# CBSE Important Questions Class 9 Maths Chapter 7

## Important Questions Class 9 Mathematics Chapter 7 – Triangles

The word “triangle” itself conveys its meaning. It is  a closed figure made up of three crossing lines since the prefix “tri” implies “three”. The major objective of Chapter 7 is to study triangles and their congruence or similarity of triangles. You will learn in-depth information on triangle congruence, congruence rules, various triangle properties, and triangle inequalities in this chapter.

With the help of Extramarks’ Class 9 Mathematics NCERT Solutions, students can more easily and effectively prepare for all the concepts covered in the CBSE Syllabus. Revision Notes, which include a thorough explanation, key formulas, and time-saving advice, are also offered to students to assist them in getting a quick review of all the topics. By practising the NCERT Important Questions Class 9 Mathematics Chapter 7, students can improve their test preparation.

Students’ grasp of the fundamental ideas in Mathematics is the main goal of the Important Questions Class 9 Mathematics Chapter 7. The NCERT Class 9 Mathematics Chapter 7 Important Questions should be correctly and precisely practised by the students in order to effectively assimilate its conceptual meaning. Our subject specialists created these sample answers for Class 9 students. These answers present useful recommendations along with many beneficial tips and tactics for correctly answering the problems.

Students should practise these problems in order to achieve high grades on the class 9 Math final exam. To get better practice and a quick revision before the test, solve extra problems in addition to the Important Questions Class 9 Mathematics Chapter 7. Triangle congruency or resemblance is a prominent topic in the questions in Chapter 7, Triangles. Students who can complete these problems will have a better understanding of the questions that will be on the test.

## Important Questions Class 9 Mathematics Chapter 7 – With Solutions

The Important Questions Class 9 Mathematics Chapter 7 have been compiled by the Extramarks team from various sources. Students must be familiar with the congruence rule in order to solve questions on this subject. Therefore, read through the theory first and then have a look at the already-solved cases in the book. Start working on the exercise problems after that.

A few Chapter 7 Class 9 Mathematics Important Questions are provided here, along with their answers:

Question 1:  If AD = BC and ∠ BAD = ∠ ABC, then ∠ ACB is equal to

• ∠ABD
• ∠BAC
• ∠BDA

Solution 1: (D)

Explanation:  In △ABC and △ABD

∠ BAD = ∠ ABC    (Given)

AB = AB   (Common side)

∴ △ABC ≅ △ABD

By CPCT theorem, ∠ACB=∠BDA ( By SAS Congruency )

Question 2: If ABCD is a quadrilateral where AD= CB, AB=CD, and ∠ D= ∠ B, then ∠CAB is equal to

•  ∠ ACD
•  ∠ ACD

Solution 2: (C)

Explanation: In △ABC and △CDA

AB=CD         (Given)

∠B = ∠D         (Given)

∴△ABC≅△CDA       ( By SAS Congruency )

By CPCT theorem

∠CAB=∠ACD

Question 3:  If O is a midpoint of AB and ∠ BQO = ∠ APO, then ∠ OAP is equal to

•  ∠ QPA
•  ∠ OQB
•  ∠ QBO
•  ∠ BOQ

Solution 3: (C)

Explanation: In △AOP and △BOQ

AO=BO…..(O is the midpoint of AB)

∠APO=∠BQO(Given)

∠AOP=∠BOQ(Vertically opposite Angles)

∴△AOP≅△BOQ         ( By AAS Congruency )

By CPCT ∠OAP=∠QBO
Question 4:  If AB ⊥BC and ∠ A =∠ C, then the correct statement will

• AB≠AC
• AB=BC
• AB=AC

Solution 4: (B) AB=BC

Explanation: In △ABC, ∠A=∠C

Opposite sides to equal angles are also equal

AB = BC

Question 5: If △ABC is an isosceles triangle, ∠ B = 650, find ∠ A.

