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#### Question 1:

In a ∆ABC, if AB = AC and ∠A = 70°, find ∠B and ∠C.

#### Question 2:

The vertical angle of an isosceles triangle is 100°. Find its base angles.

Let the base angles be $x°$.
The sum of the angles of a triangle is 180$°$.

∴ The base angles measure $40°$.

#### Question 3:

In a ∆ABC, if AB = AC and ∠B = 65°, find ∠C and ∠A.

#### Question 4:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

In an isosceles triangle, the base angles are equal to each other.
Let the base angles be $x$$°$.

The vertex angle is $2\left(x+x\right)=2\left(2x\right)=4x$o
Since the sum of the angles of a triangle is 180o, we have:
$x+x+4x=180°\phantom{\rule{0ex}{0ex}}⇒6x=180°\phantom{\rule{0ex}{0ex}}⇒x=\frac{180}{6}=30°\phantom{\rule{0ex}{0ex}}\therefore 4x=4\left(30\right)=120°$

Thus, the base angles measure 30o and the vertex angle measures 120o.

#### Question 5:

What is the measure of each of the equal angles of a right-angled isosceles triangle?

In a right isosceles triangle, one angle is a right angle and the the other two angles are equal.
Let the equal angles measure x.

Thus, the equal angles of the right isosceles triangle measure 45o​.

#### Question 6:

If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Given:
Triangle ABC is an isosceles triangle.
Or AB =AC
Or
Now, we have:

Subtracting (1) and (2), we get:
$y-z=0\phantom{\rule{0ex}{0ex}}⇒y=z$
Thus, the exterior angles so formed are equal to each other.
Hence, proved.

#### Question 7:

Find the measure of each exterior angle of an equilateral triangle.

Let the exterior angle of .

Similarly, each exterior angle would be equal to $120°$.
Thus, all the exterior angles of an equilateral triangle are $120°$.

#### Question 8:

In the given figure, O is the midpoint of each of the line segments AB and CD. Prove that AC = BD and AC || BD.

To prove:

$AC=BD\phantom{\rule{0ex}{0ex}}AC\parallel BD$

Here,
∴ AC = BD
Hence, proved.

#### Question 9:

In the adjoining figure, PA AB, QBAB and PA = QB. If PQ intersects AB at O, show that O is the midpoint of AB as well as that of PQ.

So, O is midpoint of AB and PQ.

#### Question 10:

Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.

However, $\angle OAC$ is not equal to $\angle OBD$ and $\angle OCA$ is not equal to $\angle ODB$.
Thus, AC may not be parallel to BD.
Hence, proved.

#### Question 11:

In the given figure, l || m and M is the midpoint of AB. Prove that M is also the midpoint of any line segment CD having its end points at l and m respectively.

Consider the triangles AMC and BMD.
Since lines l and m are parallel, we have:

It is also given that AM = MB.

$∆AMC\cong ∆BMD$ (AAS criterion)
i.e., DM = MC  (CPCT)
∴ M is the mid point of segment CD.
Hence, proved.

#### Question 12:

In the given figure, AB = AC and OB = OC. Prove that ABO = ∠ACO.

Given: AB  = AC
$⇒\angle B=\angle C$     ...(i)

Subtracting (ii) from (i), we get:
$\angle B-\angle OBC=\angle C-\angle OCB$
∴ $\angle ABO=\angle ACO$
Hence, proved.

#### Question 13:

In the given figure, ABC is a triangle in which AB = AC and D is a point on AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.

In the given figure, $DE||BC$.
$\angle ADE=\angle ABC\phantom{\rule{0ex}{0ex}}\angle AED=\angle ACB$       (Corresponding angles)   ...(i)

But in triangle ABC, AB = AC.
$\therefore \angle ABC=\angle ACB$   ...(ii)

From (i) and (ii), we have:

$\angle ADE=\angle ABC\phantom{\rule{0ex}{0ex}}\angle AED=\angle ABC$
$⇒\angle ADE=\angle AED$

Therefore, triangle ADE is also isosceles.
Hence, proved.

#### Question 14:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ABC such that AX = AY. Prove that CX = BY.

Hence, proved.

#### Question 15:

In the given figure, C is the midpoint of AB. If DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.

