∠F = 180â (Angle sum property of triangle)
⇒ ∠F = 180â − 48â − 57â = 75â
Page No 447:
Question 23:
If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, â³DEF ∼ â³GHK
∴∠E = ∠H = 57â
Now, In â³DEF
∠D + ∠E + ∠F = 180â (Angle sum property of triangle)
⇒ ∠F = 180â − 48â − 57â = 75â
Answer:
We have
AM : MB = 1 : 2
Adding 1 to both sides, we get
Now, In â³AMN and â³ABC
∠AMN = ∠ABC (Corresponding angles in MNâ¥BC)
∠ANM = ∠ACB (Corresponding angles in MNâ¥BC)
By AA similarity criterion, â³AMN ∼ â³ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Page No 447:
Question 24:
We have
AM : MB = 1 : 2
Adding 1 to both sides, we get
Now, In â³AMN and â³ABC
∠AMN = ∠ABC (Corresponding angles in MNâ¥BC)
∠ANM = ∠ACB (Corresponding angles in MNâ¥BC)
By AA similarity criterion, â³AMN ∼ â³ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Answer:
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, â³BMP ∼ â³CNR
Perimeter of â³CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm
Page No 447:
Question 25:
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, â³BMP ∼ â³CNR
Perimeter of â³CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm
Answer:
We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
Page No 448:
Question 26:
We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
Answer:
In right triangle SOW
By using Pythagoras theorem, we have
Hence, the man is 37 m away from the starting point.
Page No 448:
Question 27:
In right triangle SOW
By using Pythagoras theorem, we have
Hence, the man is 37 m away from the starting point.
Answer:
Let DC = x
∴ BD = a − x
By using angle bisector theore in â³ABC, we have
Now,
Page No 448:
Question 28:
Let DC = x
∴ BD = a − x
By using angle bisector theore in â³ABC, we have
Now,
Answer:
In â³AMN and â³ABC
∠AMN = ∠ABC = 760 (Given)
∠A = ∠A (Common)
By AA similarity criterion, â³AMN ∼ â³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
Page No 448:
Question 29:
In â³AMN and â³ABC
∠AMN = ∠ABC = 760 (Given)
∠A = ∠A (Common)
By AA similarity criterion, â³AMN ∼ â³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
Answer:
Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + OB2
= 202 + 212
= 400 + 441
= 841
∴AB2 = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm
Page No 448:
Question 30:
Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + OB2
= 202 + 212
= 400 + 441
= 841
∴AB2 = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm
Answer:
(ii) True
Two circles of any radii are similar to each other.
(i) False
Two rectangles are similar if their corresponding sides are proportional.
(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.
(iv) True
Suppose ABC is a triangle and M, N are
Construction: DE is expanded to F sich that EF = DE
To Prove =
Proof: In â³ADE and â³CEF
AE = EC (E is the mid point of AC)
DE = EF (By construction)
AED = CEF (Vertically Opposite angle)
By SAS criterion, â³ADE ≅ â³CEF
CF = AD (CPCT)
⇒ BD = CF
∠ADE = ∠EFC (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ⥠CF and BD ⥠CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
âµDF = BC and BD ⥠CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
(v) False
In â³ABC, AB = 6 cm, ∠A = 45â and AC = 8 cm and in â³DEF, DF = 9 cm, ∠D = 45â and DE = 12 cm, then â³ABC ∼ â³DEF.
In â³ABC and â³DEF
∠A = ∠D = 45â
So â³ABC is not similar to â³DEF
(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.
(vii) True
Given: â³ABC ∼ â³DEF
To Prove =
Proof: In â³ABP and â³DEQ
∠BAP = ∠EDQ (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E (â³ABC ∼ â³DEF)
By AA criterion, â³ABP ∼ â³DEQ
.....(1)
Since, â³ABC ∼ â³DEF
(viii) True
Given: â³ABC ∼ â³DEF
To Prove =
Proof: In â³ABP and â³DEQ
∠B = ∠E (âµâ³ABC ∼ â³DEF)
âµâ³ABC ∼ â³DEF
By SAS criterion, â³ABP ∼ â³DEQ
.....(1)
Since, â³ABC ∼ â³DEF
(ix) True
Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ⥠AB ⥠DC and PR ⥠BC ⥠AD
To prove: OA2 + OC2 = OB2 + OD2
Proof:
OA2 + OC2 = (AS2 + OS2) + (OQ2 + QC2) [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ2 + OS2) + (OQ2 + DS2)
= (BQ2 + OQ2) + (OS2 + DS2) [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB2 + OD2
Hence, LHS = RHS
(x) True
Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
Page No 451:
Question 1:
(ii) True
Two circles of any radii are similar to each other.
