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Page No 5:
Question 1:
Find the distances with the help of the number line given below.
(vi) d(O, E) (vii) d(P, J) (viii) d(Q, B)
Answer:
It is known that, distance between the two points is obtained by subtracting the smaller co-ordinate from larger co-ordinate.
(i) The co-ordinates of points B and E are 2 and 5 respectively. We know that 5 > 2.
∴ d(B, E) = 5 − 2 = 3
(ii) The co-ordinates of points J and A are −2 and 1 respectively. We know that 1 > −2.
∴ d(J, A) = 1 − (−2) = 1 + 2 = 3
(iii) The co-ordinates of points P and C are −4 and 3 respectively. We know that 3 > −4.
∴ d(P, C) = 3 − (−4) = 3 + 4 = 7
(iv) The co-ordinates of points J and H are −2 and −1 respectively. We know that −1 > −2.
∴ d(J, H) = −1 − (−2) = −1 + 2 = 1
(v) The co-ordinates of points K and O are −3 and 0 respectively. We know that 0 > −3.
∴ d(K, O) = 0 − (−3) = 0 + 3 = 3
(vi) The co-ordinates of points O and E are 0 and 5 respectively. We know that 5 > 0.
∴ d(O, E) = 5 − 0 = 5
(vii) The co-ordinates of points P and J are −4 and −2 respectively. We know that −2 > −4.
∴ d(P, J) = −2 − (−4) = −2 + 4 = 2
(viii) The co-ordinates of points Q and B are −5 and 2 respectively. We know that 2 > −5.
∴ d(Q, B) = 2 − (−5) = 2 + 5 = 7
Page No 5:
Question 2:
(i) x = 1, y = 7 (ii) x = 6, y = 2 (iii) x = 3, y = 7
(iv) x = 4, y = 5 (v) x = 3, y = 6 (vi) x = 4, y = 8
Answer:
It is known that, distance between the two points is obtained by subtracting the smaller co-ordinate from larger co-ordinate.
(i) The coordinates of A and B are x and y respectively. We have, x = 1 and y = 7. We know that 7 > 1.
∴ d(A, B) = y − x = 7 − 1 = 6
(ii) The coordinates of A and B are x and y respectively. We have, x = 6 and y = −2. We know that 6 > −2.
∴ d(A, B) = x − y = 6 − (−2) = 6 + 2 = 8
(iii) The coordinates of A and B are x and y respectively. We have, x = −3 and y = 7. We know that 7 > −3.
∴ d(A, B) = y − x = 7 − (−3) = 7 + 3 = 10
(iv) The coordinates of A and B are x and y respectively. We have, x = −4 and y = −5. We know that −4 > −5.
∴ d(A, B) = x − y = −4 − (−5) = −4 + 5 = 1
(v) The coordinates of A and B are x and y respectively. We have, x = −3 and y = −6. We know that −3 > −6.
∴ d(A, B) = x − y = −3 − (−6) = −3 + 6 = 3
(vi) The coordinates of A and B are x and y respectively. We have, x = 4 and y = −8. We know that 4 > −8.
∴ d(A, B) = x − y = 4 − (−8) = 4 + 8 = 12
Page No 5:
Question 3:
(i) d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
(ii) d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
(iii) d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
(iv) d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
(v) d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
(vi) d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Answer:
(i) We have, d(P, R) = 7; d(P, Q) = 10; d(Q, R) = 3
Now, d(P, R) + d(Q, R) = 7 + 3
Or, d(P, R) + d(R, Q) = 10
∴ d(P, Q) = d(P, R) + d(Q, R)
Hence, the points P, R and Q are collinear.
The point R is between P and Q i.e., P-R-Q.
(ii) We have, d(R, S) = 8; d(S, T) = 6; d(R, T) = 4
Now, 8 + 6 = 14, so 8 + 6 ≠ 4; 6 + 4 = 10, so 6 + 4 ≠ 8 and 8 + 4 = 12, so 8 + 4 ​≠ 6
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points R, S and T are non-collinear.
(iii) We have, d(A, B) = 16; d(C, A) = 9; d(B, C) = 7
Now, d(C, A) + d(B, C) = 9 + 7
Or, d(A, C) + d(C, B) = 16
∴ d(A, B) = d(A, C) + d(C, B)
Hence, the points A, C and B are collinear.
The point C is between A and B i.e., A-C-B.