Given: C is the midpoint of AB.
Or AC = CB

∴ DC = EC         (CPCT)
Hence, proved.

#### Question 16:

In the given figure, BA AC and DEEF such that BA =DE and BF = DC. Prove that AC = EF.

Hence, proved.

#### Question 17:

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Consider the triangles AEB and CDB.

$\angle EBA=\angle DBC$  (Common angle)  ...(i)

Further, we have:

AB = CB     (Given) ...(iii)
From (i), (ii) and (iii), we have:
$△BDC\cong △BEA$   (AAS criterion)
∴ AE = CD (CPCT)
Hence, proved.

#### Question 18:

ABC is a triangle in which AB = AC. If the bisectors of B and ∠C meet AC and AB in D and E respectively, prove that BD = CE.

In triangle ABC, we have:
AB = AC    (Given)

Hence, proved.

#### Question 19:

In the adjoining figure, AD is a median of ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.

Hence, proved.

#### Question 20:

In ABC, D is the midpoint of BC. If DLAB and DMAC such that DL = DM, prove that AB = AC.

$\angle B=\angle C$
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

#### Question 21:

In ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠A.

In triangle ABC, we have:
AB = AC   (Given)
$⇒\angle B=\angle C\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\angle B=\frac{1}{2}\angle C\phantom{\rule{0ex}{0ex}}⇒\angle OBC=\angle OCB\phantom{\rule{0ex}{0ex}}⇒BO=CO$

i.e.,

So, it shows that ray AO is the bisector of $\angle A$.
Hence, proved.

#### Question 22:

In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS, (ii) PSR = 15°.

Given:
Triangle PQR is an equilateral triangle, i.e., each angle is $60°$.
QRST is a square, i.e., each angle is $90°$.

Similarly, $\angle PRS=150°$
Consider the triangles PQT and PRS.

Further, the side of the square coincides with the triangle, PR = RS, which makes PRS isosceles.
Let $\angle PSR=\angle SPR=x$.
Using angle sum property of a triangle, we get:
$150°+x+x=150°+2x=180°\phantom{\rule{0ex}{0ex}}⇒2x=30°\phantom{\rule{0ex}{0ex}}⇒x=15°$

Hence, proved.

#### Question 23:

In the given figure,, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.

∴ $\angle ACD=\angle FCB$

Hence, proved.

#### Question 24:

Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.

​The triangle ABC shown in the figure is an isosceles triangle.
∴ AB = AC and $\angle ABC=\angle ACB$

AD is the median drawn from vertex A to BC.
i.e., BD = DC

To prove: $\angle BAD=\angle CAD$

$△ABD\cong △ACD$ (SAS criterion)

Since the corresponding angles of the congruent triangles are congruent, $\angle BAD=\angle CAD$.

∴ The median from the vertex of an isosceles triangle bisects that angle.

#### Question 25:

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

∴ AB = CQ            (CPCT)

Hence, proved.

#### Question 26:

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.

Proof:

Now, consider triangles BQX and APX.

Further,
Also, we have proven that $\angle QBX=\angle PAX$.

Hence, proved.

#### Question 27:

In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

$\therefore \angle \mathrm{APD}+\angle \mathrm{CPD}=\angle \mathrm{APB}+\angle \mathrm{CPB}\phantom{\rule{0ex}{0ex}}\mathrm{But}\angle \mathrm{APD}+\angle \mathrm{CPD}+\angle \mathrm{APB}+\angle \mathrm{CPB}=360°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{APD}+\angle \mathrm{CPD}=180°$

So, CPA is a straight line.
Hence, proved.

#### Question 28:

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

#### Question 29:

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that (i) AC bisects A and ∠C, (ii) AC is the perpendicular bisector of BD.

Consider the triangles ABC and ADC.

Thus, AC bisects angles A and C.

Let AC intersect BD at O. Consider the triangles AOB and AOD.

Thus, AC bisects BD at O.

Thus, AC is the perpendicular bisector of BD.

#### Question 30:

In the given figure, the bisectors of B and ∠C of ∆ABC meet at I. If IPBC, IQCA and IRAB, prove that (i) IP = IQ = IR, (ii) IA bisects ∠A.

From (1) and (2), we have,

IP = IQ = IR

Now consider the right triangles, ARI and AQI.