(i) False
Two rectangles are similar if their corresponding sides are proportional.
(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.
(iv) True
Suppose ABC is a triangle and M, N are
Construction: DE is expanded to F sich that EF = DE
To Prove =
Proof: In â³ADE and â³CEF
AE = EC (E is the mid point of AC)
DE = EF (By construction)
AED = CEF (Vertically Opposite angle)
By SAS criterion, â³ADE ≅ â³CEF
CF = AD (CPCT)
⇒ BD = CF
∠ADE = ∠EFC (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ⥠CF and BD ⥠CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
âµDF = BC and BD ⥠CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
(v) False
In â³ABC, AB = 6 cm, ∠A = 45â and AC = 8 cm and in â³DEF, DF = 9 cm, ∠D = 45â and DE = 12 cm, then â³ABC ∼ â³DEF.
In â³ABC and â³DEF
∠A = ∠D = 45â
So â³ABC is not similar to â³DEF
(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.
(vii) True
Given: â³ABC ∼ â³DEF
To Prove =
Proof: In â³ABP and â³DEQ
∠BAP = ∠EDQ (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E (â³ABC ∼ â³DEF)
By AA criterion, â³ABP ∼ â³DEQ
.....(1)
Since, â³ABC ∼ â³DEF
(viii) True
Given: â³ABC ∼ â³DEF
To Prove =
Proof: In â³ABP and â³DEQ
∠B = ∠E (âµâ³ABC ∼ â³DEF)
âµâ³ABC ∼ â³DEF
By SAS criterion, â³ABP ∼ â³DEQ
.....(1)
Since, â³ABC ∼ â³DEF
(ix) True
Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ⥠AB ⥠DC and PR ⥠BC ⥠AD
To prove: OA2 + OC2 = OB2 + OD2
Proof:
OA2 + OC2 = (AS2 + OS2) + (OQ2 + QC2) [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ2 + OS2) + (OQ2 + DS2)
= (BQ2 + OQ2) + (OS2 + DS2) [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB2 + OD2
Hence, LHS = RHS
(x) True
Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
Answer:
(c) 26 m
Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
Page No 451:
Question 2:
(c) 26 m
Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
Answer:
(b) 10 m
Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:
Page No 451:
Question 3:
(b) 10 m
Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:
Answer:
(c) 1.5 m
Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
Page No 451:
Question 4:
(c) 1.5 m
Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
Answer:
(d) 30 m
Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
âDE = ?
Now, in right-angled triangles ABC and DEF, we have:
Page No 451:
Question 5:
(d) 30 m
Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
âDE = ?
Now, in right-angled triangles ABC and DEF, we have:
Answer:
Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
Now, In â³ABC and â³ADE
∠ABC= ∠ADE = 900
∠A = ∠A (Common)
By AA-similarity criterion
â³ABC ∼ â³ADE
If two triangles are similar, then the the ratio of their corresponding sides are equal.
Hence, the correct answer is option (d).
Page No 451:
Question 6:
Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
Now, In â³ABC and â³ADE
∠ABC= ∠ADE = 900
∠A = ∠A (Common)
By AA-similarity criterion
â³ABC ∼ â³ADE
If two triangles are similar, then the the ratio of their corresponding sides are equal.
Hence, the correct answer is option (d).
Answer:
(a) 7 m
Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
Page No 451:
Question 7:
(a) 7 m
Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
Answer:
Now, In right triangle MOP
By using Pythagoras theorem, we have
Now, In right triangle MPN
By using Pythagoras theorem, we have
Hence, the correct answer is option (b).