• 600
• 700
• 500
• none of these

Solution 5: (c)

Explanation: Since △ABC is on an isosceles triangle

∴∠B=∠C

∴∠B=65∘

∴∠C=65∘

∴∠A+∠B+∠C=180∘

∴∠A+130∘=180∘

∴∠A=180∘−130∘

∴∠A=50∘

Question 6: If AB=AC and ∠ ACD= 1200, find ∠A.

• 500
• 600
• 700
• none of these

Solution 6: (b)600

Explanation: Since AB=AC

⇒∠ABC=∠ACB=x (say)

Let ∠BAC=y

We know,

Exterior angles = sum of interior opposite angles

120∘ =∠ABC+∠BAC

120∘=x+y−−−(1)

Again,∠ACB+∠ACD=180∘

x+120∘=180∘

∴x=60∘

From(1),

60∘+y=120∘

⇒y=60∘

⇒∠A=60∘

Question 7: An angle is 140 more than its complement. Find its measure.

• 42
• 32
• 52
• 62

Solution 7:  (C) 52

Explanation: Two angles whose sum equals 90 degrees are called complementary angles.

let first angle =x

it’s Complement = 90∘−x

According to the question,

x=14∘ + 90∘−x

x=104−x

⇒2x=104∘

⇒x= 1042

∴x=52∘

Question 8:  An angle is four times its complement. Find measure.

• 62
• 72
• 52
• 42

Solution 8:  (B) 72

Explanation: Two angles, when summed up to 90 degrees, are called complementary angles.

let  angle = x

Therefore, its complement = 90∘−x

According to the question,

x=4(90∘−x)

x=360∘−4x

x+4x=360∘

⇒5x=360∘

⇒x= 360∘5

∴x=72∘

Question 9:  Find the measure of angles which are supplementary.

• 1200
• 600
• 450
• 900

Solution 9: (D)

Explanation: Two angles which sum up to 180 degrees are called supplementary angles.

x=180∘−x

2x=180∘

⇒x= 180∘2

∴x=90∘

Question 10:  Which of the following pair of angles will be supplementary?

• 300, 1200
• 450, 1350
• 1200,30
• None of these.

Solution 10: (B) 450, 1350

Explanation: Because 45∘+135∘=180∘

Question 11: In an isosceles △ ABC, if AB=AC and ∠A=900 , Find ∠ B.

•  450
• 800
• 95
• 600

Solution 11:  (A)

Explanation: Since AB = AC

⇒∠C=∠B     (angles opposite equal sides are also equal )

In △ABC

∠A+∠B+∠C=180∘

90∘ +2∠B=180∘

⇒2∠B=180∘−90∘

⇒∠B= 90∘2

⇒∠B=45∘

Question 12: In an △ABC, if ∠B= ∠C=450, Which is the longest side?

• BC
• AC
• CA
• None of these.

Solution 12: (A) BC

Explanation: In △ABC,

∠A+∠B+∠C=180∘

∠A+45∘+45∘=180∘

⇒∠A=180∘−90∘=90∘

This is a right-angled triangle in which the right angle is at ∠A

Therefore, the Side opposite to ∠A is the longest side (hypotenuse)

Question 13: In an △ABC, if AB=AC and ∠ B= 700, Find A.

•  400
•  500
•  450
•  600

Solution 13: (A)

Explanation:  In △ABC,  AB=AC

∠C=∠B= 70∘

∠A+∠B+∠C=180∘

∠A+70∘+70∘=180∘

⇒∠A=180∘−140∘=40∘

Question 14: In an △ABC, if ∠A = 450  and ∠B =700, determine the longest sides of the triangle.

• AC
• AB
• BC
• none of these

Solution 14: (a) AC

Explanation: The angle opposite to the longest side is the largest

The side opposite  ∠B is AC

The correct option is (a) AC

Question 15:  Which of the following is not a criterion for the congruence of triangles?

(A) SAS

(B) SSS

(C) SSA

(D) ASA

Solution 15: (C) SSA

Explanation:

We know,

Two triangles are congruent when the side(S) and angles (A) of one triangle are equal to another.