Thus, IA bisects angle A.

#### Question 31:

In the adjoining figure, P is a point in the interior of AOB. If PLOA and PMOB such that PL = PM, show that OP is the bisector of ∠AOB.

i.e., $\angle POM=\angle POL$

Thus, OP is the bisector of $\angle AOB$.

#### Question 32:

In the given figure, ABCD is a square, M is the midpoint of AB and PQ CM meets AD at P and CB produced at Q. Prove that (i) PA = BQ, (ii) CP = AB + PA.

Proof:

and $MP=MQ$     (CPCT)

Thus,
But

$CP=AB+PA$
Hence, proved.

#### Question 33:

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Let AB be the breadth of the river.
M is any point situated on the bank of the river.
Let O be the mid point of BM.
​Moving along perpendicular to point such that A,O and N are in a straight line.
Then MN is the required breadth of the river.

Thus, MN = AB (CPCT)
If MN is known, one can measure the width of the river without actually crossing it.

#### Question 34:

In ABC, if ∠A = 36° and ∠B = 64°, name the longest and shortest sides of the triangle.

In triangle ABC, .

The side opposite the largest angle is the greatest and the side opposite the smallest angle is the shortest.
The side opposite to $\angle A$, BC, is the shortest side of the triangle and the side opposite to $\angle C$, AB, the longest side of the triangle.

#### Question 35:

In ABC, if A = 90°, which is the longest side?

.
The sum of the other two angles must be such that

Thus, angles B and C both measure less than 90$°$.
Angle A is the largest in the triangle and the side opposite to it, i.e., BC is the longest.
∴ BC is the hypotenuse of the right triangle ABC.

#### Question 36:

In ABC, if A = ∠B = 45° name the longest side.

In$∆ABC$, we have:

Thus, the side opposite to $\angle C$, i.e., AB, is the longest side.

The side AB is the hypotenuse of triangle ABC.

#### Question 37:

In ABC, side AB is produced to D such that BD = BC. If B = 60° and ∠A = 70°, prove that (i) AD > CD and (ii) AD > AC.

In triangle CBA, CBD is an exterior angle.

Triangle BCD is isosceles and BC = BD.
Let .

#### Question 38:

In ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

To arrange AX, BX and CX in descending order, we have to find the measure of the angle opposite of which these sides lie.

In triangle ABC, we have:

$\angle A+\angle B+\angle C=180°\phantom{\rule{0ex}{0ex}}⇒\angle A+35°+65°=180°\phantom{\rule{0ex}{0ex}}⇒\angle A=80°$

Since AX bisects angle BAC, we have:
$\angle BAX=40°$

So, in triangle AXC, we have:
$\angle AXC+40°+65°=180°\phantom{\rule{0ex}{0ex}}⇒\angle AXC=75°$
Thus, AX > CX       ...(i)

In triangle AXB, we have:
$\angle AXB+35°+40°=180°\phantom{\rule{0ex}{0ex}}⇒\angle AXB=105°$
Thus, BX > AX       ...(ii)

From (i) and (ii), we get:
$BX>AX>CX$

#### Question 39:

In ABC, if AD is the bisector of ∠A, show that AB > BD and AC > DC.

Thus, AB > BD

Thus, AC > DC
Hence, proved.

#### Question 40:

In the given figure, ABC is a triangle in which AB = AC. If D be a point on BC produced, prove that AD > AC.

$AB=AC\phantom{\rule{0ex}{0ex}}\angle ABC=\angle ACB$

AB =AC (Given)
Hence, proved.

#### Question 41:

Given:

In triangles ABD and ADC, the sums of the respective angles will be 180$°$.

Using (i), we get:

Hence, proved.

#### Question 42:

In PQR, if S is any point on the side QR, show that PQ + QR + RP > 2PS.

On adding (1) and (2) we get:
$PQ+QS+PR+SR>2PS\phantom{\rule{0ex}{0ex}}⇒PQ+QR+PR>2PS$

Hence, proved.

#### Question 43:

In the given figure, O is the centre of the circle and XOY is a diameter. If XZ is any other chord of the circle, show that XY > XZ.

Hence Proved.