(iv) We have, d(L, M) = 11; d(M, N) = 12; d(N, L) = 8
Now, 11 + 12 = 23, so 11 + 12 ≠ 8; 12 + 8 = 20, so 12 + 8 ≠ 11 and 11 + 8 = 19, so 11 + 8 ​≠ 12
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points L, M and N are non-collinear.
(v) We have, d(X, Y) = 15; d(Y, Z) = 7; d(X, Z) = 8
Now, d(X, Z) + d(Y, Z) = 7 + 8
Or, d(X, Z) + d(Z, Y) = 15
∴ d(X, Y) = d(X, Z) + d(Z, Y)
Hence, the points X, Z and Y are collinear.
The point Z is between X and Y i.e., X-Z-Y.
(vi) We have, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Now, 5 + 8 = 13, so 5 + 8 ≠ 6; 8 + 6 = 14, so 8 + 6 ≠ 5 and 5 + 6 = 11, so 5 + 6 ​≠ 8
Since, the sum of the distances between two pairs of points is not equal to the distance between the third pair of points, so the given points D, E and F are non-collinear.
Page No 5:
Question 4:
On a number line, points A, B and C are such that d(A,C) = 10, d(C,B) = 8 . Find d(A, B) considering all possibilities.
Answer:
There are only two possibilities.
Case 1 : When point C is between the points A and B.
We have, d(A, C) = 10; d(C, B) = 8
Now, d(A, B) = d(A, C) + d(C, B) = 10 + 8
∴ d(A, B) = 18
Case 2 : When point B is between the points A and C.
We have, d(A, C) = 10; d(C, B) = 8
Now, d(A, C) = d(A, B) + d(B, C)
So, d(A, B) = d(A, C) − d(B, C) = 10 − 8
∴ d(A, B) = 2
Page No 5:
Question 5:
Points X, Y, Z are collinear such that d(X,Y) = 17, d(Y,Z) = 8, find d(X,Z) .
Answer:
It is given that the points X, Y and Z are collinear.
We have d(X,Y) = 17; d(Y,Z) = 8.
Now, d(X,Z) = d(X,Y) + d(Y,Z) = 17 + 8
∴ d(X,Z) = 25
Page No 5:
Question 6:
(ii) If R - S - T and l(ST) = 3.7, l(RS) = 2.5, then l(RT) =?
(iii) If X - Y - Z and l(XZ) = 3 , l(XY) = , then l(YZ) =?
Answer:
(i)
We have, l(AC) = 11; l(BC) = 6.5.
Now, l(AC) = l(AB) + l(BC)
So, l(AB) = l(AC) − l(BC) = 11 − 6.5
∴ l(AB) = 4.5
(ii)
We have, l(ST) = 3.7; l(RS) = 2.5.
Now, l(RT) = l(RS) + l(ST) = 3.7 + 2.5
∴ l(RT) = 5.6
(iii)
We have, l(XZ) = ; l(XY) = .
Now, l(XZ) = l(XY) + l(YZ)
So, l(YZ) = l(XZ) − l(XY) = −
∴ l(YZ) =
Page No 5:
Question 7:
Which figure is formed by three non-collinear points ?
Answer:
A triangle is formed by three segments joining three non-collinear points.
A, B and C are three non-collinear points. When A, B and C are joined, we get a ∆ABC.
Page No 7:
Question 1:
| Point | A | B | C | D | E |
| Co-ordinate | 3 | 5 | 2 | 7 | 9 |
(i) seg DE and seg AB (ii) seg BC and seg AD (iii) seg BE and seg AD
Answer:
The given table is,
| Point | A | B | C | D | E |
| Co-ordinate | 3 | 5 | 2 | 7 | 9 |
(i) The co-ordinates of points D and E are −7 and 9 respectively. We know that 9 > −7.
∴ l(DE) = 9 − (−7) = 9 + 7 = 16
The co-ordinates of points A and B are −3 and 5 respectively. We know that 5 > −3.
∴ l(AB) = 5 − (−3) = 5 + 3 = 8
Since, l(DE) ≠ l(AB), so seg DE ≇ seg AB.
(ii) The co-ordinates of points B and C are 5 and 2 respectively. We know that 5 > 2.
∴ l(BC) = 5 − 2 = 3
The co-ordinates of points A and D are −3 and −7 respectively. We know that −3 > −7.
∴ l(AD) = −3 − (−7) = −3 + 7 = 4
Since, l(BC) ≠ l(AD), so seg BC ≇ seg AD.