And the criterion for congruence of triangles is SAS, ASA, SSS, and RHS.

So, SSA is not the criterion for the congruency of the triangle.

Question 16:  If AB = QR, BC = PR and CA = PQ, then

(A) Δ PQR ≅ Δ BCA

(B) ΔCBA ≅ ΔPRQ

(C) ΔBAC ≅ Δ RPQ

(D) Δ ABC ≅ ΔPQR

Solution 16: (B) ΔCBA ≅ ΔPRQ

Explanation:

As per the question,

AB = QR, BC = PR and CA = PQ

Also, AB = QR, BC = PR and CA = PQ

Thus,

A corresponds to Q, B corresponds to R, and C corresponds to P.

Therefore, (B) ΔCBA ≅ ΔPRQ

Question 17:  In Δ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to

(A)130° (B) 50° (C) 80° (D) 40°

Solution 17: (B) 50°

Explanation:

As per the question,

Δ ABC, AB = AC and ∠B = 50°.

Image Source: NCERT textbook

Since, AB = AC

Δ ABC is an isosceles triangle.

Hence, ∠B = ∠C

We know ∠B = 50°

⇒ ∠C = 50°

Question 18:  In Δ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to

(A) 80° (B) 100° (C) 50° (D) 40°

Solution 18: (C) 50°

Explanation:

Given: Δ ABC, BC = AB and ∠B = 80°

Image source: NCERT textbook

Since, BC = AB

Δ ABC is an isosceles triangle.

Let, ∠C = ∠A = x

∠B = 80° (given)

We know that,

Using the angle sum property,

The sum of the interior angles of a triangle is= 180o

∠A + ∠B + ∠C = 180°

⇒ x + 80° + x = 180°

⇒ 2x = 180° – 80°

⇒ 2x = 100°

⇒ x = 50°

Thus, ∠C = ∠A = 50°

Question 19:  In Δ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is

(A) 4 cm (B)2.5 cm  (C) 2 cm (D) 5 cm

Solution 19: (A) 4 cm

Explanation: Given: In ΔPQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm

As given in the question, ∠R = ∠P

Therefore, Δ PQR is an isosceles triangle.

and PQ = QR

⇒ PQ = 4cm

Question 20:  In quadrilateral ACBD, AC = AD and AB are bisecting ∠A (see figure given below). Show that ΔABC≅ ΔABD. What can you depict about BC and BD?

Image source: NCERT textbook

Solution 20:

As given in the question, AC and AD are equal, i.e. AC = AD, and the line segment AB is bisecting ∠A.

We have to prove that the two triangles ABC and ABD are similar, i.e. ΔABC ≅ ΔABD

Proof:

We consider the triangles ΔABC and ΔABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common side)

(iii) ∠CAB = ∠DAB (Since AB is a bisector of angle A)

So, by SAS congruency rule criterion, ΔABC ≅ ΔABD.

Again, for the 2nd part of the question, BC and BD are of equal lengths by the rule of CPCT.

Question 21:  ABCD is a quadrilateral where AD = BC and ∠DAB = ∠CBA (see figure given below). Prove that

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Image source: NCERT textbook

Solution 21:

Given in the question are ∠DAB = ∠CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruency rule as

AB = BA (It is the common arm)

∠DAB = ∠CBA and AD = BC (given in the question)

So, triangles ABD and BAC are quite similar, i.e. ΔABD ≅ ΔBAC. (Hence proved).

(ii) Again, we know that ΔABD ≅ ΔBAC, so,

BD = AC (by the rule of CPCT).

(iii) As ΔABD ≅ ΔBAC therefore,

Angles ∠ABD = ∠BAC (by the CPCT rule).

Question 22: AD and BC are equal perpendiculars to line segment AB (see figure given below). Show that CD is bisecting AB.

Image source: NCERT textbook

Solution 22:

Given in the question that AD and BC are the two equal perpendiculars to AB.