Page No 451:
Question 8:
Now, In right triangle MOP
By using Pythagoras theorem, we have
Now, In right triangle MPN
By using Pythagoras theorem, we have
Hence, the correct answer is option (b).
Answer:
(b) 15 cm, 20 cm
It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x5) cm.
Applying Pythagoras theorem, we get:
Now,
x 5 = 20 5 = 15 cm
Page No 451:
Question 9:
(b) 15 cm, 20 cm
It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x5) cm.
Applying Pythagoras theorem, we get:
Now,
x 5 = 20 5 = 15 cm
Answer:
(b)
Let ABC be the equilateral triangle with AD as its altitude from A.
In right-angled triangle ABD, we have:
Page No 452:
Question 10:
(b)
Let ABC be the equilateral triangle with AD as its altitude from A.
In right-angled triangle ABD, we have:
Answer:
(d) 24 cm
In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
Therefore, BC = 2BD = 24 cm
Page No 452:
Question 11:
(d) 24 cm
In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
Therefore, BC = 2BD = 24 cm
Answer:
(a) 3 : 4
In â ABD and âACD, we have:
Now,
Page No 452:
Question 12:
(a) 3 : 4
In â ABD and âACD, we have:
Now,
Answer:
(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:
Hence, AC = 7.5 cm
Page No 452:
Question 13:
(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:
Hence, AC = 7.5 cm
Answer:
By using angle bisector theore in â³ABC, we have
Hence, the correct answer is option (b).
Page No 452:
Question 14:
By using angle bisector theore in â³ABC, we have
Hence, the correct answer is option (b).
Answer:
(b) isosceles
In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
Page No 452:
Question 15:
(b) isosceles
In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
Answer:
(c) 3AB2 = 4AD2
Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:
Page No 452:
Question 16:
(c) 3AB2 = 4AD2
Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:
Answer:
(c) 16 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:
Hence, the length of the second diagonal BD is 16 cm.
Page No 452:
Question 17:
(c) 16 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:
Hence, the length of the second diagonal BD is 16 cm.
Answer:
(b) 13 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the rhombus is 13 cm.
Page No 453:
Question 18:
(b) 13 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the rhombus is 13 cm.
Answer:
(b) trapezium
Diagonals of a trapezium divide each other proportionally.
Page No 453:
Question 19:
(b) trapezium
Diagonals of a trapezium divide each other proportionally.
Answer:
(a) 2
We know that the diagonals of a trapezium are proportional.
Therefore,
Page No 453:
Question 20:
(a) 2
We know that the diagonals of a trapezium are proportional.
Therefore,
Answer:
(a) a parallelogram
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Page No 453:
Question 21:
(a) a parallelogram
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Answer:
(c) isosceles
Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
It is given that AD bisects BC.
Therefore, BD = DC
Therefore, the triangle is isosceles.
Page No 453:
Question 22:
(c) isosceles
Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
It is given that AD bisects BC.
Therefore, BD = DC
Therefore, the triangle is isosceles.
Answer:
(a) 30
We have:
Applying angle bisector theorem, we can conclude that AD bisects A.
Page No 453:
Question 23:
(a) 30
We have:
Applying angle bisector theorem, we can conclude that AD bisects A.
Answer:
(b) 6 cm
It is given that DEBC.
Applying basic proportionality theorem, we have:
Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm
Page No 453:
Question 24:
(b) 6 cm
It is given that DEBC.
Applying basic proportionality theorem, we have:
Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm
Answer:
(b) 4 cm
It is given that DEBC.
Applying basic proportionality theorem, we get:
Page No 454:
Question 25:
(b) 4 cm
It is given that DEBC.
Applying basic proportionality theorem, we get:
Answer:
(c) x = 4
It is given that .
Applying Thales' theorem, we get:
Page No 454:
Question 26:
(c) x = 4
It is given that .
Applying Thales' theorem, we get:
Answer:
(d) 2.1 cm
It is given that DE.
Applying Thales' theorem, we get:
Page No 454:
Question 27:
(d) 2.1 cm
It is given that DE.