#### Question 44:

If O is a point within ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) $OA+OB+OC>\frac{1}{2}\left(AB+BC+CA\right)$

Given:
In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.​

Adding the above three equations, we get:
...(6)

OA cannot be longer than both AB and CA.​

Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:

#### Question 45:

Can we draw a triangle ABC with AB = 3 cm, BC = 3.5 cm and CA = 6.5 cm? Why?

No.

A triangle can be drawn when the sum of any two sides is greater than the third side.
Here, $AB+BC=AC$
i.e., 3+3.5 = 6.5
So, we cannot draw the triangle ABC.

#### Question 1:

Which of the following is not a criterion for congruence of triangles?
(a) SSA
(b) SAS
(c) ASA
(d) SSS

(a) SSA
SSA is not a criterion for congruence of triangles.

#### Question 2:

If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ABC ≅ ∆PQR
(b) CBA ≅ ∆PQR
(c) CAB ≅ ∆PQR
(d) BCA ≅ ∆PQR

(c) $∆CAB\cong ∆PQR$
As,
AB = QR,   (given)
BC = RP,     (given)
CA = PQ     (given)

$∆CAB\cong ∆PQR$

#### Question 3:

If ABC ≅ ∆ PQR and ∆ABC is not congruent to ∆RPQ, then which of the following is not true?
(a) BC = PQ
(b) AC = PR
(c) BC = QR
(d) AB = PQ

(a) BC = PQ

If $∆ABC\cong ∆PQR$ and $∆ABC$ is not congruent to $∆RPQ$, then
BC $\ne$ PQ
Hence, it is false.

#### Question 4:

It is given that ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, A = 80° and FD = 5 cm. Then, which of the following is true?
(a) ∠D = 60°
(b) ∠E = 60°

(c) ∠F = 60°
(d) D = 80°

(b)​ $\angle E=60°$

$∆ABC\cong ∆FDE$
AB = 5cm, $\angle B=40°,\angle A=80°$ and FD = 5cm

$\angle E=60°$

#### Question 5:

In ABC, AB = 2.5 cm and BC = 6 cm. Then, the length of AC cannot be
(a) 3.4 cm
(b) 4 cm
(c) 3.8 cm
(d) 3.6 cm

(a) 3.4 cm
In triangle ABC, AB = 2.5 cm and BC = 6 cm.
​The length of a side must be less than the sum of the other two sides.
Let the side AC be x cm.
i.e., $x<2.5+6⇒x<8.5$
In addition, the length of a side must be greater then the difference between the other two sides.
i.e., $x>6-2.5⇒x>3.5$
Hence, the limits for the value of x is $3.5.
Thus, the length of AC cannot be 3.4 cm

#### Question 6:

In ABC, ∠A = 40° and ∠B = 60°. Then the longest side of ∆ABC is
(a) BC
(b) AC
(c) AB
(d) cannot be determined

(c) AB

In triangle ABC, we have:
$\angle A=40°,\angle B=60°$        ...(Given)

∴ The side opposite to$\angle C$, i.e., AB, is the longest side of triangle ABC.

#### Question 7:

In ABC, ∠B = 35°, ∠C = 65° and the bisector AD of ∠BAC meets BC at D. Then, which of the following is true?
(a) AD > BD > CD
(b) BD > AD > CD
(c) AD > CD > BD
(d) None of these

(b)$BD>AD>CD$

In $∆ABC$, we have:
$\angle B=35°,\angle C=65°$ and the bisector AD of $\angle BAC$ meets BC at D.
Then we have:
$\angle A+\angle B+\angle C=180°\phantom{\rule{0ex}{0ex}}⇒\angle A+35°+65°=180°\phantom{\rule{0ex}{0ex}}⇒\angle A=80°\phantom{\rule{0ex}{0ex}}$

AD is the angle bisector of $\angle BAC$
$\angle BAD=\angle CAD=40°$

Now, in triangle ABD, we have:
$\angle BAD>\angle ABD$
$⇒BD>AD$
Also, in triangle ACD, we have:
$\angle ACD>\angle CAD$
$⇒AD>CD$
$\therefore BD>AD>CD$

#### Question 8:

In the given figure, AB > AC. Then which of the following is true?
(d) Cannot be determined

$AB>AC$ is given.