We are to prove that CD is the bisector of AB

Now, as per the question,

Triangles ΔAOD and ΔBOC are similar by AAS congruency rule since:

(i) ∠A = ∠B (Since perpendiculars)

(ii) AD = BC (Since given in the question)

(iii) ∠AOD = ∠BOC (Since vertically opposite angles)

∴ ΔAOD ≅ ΔBOC.

Thus, AO = OB (by CPCT rule).

Therefore, CD bisects AB (Hence proved).

Question 23:  l and m are the two parallel lines that are intersected by another pair of parallel lines, such as p and q (see figure given below). Show that ΔABC ≅ ΔCDA.

Image source: NCERT textbook

Solution 23:

Given in the question that p || q and l || m

To prove:

Triangles ABC and CDA are similar, i.e. ΔABC ≅ ΔCDA

Proof:

We consider ΔABC and ΔCDA,

(i) ∠BCA = ∠DAC and ∠BAC = ∠DCA (Since alternate interior angles)

(ii) AC = CA (as common arm)

Thus, by ASA congruency rule criterion, ΔABC ≅ ΔCDA.

Question 24: Line l is the bisector of angle ∠A, and B is any point on l. BP and BQ are the perpendiculars from B to the arms of ∠A (see figure given below). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Image source: NCERT textbook

Solution 24:

As given in the question that the line “l” is the bisector of angle ∠A, and the line segments BP and BQ are perpendiculars drawn from l.

(i) ΔAPB and ΔAQB are similar by AAS congruency rule because:

∠P = ∠Q (Since they are two right angles)

AB = AB (Since common arm)

∠BAP = ∠BAQ (As l is the bisector of angle A)

So, ΔAPB ≅ ΔAQB.

(ii) By the CPCT rule, BP = BQ. So, it can be said that point B is equidistant from the arms of ∠A.

Question 25:  In figure given below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Image source: NCERT textbook

Solution 25:

As given in the question, AB = AD, AC = AE, and ∠BAD = ∠EAC

To prove:

Line segments BC and DE are similar, i.e. BC = DE

Proof:

Now, by adding ∠DAC on both sides, we get,

∠BAD + ∠DAC = ∠EAC +∠DAC

Now, ΔABC and ΔADE are similar by the SAS congruency rule:

(i) AC = AE (given in the question)

(iii) AB = AD (given in the question)

Thus the triangles ABC and ADE are similar, i.e. ΔABC ≅ ΔADE.

Hence, by the CPCT rule, BC = DE.

Question 26:  AB is a line segment, and P is the mid-point. D and E are the points on the same side of AB in such a way that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure given below). Show that

(i) ΔDAP ≅ ΔEBP

Image source: NCERT textbook

Solution 26:

Given that P is the mid-point of line segment AB. And, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) Also, given that ∠EPA = ∠DPB

Now, add ∠DPE on both sides,

∠EPA +∠DPE = ∠DPB+∠DPE

Hence, angles DPA and EPB are equal, i.e. ∠DPA = ∠EPB

Now, considering the triangles DAP and EBP.

∠DPA = ∠EPB

AP = BP (As P is the mid-point of AB)

∠BAD = ∠ABE (as per the question)

So, by ASA congruency rule, ΔDAP ≅ ΔEBP.

(ii) By CPCT rule, AD = BE.

Question 27:  In the right triangle ABC, right-angled at C, and M is the mid-point of hypotenuse AB. Point C is joined to M and then produced to point D so that DM = CM. Point D is joined to point B (see figure given below). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = ½ AB

Image source: NCERT textbook

Solution 27:

Given in the question that M is a mid-point of the line segment AB, ∠C = 90°, and DM = CM

(i) We consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (As given in the question)

∠CMA = ∠DMB (Since vertically opposite angles)

So, by SAS congruency rule criterion, ΔAMC ≅ ΔBMD.

(ii) ∠ACM = ∠BDM (by CPCT rule)

∴ AC || BD since alternate interior angles are equal.