Applying Thales' theorem, we get:
Answer:
(b) 5.4 cm
âABC ∼ âDEF
Therefore,
Page No 454:
Question 28:
(b) 5.4 cm
âABC ∼ âDEF
Therefore,
Answer:
(a) 35 cm
âABC ∼ âDEF
Page No 454:
Question 29:
(a) 35 cm
âABC ∼ âDEF
Answer:
(d) 30 cm
Perimeter of ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm
âDEF ∼ âABC
Page No 454:
Question 30:
(d) 30 cm
Perimeter of ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm
âDEF ∼ âABC
Answer:
Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
Now, In â³ABC and â³EBD
∠BED = ∠BAC (Corresponding angles)
∠B = ∠B (Common)
By AA-similarity criterion
â³ABC ∼ â³EBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the correct answer is option (d).
Page No 454:
Question 31:
Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
Now, In â³ABC and â³EBD
∠BED = ∠BAC (Corresponding angles)
∠B = ∠B (Common)
By AA-similarity criterion
â³ABC ∼ â³EBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the correct answer is option (d).
Answer:
(b) DE = 12 cm, F = 100°
Disclaimer: In the question, it should be âABC ∼ âDFE instead of âABC ∼ âDEF.
In triangle ABC,
âABC ∼ âDFE
Page No 454:
Question 32:
(b) DE = 12 cm, F = 100°
Disclaimer: In the question, it should be âABC ∼ âDFE instead of âABC ∼ âDEF.
In triangle ABC,
âABC ∼ âDFE
Answer:
(c) BD ⋅ CD = AD2
Page No 454:
Question 33:
(c) BD ⋅ CD = AD2
Answer:
Since, the square of the longest side is equal to the sum of the square of two sides, so â³ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)
Page No 455:
Question 34:
Since, the square of the longest side is equal to the sum of the square of two sides, so â³ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)
Answer:
(c) B = D
Page No 455:
Question 35:
(c) B = D
Answer:
(b)
In âDEF and âPQR, we have:
Page No 455:
Question 36:
(b)
In âDEF and âPQR, we have:
Answer:
(c) BC ⋅ DE = AB ⋅ EF
âABC ∼ âEDF
Therefore,
Page No 455:
Question 37:
(c) BC ⋅ DE = AB ⋅ EF
âABC ∼ âEDF
Therefore,
Answer:
(b) similar but not congruent
In âABC and âDEF, we have:
Page No 455:
Question 38:
(b) similar but not congruent
In âABC and âDEF, we have:
Answer:
(a) âPQR ∼ âCAB
In âABC and âPQR, we have:
Page No 455:
Question 39:
(a) âPQR ∼ âCAB
In âABC and âPQR, we have:
Answer:
(d) 100°
Page No 455:
Question 40:
(d) 100°
Answer:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the correct answer is option (d).
Page No 455:
Question 41:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
Hence, the correct answer is option (d).
Answer:
(d) 9 : 4
It is given that â ABC ∼ âPQR and .
Therefore,
Page No 455:
Question 42:
(d) 9 : 4
It is given that â ABC ∼ âPQR and .
Therefore,
Answer:
(b) 4:1
In âABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DEBC.
Also, by Basic Proportionality Theorem,
Page No 456:
Question 43:
(b) 4:1
In âABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DEBC.
Also, by Basic Proportionality Theorem,
Answer:
(b) 25 : 49
Page No 456:
Question 44:
(b) 25 : 49
Answer:
(b) 6:7
âABC ∼ âDEF
Page No 456:
Question 45:
(b) 6:7
âABC ∼ âDEF
Answer:
(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
Page No 456:
Question 46:
(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
Answer:
(b) similar to the original triangle
The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.
Page No 456:
Question 47:
(b) similar to the original triangle
The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.
Answer:
(b) 10 cm
Page No 456:
Question 48:
(b) 10 cm
Answer:
(c) isosceles and similar
In âAOC and âODB, we have:
Page No 456:
Question 49:
(c) isosceles and similar
In âAOC and âODB, we have:
Answer:
(d) 90°
Given:
AC = BC
Applying Pythagoras theorem, we conclude that âABC is right angled at C.
or,
Page No 456:
Question 50:
(d) 90°
Given:
AC = BC
Applying Pythagoras theorem, we conclude that âABC is right angled at C.
or,
Answer:
(b) right-angled
We have:
Hence, âABC is a right-angled triangle.