(iii) The co-ordinates of points B and E are 5 and 9 respectively. We know that 9 > 5.
∴ l(BE) = 9 − 5 = 4
The co-ordinates of points A and D are −3 and −7 respectively. We know that −3 > −7.
∴ l(AD) = −3 − (−7) = −3 + 7 = 4
Since, l(BE) = l(AD), so seg BE ≅ seg AD.
Page No 7:
Question 2:
Point M is the midpoint of seg AB. If AB = 8 then find the length of AM.
Answer:
We have l(AB) = 8.
Since, M is the midpoint of seg AB, then
l(AM) = of l(AB)
∴ l(AM) = × 8 = 4
So, length of AM is 4.
Page No 7:
Question 3:
Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).
Answer:
We have l(CP) = 2.5.
Since, P is the midpoint of seg CD, then
l(CP) = of l(CD)
∴ l(CD) = 2 × l(CP) = 2 × 2.5 = 5
So, length of CD is 5.
Page No 7:
Question 4:
If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Answer:
We have l(AB) = 5 cm; l(BP) = 2 cm; l(AP) = 3.4 cm
We know that 5 > 3.4 > 2.
So, l(AB) > l(AP) > l(BP).
∴ seg AB > seg AP > seg BP.
Page No 8:
Question 5:
(i) Write the name of the opposite ray of ray RP.
(iii) Write the union set of ray PQ and ray QR.
Answer:
(i) Ray RS or Ray RT
(ii) Ray PQ
(iii) Ray QR
(iv) Ray QR, Ray RQ etc.
(v) Ray RQ and Ray RT etc.
(vi) Ray ST and Ray SR etc.
(vii) Point S
Page No 8:
Question 6:
Answer the questions with the help of a given figure.
Answer:
(i) The co-ordinates of points B and C are 2 and 4 respectively. We know that 4 > 2.
∴ d(B, C) = 4 − 2 = 2
The co-ordinates of points B and A are 2 and 0 respectively. We know that 2 > 0.
∴ d(B, A) = 2 − 0 = 2
Since d(B, A) = d(B, C), then points A and C are equidistant from point B.
The co-ordinates of points B and D are 2 and 6 respectively. We know that 6 > 2.
∴ d(B, D) = 6 − 2 = 4
The co-ordinates of points B and P are 2 and −2 respectively. We know that 2 > −2.
∴ d(B, P) = 2 − (−2) = 2 + 2 = 4
Since d(B, D) = d(B, P), then points D and P are equidistant from point B.
(ii) The co-ordinates of points Q and U are −4 and −5 respectively. We know that −4 > −5.
∴ d(Q, U) = −4 − (−5) = −4 + 5 = 1
The co-ordinates of points Q and L are −4 and −3 respectively. We know that −3 > −4.
∴ d(Q, L) = −3 − (−4) = −3 + 4 = 1
Since d(Q, U) = d(Q, L), then points U and L are equidistant from point Q.
The co-ordinates of points Q and R are −4 and −6 respectively. We know that −4 > −6.
∴ d(Q, R) = −4 − (−6) = −4 + 6 = 2
The co-ordinates of points Q and P are −4 and −2 respectively. We know that −2 > −4.
∴ d(Q, P) = −2 − (−4) = −2 + 4 = 2
Since d(Q, R) = d(Q, P), then points R and P are equidistant from point Q.
(iii) The co-ordinates of points U and V are −5 and 5 respectively. We know that 5 > −5.
∴ d(U, V) = 5 − (−5) = 5 + 5 = 10
The co-ordinates of points P and C are −2 and 4 respectively. We know that 4 > −2.
∴ d(P, C) = 4 − (−2) = 4 + 2 = 6
The co-ordinates of points V and B are 5 and 2 respectively. We know that 5 > 2.
∴ d(V, B) = 5 − 2 = 3
The co-ordinates of points U and L are −5 and −3 respectively. We know that −3 > −5.
∴ d(U, L) = −3 − (−5) = −3 + 5 = 2
Page No 11:
Question 1:
Answer:
(i) If a quadrilateral is a parallelogram, then the opposite angles of that quadrilateral are congruent.
(ii) If a quadrilateral is a rectangle, then the diagonals of that quadrilateral are congruent.
(iii) If a triangle is an isosceles, then the segment joining the vertex and the mid point of the base is perpendicular to the base.