Now, ∠ACB +∠DBC = 180° (i.e., co-interiors angles)

⇒ 90° +∠B = 180°

∴ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Since common side)

∠ACB = ∠DBC (Since they are right angles)

DB = AC (by CPCT rule)

Thus, ΔDBC ≅ ΔACB by SAS congruency rule.

(iv) DC = AB (Since ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (As M the is mid-point)

So, DM + CM = BM+AM

Therefore, CM + CM = AB

⇒ CM = (½) AB

Question 28: In the isosceles triangle ABC, AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

Solution 28:

Given:

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is isosceles in which AB = AC,

∠B = ∠C

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB (i.e., Angle bisectors)

∴ OB = OC (opposite sides to equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved earlier)

So, ΔAOB ≅ ΔAOC by SSS congruence rule.

BAO = CAO (by CPCT rule)

Therefore, AO bisects ∠A.

Question 29: In ΔABC, AD is the perpendicular bisector of BC (see figure given below). Show that ΔABC is an isosceles triangle where AB = AC.

Image source: NCERT textbook

Solution 29:

Given AD is the perpendicular bisector of BC

To prove:

AB = AC

Proof:

BD = CD (Because AD is the perpendicular bisector)

Therefore,

AB = AC (by CPCT rule)

Question 30:  ABC is an isosceles triangle in which altitudes BE, and CF is drawn to equal sides AC and AB, respectively (see figure given below). Show that the two altitudes are equal.

Image source: NCERT textbook

Solution 30:

Given:

(i) BE and CF are the two altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar to the AAS congruency rule as

∠A = ∠A (common arm)

∠AEB = ∠AFC (Since right angles)

AB = AC (Given in the question)

∴ ΔAEB ≅ ΔAFC and so, BE = CF (by CPCT).

Question 31: ABC is a triangle where the altitudes BE and CF to the sides AC and AB are equal (see  figure given below). Show that

(i) ΔABE ≅ ΔACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Image source: NCERT textbook

Solution 31:

We know BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (Since common angle)

∠AEB = ∠AFC (Since right angles)

BE = CF (Given)

Therefore, ΔABE ≅ ΔACF by AAS congruency condition.

(ii) AB = AC by CPCT rule, and thus, ABC is an isosceles triangle.

Question 32:  ABC and DBC are the two isosceles triangles lying on the same base BC (see figure given below). Show that ∠ABD = ∠ACD.

Image source: NCERT textbook

Solution 32:

We have ABC and DBC are two isosceles triangles.

We are to show that ∠ABD = ∠ACD

Proof:

Triangles ΔABD and ΔACD are similar to the SSS congruency rule since

AB = AC (ABC is an isosceles triangle)

BD = CD (BCD is an isosceles triangle)

As, ΔABD ≅ ΔACD.

Therefore, ∠ABD = ∠ACD by the rule of CPCT.

Question 33:  ΔABC is an isosceles triangle where AB = AC. Side BA is produced to D so that AD = AB (see figure given below). Show that ∠BCD is a right angle.

Image source: NCERT textbook

Solution 33:

Given, AB = AC and AD = AB

We are to prove ∠BCD is a right angle.

Proof:

We consider ΔABC,

AB = AC (given in the question)

And ∠ACB = ∠ABC (angles opposite to the equal sides are equal)

Now, we consider ΔACD,

Also, ∠ADC = ∠ACD (angles opposite to the equal sides are equal)

Again,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

So, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — eq.(i)

∠CAD = 180° – 2∠ACD — eq.(ii)

and,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding equations (i) and (ii), we have,

∠CAB + ∠CAD = 180° – 2∠ACB+180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB-2∠ACD

⇒ 2(∠ACB+∠ACD) = 180°

⇒ ∠BCD = 90°

Question 34:  ABC is a right-angled triangle where ∠A = 90° and AB = AC. Find ∠B and ∠C.