$\angle ACB>\angle ABC$

#### Question 9:

In the given figure, AB > AC. If BO and CO are the bisectors of B and ∠C respectively, then
(a) OB < OC
(b) OB = OC
(c) OB > OC

(c) OB > OC

AB >AC    (Given)
$⇒\angle C>\angle B$
$⇒\frac{1}{2}\angle C>\frac{1}{2}\angle B$
$⇒\angle OCB>\angle OBC$  (Given)
$⇒OB>OC$

#### Question 10:

In the given figure, AB = AC and OB = OC. Then, ABO : ∠ACO = ?
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) None of these

(a) 1:1

​In

i.e., $\angle ABO=\angle ACO$
∴

#### Question 11:

In ABC, if ∠C > ∠B, then
(a) BC > AC
(b) AB > AC

(c) AB > AC
(d) BC > AC

(b) AB > AC

In $∆ABC$, we have:
$\angle C>\angle B$
The side opposite to the greater angle is larger.
$\therefore AB>AC$

#### Question 12:

O is any point in the interior of ABC. Then, which of the following is true?
(a) $\left(OA+OB+OC\right)>\left(AB+BC+CA\right)$
(b) $\left(OA+OB+OC\right)>\frac{1}{2}\left(AB+BC+CA\right)$
(c) $\left(OA+OB+OC\right)<\frac{1}{2}\left(AB+BC+CA\right)$
(d) None of these

(c)

In

#### Question 13:

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

(b) isosceles

In

To prove: AB = AC

#### Question 14:

In the given figure, AE = DB, CB = EF and ABC = ∠FED. Then, which of the following is true?
(a) ∆ABC ≅ ∆DEF
(b) ∆ABC ≅ ∆EFD

(c) ∆ABC ≅ ∆FED
(d) ∆ABC ≅ ∆EDF

(a)

AE = DB          ...(Given)
CB = EF           ...(Given)

#### Question 15:

In the given figure, BE CA and CFBA such that BE = CF. Then, which of the following is true?
(a) ∆ABE ≅ ∆ACF
(b) ∆ABE ≅ ∆AFC

(c) ∆ABE ≅ ∆CAF
(d) ∆ABE ≅ ∆FAC

(a)

#### Question 16:

In the given figure, D is the midpoint of BC, DE AB and DFAC such that DE = DF. Then, which of the following is true?
(a) AB = AC
(b) AC = BC
(c) AB = BC
(d) None of these

(a) AB = AC

#### Question 17:

In ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have
(a) ∠A = ∠D
(b) ∠B = ∠E
(c) ∠C = ∠F
(d) none of these

(b) $\angle B=\angle E$

In
AB = DE      (Given)
BC = EF       (Given)
In order that $△ABC\cong DEF$, we must have $\angle B=\angle E$.

#### Question 18:

In ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABCDEF, we must have
(a) AB = DF
(b) AC = DE
(c) BC = EF
(d) A = ∠D

(c) BC = EF
In order that $△ABC\cong △DEF$, we must have BC = EF.

#### Question 19:

In ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

(a) isosceles but not congruent

Thus, both the triangles are isosceles but not congruent.

#### Question 20:

Which is true?
(a) A triangle can have two right angles.
(b) A triangle can have two obtuse angles.
(c) A triangle can have two acute angles.
(d) An exterior angle of a triangle is less than either of the interior opposite angles.

(c) A triangle can have two acute angles.
The sum of two acute angles is always less than 180o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

#### Question 21:

Three statements are given below:
I. In a ABC in which AB = AC, the altitude AD bisects BC.
II. If the altitudes AD, BE and CF of ∆ABC are equal, then ∆ABC is equilateral.
III. If D is the midpoint of the hypotenuse AC of a right ∆ABC, then BD = AC.
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

(c) I and II are true
(I)

∴ AD bisects BC, which is true.

(II)

Similarly, if AD is perpendicular to BC and AD = BE, then we can prove that BC = AC.
Therefore, triangle ABC is an equilateral triangle.
Hence, (II) is true.

(III)

Then we know that AD = BD = CD
Therefore, BD = AC is not true.
So, only (I) and (II) are true.