Page No 11:
Question 2:
(i) The alternate angles formed by two parallel lines and their transversal are congruent.
Answer:
(i) If the alternate angles made by the transversal with the two lines are congruent, then the lines are parallel.
(ii) If the two parallel lines are intersected by a transversal, then the pair of interior angles are supplementary.
(iii) If the diagonals of a quadrilateral are congruent, then that quadrilateral is a rectangle.
Page No 11:
Question 1:
(A) 12 (B) 8 (C) (D) 20
Answer:
(i) Every segment has one and only one midpoint.
Hence, the correct answer is option (A).
(ii) It is known that, two distinct lines intersect at one point.
Hence, the correct answer is option (C).
(iii) Consider the 3 distinct points as P, Q and R.
Suppose the points P, Q and R are collinear.
So, only one line is determined by the points P, Q and R.
Suppose the points P,Q and R are non collinear.
So, three lines can be determined by the points P, Q and R.
Hence, the correct answer is option(C).
(iv) The co-ordinates of points A and B are 2 and 5 respectively. We know that 5 > ​2.
∴ d(A, B) = 5 − (−2) = 5 + 2 = 7
Hence, the correct answer is option(C).
(v) It is given that, point Q is between point P and point R.
We have, d(P,Q) = 2; d(P,R) = 10
Now, d(P,R) = d(P,Q) + d(Q,R)
∴ d(Q,R) = d(P,R) − d(P,Q) = 10 − 2 = 8
Hence, the correct answer is option (B).
Page No 11:
Question 2:
(i) d(P,Q) + d(Q,R) = d(P,R)
(ii) d(P,R) + d(R,Q) = d(P,Q)
Answer:
The co-ordinates of points P and Q are 3 and −5 respectively. We know that 3 > −5.
Now, d(P, Q) = 3 − (−5) = 3 + 5 = 8
The co-ordinates of points Q and R are −5 and 6 respectively. We know that 6 > −5.
Now, d(Q, R) = 6 − (−5) = 6 + 5 = 11
The co-ordinates of points P and R are 3 and 6 respectively. We know that 6 > 3.
Now, d(P, R) = 6 − 3 = 3
(i) d(P, Q) + d(Q, R) = 8 + 11 = 19; d(P, R) = 3
So, d(P, Q) + d(Q, R) ≠ d(P, R)
Hence, the given statement is false.
(ii) d(P, R) + d(R, Q) = d(P, R) + d(Q, R) = 3 + 11 = 14; d(P, Q) = 8
So, d(P, R) + d(R, Q) ≠ d(P, Q)
Hence, the given statement is false.
(iii) d(R,P) + d(P,Q) = d(P, R) + d(P,Q) = 3 + 8 = 11; d(R,Q) = d(Q, R) = 11
So, d(R,P) + d(P,Q) = d(R,Q)
Hence, the given statement is true.
(iv) d(P,Q) d(P,R) = 8 − 3 = 5; d(Q,R) = 11
So, d(P,Q) d(P,R) ≠ d(Q,R)
Hence, the given statement is false.
Page No 11:
Question 3:
(v) x + 3, x 3
(vi) 25,47
(vii) 80, 85
Answer:
(i) Let the co-ordinates of A and B are 3 and 6 respectively. We know that 6 > 3
d(A, B) = 6 − 3 = 3
(ii) Let the co-ordinates of C and D are −9 and −1 respectively. We know that −1 > −9
d(C, D) = −1 − (−9) = −1 + 9 = 8
(iii) Let the co-ordinates of E and F are −4 and 5 respectively. We know that 5 > −4
d(E, F) = 5 − (−4) = 5 + 4 = 9
(iv) Let the co-ordinates of P and Q are x and −2 respectively. Suppose x > 0, then x > −2.
d(P, Q) = x − (−2) = x + 2
(v) Let the co-ordinates of R and S are x + 3 and x − 3 respectively. Suppose x > 0, then x + 3 > x − 3
d(R, S) = (x + 3) − (x − 3) = x + 3 − x + 3 = 2x
(vi) Let the co-ordinates of L and M are −25 and −47 respectively. We know that −25 > −47
d(L, M) = −25 − (−47) = −25 + 47 = 22
(vii) Let the co-ordinates of G and H are 80 and −85 respectively. We know that 80 > −85
d(G, H) = 80 − (−85) =80 + 85 = 165
Page No 12:
Question 4:
Co-ordinate of point P on a number line is 7. Find the co-ordinates of points on the number line which are at a distance of 8 units from point P.