Solution 34:

Image source: NCERT textbook

Given,

∠A = 90° and AB = AC

AB = AC

⇒ ∠B = ∠C (angles opposite to the equal sides are equal)

Now,

∠A+∠B+∠C = 180° (We know the sum of the interior angles of the triangle)

∴ 90° + 2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

So, ∠B = ∠C = 45°

Question 35: Show all the angles of the equilateral triangle 60° each.

Solution 35:

Let ABC be the equilateral triangle as shown below:

Here, BC = AC = AB (length of all sides are same)

⇒ ∠A = ∠B =∠C (opposite sides to equal angles are equal.)

Also, we know,

∠A+∠B+∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

∴ ∠A = ∠B = ∠C = 60°

Therefore, the angles of the equilateral triangle are 60° each.

Question 36:  ΔABC and ΔDBC are the two isosceles triangles lying on the same base BC and the vertices A and D lying on the same side of BC (see figure given below). If AD is extended so that it intersects BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A and ∠D.

(iv) AP is the perpendicular bisector of the side BC.

Image source: NCERT textbook

Solution 36:

As per the question, ΔABC and ΔDBC are the two isosceles triangles.

(i) ΔABD and ΔACD are similar as per the SSS congruency because:

AB = AC (As ΔABC is an isosceles triangle)

BD = CD (Since ΔDBC is an isosceles triangle)

∴ ΔABD ≅ ΔACD.

(ii) ΔABP and ΔACP are similar because:

AP = AP (Since common side)

∠PAB = ∠PAC (by CPCT rule since ΔABD ≅ ΔACD)

AB = AC (As ΔABC is an isosceles triangle )

Thus, ΔABP ≅ ΔACP by SAS congruency rule condition.

(iii) ∠PAB = ∠PAC by CPCT rule as ΔABD ≅ ΔACD.

AP bisects ∠A. — eq. (i)

Also, ΔBPD and ΔCPD are similar in the SSS congruency rule as

PD = PD (Since common side)

BD = CD (Since ΔDBC is an isosceles triangle.)

BP = CP (by CPCT rule as ΔABP ≅ ΔACP)

Thus, ΔBPD ≅ ΔCPD.

Thus, ∠BDP = ∠CDP by CPCT. — eq. (ii)

Now by comparing equations (i) and (ii), we can say that AP bisects ∠A and ∠D.

(iv) ∠BPD = ∠CPD (by CPCT rule as ΔBPD ΔCPD)

and BP = CP — eq.(i)

also,

∠BPD +∠CPD = 180° (Since BC is a straight line.)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90° —eq.(ii)

Now, from equations (i) and (ii), it can be concluded that

AP is the perpendicular bisector of BC.

Question 37:  AD is an altitude of an isosceles triangle ABC where AB = AC. Show that

Solution 37:

Given AD is an altitude and AB = AC.

(i) In ΔABD and ΔACD,

AB = AC (given in the question)

Therefore, ΔABD ≅ ΔACD by RHS congruence condition.

Now, by the CPCT rule,

BD = CD.

Question 38:  Two sides AB and BC and the median AM of triangle ABC are respectively equal to the sides PQ and QR and the median PN of ΔPQR (see figure given below). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR

Image source: NCERT textbook

Solution 38:

Some given parameters are:

AB = PQ,

BC = QR and

AM = PN

(i) ½ BC = BM and ½ QR = QN (As AM and PN are the medians)

Also, BC = QR

So, ½ BC = ½ QR

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN and AB = PQ (As per the question)

BM = QN (proved earlier)

∴ ΔABM ≅ ΔPQN by SSS rule congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (As per the question)

∠ABC = ∠PQR (by CPCT)

Therefore, ΔABC ≅ ΔPQR by SAS congruency.

Question 39:  BE and CF are the two equal altitudes of triangle ABC. Using the RHS congruence condition, prove that triangle ABC is isosceles.

Image source: NCERT textbook

Solution 39:

We know that BE and CF are two equal altitudes.

Now, in ΔBEC and ΔCFB,

∠BEC = ∠CFB = 90° (Since same altitudes)

BC = CB (Since common side)

BE = CF (Since common side)

Thus, ΔBEC ≅ ΔCFB by RHS congruence rule criterion.