#### Question 22:

Assertion: If AD is a median of ABC, then AB + AC > 2AD.
Reason: The angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

In triangle ABC, AD is a median.
We know that the sum of two sides of a triangle is greater than the third side.

$AB+DB>AD....\left(i\right)\phantom{\rule{0ex}{0ex}}AC+DC>AD.........\left(ii\right)\phantom{\rule{0ex}{0ex}}⇒AB+AC+\left(DB+DC\right)>2AD\phantom{\rule{0ex}{0ex}}⇒AB+AC+BC>2AD$

In any triangle, angles opposite to equal sides are equal.
Clearly, the assertion and the reason are both true, but reason (R) does not give assertion (A).

#### Question 23:

Assertion: In a quadrilateral ABCD, we have (AB + BC + CD + DA) > 2AC.

Reason: The sum of any two sides of a triangle is greater than the third side.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
We know that sum of any two sides of a triangle is greater than the third side.
Therefore, $\left(AB+BC\right)>AC$ and $\left(CD+DA\right)>AC$.
$\left(AB+BC+CD+DA\right)>2AC$
Thus, the assertion is true, and the reason gives the assertion.

#### Question 24:

Assertion: ABC and ∆DBC are two isosceles triangles on the same base BC. Then, ∠ABD = ∠ACD.

Reason: The angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

We know that the angles opposite to the equal sides of a triangle are equal.
So, the reason is true.

So, the assertion (A) is true and, clearly, the reason (R) gives the assertion.

#### Question 25:

Assertion: It is always possible to draw a triangle whose sides measure 4 cm, 5 cm and 10 cm respectively.
Reason: In an isosceles ABC with AB = AC, if BD and CE are bisectors of B and ∠C respectively, then BD = CE.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(d) Assertion is false and Reason is true.
Reason (R):

$AB=AC\phantom{\rule{0ex}{0ex}}⇒\angle C=\angle B\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\angle C=\frac{1}{2}\angle B\phantom{\rule{0ex}{0ex}}⇒\angle BCE=\angle CBD$

Thus, the reason (R) is true.
The assertion (A) is false, as the sum of two sides of a triangle is always greater than the third side.
Here, $\left(4+5\right)<10$
So, the reason (R) is true, but the assertion (A) is false.

#### Question 26:

Assertion: In the given figure, ABC is given with AB = AC and BA is produced to D such that AB = AD. Then, ∠BCD = 90°.

Reason: In the given figure AB = AC and D is a point on BC produced. Then, AB > AD.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(c) Assertion is true and Reason is false.
Reason (R):

The side BC of triangle ABC is produced to D.

Hence, the reason (R) is false.
Clearly, the assertion is true, but the reason (R) is false.

#### Question 27:

Match the following columns:

 Column I Column II (a) ∆ABC, if AB = AC and ∠A = 50°, then ∠C = ...... . (p) its perimeter (b) The vertical angle of an isosceles triangle is 130°. Then, each base angle is ...... . (q) 15° (c) The sum of three altitudes of a ∆ABC is less than ....... . (r) 65° (d) In the given figure, ABCD is a square and ∆EDC is an equilateral triangle. Then, ∠EBC is ...... . (s) 25°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a)$⇒$(r)
In if AB = AC and $\angle A=50°$, then

$\angle C=65°$.

(b)$⇒$(s)
If the vertical angle of an isosceles triangle is $130°$, then

Each base angle is $25°$.

(c)$⇒$(p)
The sum of three altitudes of a $△ABC$ is less than its perimeter.

(d) $⇒$(q)
In the given figure, ABCD is a square and $△EDC$ is an equilateral triangle.
Then,

Therefore, $\angle EBC=15°$.

#### Question 28:

Fill in the blanks with < or >.
(a) (Sum of any two sides of a triangle) ...... (the third side)
(b) (Difference of any two sides of a triangle) ...... (the third side)
(c) (Sum of three altitudes of a triangle) ...... (sum of its three side)
(d) (Sum of any two sides of a triangle) ...... (twice the median to the 3rd side)
(e) (Perimenter of a triangle) ...... (sum of its three medians)

a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

#### Question 29:

Fill in the blanks.
(a) Each angle of an equilateral triangle measures ...... .
(b) Medians of an equilateral triangle are ...... .
(c) In a right triangle the hypotenuse is the ...... side.
(d) Drawing a ABC with AB = 3 cm, BC = 4 cm and CA = 7 cm is ...... .

a) Each angle of an equilateral triangle measures $60°$.

b) Medians of an equilateral triangle are equal.

c) In a right triangle, the hypotenuse is the longest side.

d) Drawing a $△ABC$ with AB = 3cm, BC = 4cm and CA = 7cm is not possible.