Answer:
The co-ordinates of point P on the number line is −7. Now, there will be two points, one on the left of point P and the other on the right of point P on the number line which are at a distance of 8 units from point P.
Let the point R is on the right of point P and point Q is on the left of point P each at a distance of 8 units from point P.
The co-ordinate of point R will be larger and co-ordinate of point Q will be smaller in comparison to the co-ordinate of point P.
Now, d(P, R) = 8
So, co-ordinate of R − co-ordinate of P = 8
∴ co-ordinate of R = 8 + co-ordinate of P = 8 + (−7) = 8 − 7 = 1
Also, d(Q, P) = 8
So, co-ordinate of P − co-ordinate of Q = 8
∴ co-ordinate of Q = co-ordinate of P − 8 = −7 − 8 = −15
Hence, the co-ordinates of the required points on the number line which are at a distance of 8 units from the point P are 1 and −15.
Page No 12:
Question 5:
(i) If A - B - C and d(A,C) = 17, d(B,C) = 6.5 then d(A,B) = ?
Answer:
(i)
We have, d(A, C) = 17; d(B, C) = 6.5
Now, d(A, C) = d(A, B) + d(B, C)
So, d(A, B) = d(A, C) − d(B, C) = 17 − 6.5
∴ d(A, B) = 10.5
(ii)
We have, d(P, Q) = 3.4; d(Q, R) = 5.7
Now, d(P, R) = d(P, Q) + d(Q, R) = 3.4 + 5.7
∴ d(P, R) = 9.1
Page No 12:
Question 6:
Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the number line which are at a distance of 7 units from A ?
Answer:
The co-ordinates of point A on the number line is 1. Now, there will be two points, one on the left of point A and the other on the right of point A on the number line which are at a distance of 7 units from point A.
Let the point C is on the right of point A and point B is on the left of point A each at a distance of 7 units from point A.
The co-ordinate of point C will be larger and co-ordinate of point B will be smaller in comparison to the co-ordinate of point A.
Now, d(A, C) = 7
So, co-ordinate of C − co-ordinate of A = 7
∴ co-ordinate of C = 7 + co-ordinate of A = 7 + 1= 8
Also, d(B, A) = 7
So, co-ordinate of A − co-ordinate of B = 7
∴ co-ordinate of B = co-ordinate of A − 7 = 1 − 7 = −6
Hence, the co-ordinates of the required points on the number line which are at a distance of 7 units from the point A are 8 and −6.
Page No 12:
Question 7:
Answer:
(i) If the given quadrilateral is a square, then it must be a rhombus.
(ii) If the given two angles are forming a linear pair, then they are supplementary.
(iii) If the given figure is a triangle, then it is formed by three segments.
(iv) If the given number is having only two divisors, then it is a prime number.
Page No 12:
Question 8:
Answer:
(i) If the given figure is a triangle, then the sum of the measures of its angles is 1800.
(ii) If the given two angles are complement of each other, then the sum of the measures of two angles is 900.
(iii) If the given two lines are parallel, then the corresponding angles formed by a transversal of two lines are congruent.
(iv) If the given number is divisible by 3, then the sum of the digits of the number is divisible by 3.
Page No 12:
Question 9:
Answer:
(i) Antecedent : All the sides of a triangle are congruent.
Consequent : Its all angles are congruent.
(ii) The statement can be written in conditional form as, 'If the given quadrilateral is a parallelogram, then its diagonals bisect each other.
Antecedent : The given quadrilateral is a parallelogram.
Consequent : Its diagonals bisect each other.
Page No 12:
Question 10:
(i) Two equilateral triangles are similar.
Answer:
(i) The given statement can be written in conditional form as, ' If the given two traingles are equilateral, then they are similar.'
Antecedent : The given two triangles are equilateral.
Consequent : They are similar.
Here, ∆ABC and ∆PQR are equilateral triangles, so they are similar to each other.
(ii) Antecedent : The angles in a linear pair are congruent.
Consequent : Each of them is a right angle.
Here, ∠AOC and ∠BOC forming a linear pair are congruent to each other, so each of them is a right angle.
(iii) Antecedent : The altitudes drawn on two sides of a triangle are congruent.
Consequent : Those two sides are congruent.
Here, BL and CM are the altitudes drawn on two sides AC and AB respectively of ∆ABC and are congruent, so side AB is congruent to side AC.
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