Page No 456:
Question 51:
(b) right-angled
We have:
Hence, âABC is a right-angled triangle.
Answer:
(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:
Page No 457:
Question 52:
(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:
Answer:
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.
Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Page No 457:
Question 53:
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.
Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Answer:
(a) - (s)
Let AE be x.
Therefore, EC = 5.6 x
It is given that DE BC.
Therefore, by B.P.T., we get:
(b) - (q)
(c) - (p)
(d) - (r)
Page No 458:
Question 54:
(a) - (s)
Let AE be x.
Therefore, EC = 5.6 x
It is given that DE BC.
Therefore, by B.P.T., we get:
(b) - (q)
(c) - (p)
(d) - (r)
Answer:
(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
Hence, the man is 10 m away from the starting point.
(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
(c) - (p)
Area of an equilateral triangle with side a =
(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
Page No 462:
Question 1:
(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
Hence, the man is 10 m away from the starting point.
(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
(c) - (p)
Area of an equilateral triangle with side a =
(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
Answer:
(b) 7.5 cm
âABC ∼ âDEF
Page No 462:
Question 2:
(b) 7.5 cm
âABC ∼ âDEF
Answer:
(a) 5.6 cm
DE ⥠BC
Page No 462:
Question 3:
(a) 5.6 cm
DE ⥠BC
Answer:
(b) 13 m
Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have: â
Page No 462:
Question 4:
(b) 13 m
Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have: â
Answer:
(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,
Page No 463:
Question 5:
(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,
Answer:
âABC ∼ âDEF
Page No 463:
Question 6:
âABC ∼ âDEF
Answer:
DE ⥠BC
Page No 463:
Question 7:
DE ⥠BC
Answer:
Let the ladder be AB and BC be the height of the window from the ground.
We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:
Hence, the foot of the ladder is 6 m away from the base of the wall.
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Question 8:
Let the ladder be AB and BC be the height of the window from the ground.
We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:
Hence, the foot of the ladder is 6 m away from the base of the wall.
Answer:
Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:
Hence, the length of the altitude of an equilateral triangle of side 2a cm is cm.
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Question 9:
Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:
Hence, the length of the altitude of an equilateral triangle of side 2a cm is cm.
Answer:
âABC ∼ âDEF
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Question 10:
âABC ∼ âDEF
Answer:
In âAOB and âCOD, we have:
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Question 11:
In âAOB and âCOD, we have:
Answer:
It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.
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Question 12:
It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.
Answer:
Therefore, applying Thales' theorem, we have:
This completes the proof.
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Question 13:
Therefore, applying Thales' theorem, we have:
This completes the proof.
Answer:
Let the triangle be ABC with AD as the bisector of which meets BC at D.
We have to prove:
Draw CE DA, meeting BA produced at E.
CE DA
Therefore,
This completes the proof.
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Question 14:
Let the triangle be ABC with AD as the bisector of which meets BC at D.
We have to prove:
Draw CE DA, meeting BA produced at E.
CE DA
Therefore,
This completes the proof.
Answer:
Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have:
Therefore,
This completes the proof.
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Question 15:
Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have:
Therefore,
This completes the proof.
Answer:
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the given rhombus is 13 cm.
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Question 16:
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
Hence, the length of each side of the given rhombus is 13 cm.
Answer:
Let the two triangles be ABC and PQR.
We have:
,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
; therefore, their corresponding sides will be proportional.
This completes the proof.
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Question 17:
Let the two triangles be ABC and PQR.
We have:
,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
; therefore, their corresponding sides will be proportional.
This completes the proof.
Answer:
This completes the proof.
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Question 18:
This completes the proof.
Answer:
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Question 19:
Answer:
Applying Pythagoras theorem in right-angled triangle ADC, we get:
Applying Pythagoras theorem in right-angled triangle ADB, we get:
This completes the proof.
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Question 20:
Applying Pythagoras theorem in right-angled triangle ADC, we get:
Applying Pythagoras theorem in right-angled triangle ADB, we get:
This completes the proof.
Answer:
From equation (1) and (2), we have:
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