#### Question 1:

In an equilateral ABC, find ∠A.

In an equilateral $△ABC$, all the angles are equal.
Let the three angles .

#### Question 2:

In a ABC, if AB = AC and ∠B = 65°, find ∠A.

In $△ABC$, we have:
AB = AC
$\angle B=65°$
Since ABC is an isosceles triangle, we have:
$\angle C=\angle B$
$\angle C=65°$
In triangle ABC, we have:
$\angle A+\angle B+\angle C=180°\phantom{\rule{0ex}{0ex}}⇒\angle A+65+65=180°\phantom{\rule{0ex}{0ex}}⇒\angle A=180°-130°\phantom{\rule{0ex}{0ex}}\therefore \angle A=50°$

#### Question 3:

In a right ABC, ∠B = 90°. Find the longest side.

In right $∆ABC$, we have:
$\angle B=90°$

The side opposite to $\angle B$, i.e., side AC, is the longest.

#### Page No 205:

In a triangle, the side opposite the greater angle is longer. In triangle ABC, $\angle B>\angle C$.
Therefore, AC is longer than AB.

#### Question 5:

Can we construct a ABC in which AB = 5 cm, BC = 4 cm and AC = 9 cm? Why?

No, the sum of two side must be greater than the third side.
AC should be greater than (AB + BC).

In this case, AB + BC = 5 + 4 = 9 cm
AC = 9 cm
Thus, the sum of two sides is not greater than the third side. So, triangle ABC cannot be constructed.

#### Question 6:

Find the measure of each exterior angle of an equilateral triangle.

Here, exterior angle is $\angle AOD$.
$\angle AOD+\angle AOB=180°\phantom{\rule{0ex}{0ex}}⇒60°+\angle AOB=180°\phantom{\rule{0ex}{0ex}}⇒\angle AOB=120°$
∴ The measure of each exterior angle of an equilateral triangle is $120°$.

#### Question 7:

Show that the difference of any two sides of a triangle is less than the third side.

#### Question 8:

In a right ABC, ∠B = 90° and D is the mid-point of AC. Prove that $BD=\frac{1}{2}AC$.

AD = DC     (∵ D is the midpoint)
$\angle ADB=\angle BDC$     (Altitude = BD)
BD = BD     (Common side)
∴ $△ADB\cong △CDB$     (By SAS criterion)
$\angle A=\angle C$    (CPCT)
Now, we know that $\angle B=90°$.
$⇒\angle A={\angle }{\mathrm{ABD}}{=}{45}{°}$    (Using sum angle property)
Similarly, $\angle BDC=90°$    (BD is altitude)
$\angle C=45°$    (Proved)
$\angle DBC=45°$

(Isosceles triangle property)
$⇒BD+BD=AC\phantom{\rule{0ex}{0ex}}⇒2BD=AC\phantom{\rule{0ex}{0ex}}⇒BD=\frac{1}{2}AC$
Hence proved.

#### Question 9:

Prove that the perimeter of a triangle is greater than the sum of its three medians.

Let ABC be the triangle and D, E and F be the midpoints of BC, CA and AB, respectively.
Since the sum of two sides of a triangle is greater than twice the median bisecting the third side, we have:

On adding all opf the above inequalities, we get:

Thus, the perimeter of triangle is greater than the sum of the medians.

#### Question 10:

Which is true?
(a) A triangle can have two acute angles.
(b) A triangle can have two right angles.
(c) A triangle can have two obtuse angles.
(d) An exterior angles of a triangle is always less than either of the interior opposite angles.

(a) A triangle can have two acute angles.

This is because the sum of two acute angles is always less than 180o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

#### Question 11:

In ABC, BDAC and CE AB such that BE = CD. Prove that BD = CE.

In  we have:

$BD=CE$    (CPCT)
Hence proved.

#### Question 12:

In ABC, AB = AC. Side BA is produced to D such that AD = AB. Prove that ∠BCD = 90°.

In triangle ABC, we have:

In triangle ACD, we have:

In triangle BCD, we have:

Hence proved.

#### Question 13:

In the given figure, it is given that AD = BC and AC = BD. Prove that CAD = ∠CBD and ∠ADC = ∠BCD.

Consider the triangles DAC and CBD.

By SSS criterion of congruence, $∆ADC\cong BCD$.

By CPCT, we have:

Hence proved.

#### Question 14:

Prove that the angles opposite to equal sides of a triangle are equal.

In triangle PQR, $PQ=QR.$
To prove: $\angle Q=\angle R$
Proof:
Draw a bisector of and let S be the point of intersection.

$\angle Q=\angle R$
Hence Proved.

#### Question 15:

In an isosceles ABC, AB = AC and the bisectors of are joint.
Prove that: (i) OB = OC (ii) ∠OAB = ∠OAC

In , we have:

⇒ OB = OC     (CPCT)

Hence proved.

#### Question 16:

Prove that, of all line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

Given: l is a straight line and A is a point not lying on l.
AB is perpendicular to line l.
C is a point on l.
To prove: $AB
Proof:

Since, C can lie anywhere on l, AB is the shortest of all line segments drawn from A to l.

#### Question 17:

Assertion: Each angle of an equilateral triangle is 60°.
Reason: Angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) The Assertion and Reason are both true and the Reason is a correct explanation of the Assertion.
We know that each angle of an equilateral triangle is 60°.
Also, angles opposite to the equal sides of a triangle are equal.
Clearly, both the Assertion and the Reason are true .

#### Question 18:

Assertion: If AD is a median of ABC, then AB + AC > 2AD.

Reason: In a triangle the sum of two sides is greater than the third side.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) The Assertion and Reason are both true and the Reason is a correct explanation of the Assertion.
So, the Reason is obviously true. Let us consider the Assertion.

In triangle ABD, $AB+BD>AD$.

Similarly, in triangle ADC, $AC+CD>AD$.

Adding the above two expressions, we get:
$AB+AC+BD+CD>AD+AD\phantom{\rule{0ex}{0ex}}⇒AB+AC+BD+DC>2AD\phantom{\rule{0ex}{0ex}}⇒AB+AC+BC>2AD$

Thus, the Assertion is true and the Reason is true. The Reason is an explanation of the Assertion.

#### Question 19:

Match the following columns:

 Column I Column II (a) In ∆ABC, if AB = AC and ∠A = 70°, then ∠C = ...... . (p) less (b) The vertical angle of an isosceles triangle is 120°. Each base angle is ...... . (q) greater (c) The sum of three medians of a triangle is ...... than the perimeter. (r) 30° (d) In a triangle, the sum of any two sides is always ...... than the third side. (s) 55°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a) $⇒$ (s)
In

(b) $⇒$(r)
The vertical angle of an isosceles triangle is $120°$. Each base angle is $30°$.

(c) $⇒$(p)
The sum of three medians of a triangle is less than the perimeter.

(d) $⇒$(q)
In a triangle, the sum of any two sides is always greater than the third side.

#### Question 20:

In the given figure, PQ > PR and QS and RS are the bisectors of Q and ∠R respectively. Show that SQ > SR.

Since the angle opposite to the longer side is greater, we have:

$PQ>PR\phantom{\rule{0ex}{0ex}}⇒\angle R>\angle Q\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\angle R>\frac{1}{2}\angle Q\phantom{\rule{0ex}{0ex}}⇒\angle SRQ>\angle RQS\phantom{\rule{0ex}{0ex}}⇒QS>SR$

$SQ>SR$

#### Question 21:

In the given figure, ABC is a triangle right-angled at B such that BCA = 2 BAC. Show that AC = 2BC.

Produce CB to D such that BD = BC.

But $AC=AD$

Therefore, triangle ACD is an equilateral triangle.

#### Question 22:

S is any point in the interior of PQR. Show that ABCD AB + BC + CD + DA > AC + BD.
Figure

Produce QS so that it meets PR at T.

#### Question 23:

Show that in a quadrilateral ABCD AB + BC + CD + DA > AC + BD.
Figure