And, ∠C = ∠B (by CPCT rule)

Therefore, AB = AC as opposite sides to the equal angles is always equal.

Question 40: ABC is an isosceles triangle in which AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution 40:

Given in the question that AB = AC

Now, ΔABP and ΔACP are similar by RHS congruency rule as

∠APB = ∠APC = 90° (AP is altitude)

AB = AC (Given in the question)

AP = AP (Common side)

So, ΔABP ≅ ΔACP.

∴ ∠B = ∠C (by CPCT)

Question 41:  Show that in any right-angled triangle, the hypotenuse is the longest side.

Image source: NCERT textbook

Solution 41:

We suppose ABC is a triangle with the right angled at B.

Thus we know,

∠A +∠B+∠C = 180°

Now, when ∠B+∠C = 90°, then ∠A has to be 90°.

Again, if A is the largest angle of the triangle, the side opposite to it has to be the largest.

So, AB is the hypotenuse and is the largest side of the right-angled triangle, i.e. ΔABC.

Question 42:  In figure given below, both the sides AB and AC of ΔABC are extended to the points P and Q, respectively. Also, ∠PBC < ∠QCB. Show that AC> AB.

Image source: NCERT textbook

Solution 42:

Given in the question, ∠PBC < ∠QCB

We know ∠ABC + ∠PBC = 180°

Thus, ∠ABC = 180°-∠PBC

Also,

∠ACB +∠QCB = 180°

So, ∠ACB = 180° -∠QCB

Now, since ∠PBC < ∠QCB,

∴ ∠ABC > ∠ACB

Therefore, AC > AB as opposite sides to the larger angle are larger.

Question 43:  In figure given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Image source: NCERT textbook

Solution 43:

Given in the question that angles B and angle C are smaller than angles A and D respectively, i.e. ∠B < ∠A and ∠C < ∠D.

Now,

Since the opposite side to the smaller angle is always smaller

AO < BO — eq.(i)

And OD < OC —eq.(ii)

By adding equations (i) and (ii), we get

AO+OD < BO + OC

Question 44: △ABC is an isosceles triangle with AB=AC and ∠B = 450. Find ∠A.

Solution 44: In △ABC,  AB=AC

∠C=∠B=45∘

∠A+∠B+∠C=180∘

∠A+45∘+45∘=180∘

⇒∠A=180∘−90∘=90∘

Question 45:  AB and CD are the smallest and longest sides of a quadrilateral ABCD, respectively.

Show that ∠A > ∠C and ∠B > ∠D.

Image source: NCERT textbook

Solution 45:

In ΔABD, we see,

So, ∠ADB < ∠ABD — equation(i) (as the angle opposite to the longer side is always larger)

Now, in ΔBCD,

BC < DC < BD

Therefore, it can be said that

∠BDC < ∠CBD — equation(ii)

Now, by adding equations (i) and (ii), we get,

∠ADB + ∠BDC < ∠ABD + ∠CBD

∠B > ∠D

In a similar manner in triangle ABC,

∠ACB < ∠BAC — equation(iii) (Since the angle opposite to the larger side is always larger)

∠DCA < ∠DAC — equation(iv)

By adding equations (iii) and (iv), we have,

∠ACB + ∠DCA < ∠BAC+∠DAC

Therefore, ∠A > ∠C

Question 46:  In figure given below, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR> ∠PSQ.

Image source: NCERT textbook

Solution 46:

Given in the question PR> PQ and PS bisects ∠QPR

We have to prove that angle PSR is smaller than PSQ, i.e. ∠PSR> ∠PSQ

Proof:

∠PQR> ∠PRQ — equation(i) (Since PR > PQ as angle opposite to the larger side is larger)

∠QPS = ∠RPS — equation (ii) (Since PS bisects ∠QPR)

∠PSR = ∠PQR + ∠QPS — equation (iii) (Since the exterior angle of a triangle equals the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS —equation (iv) (As the exterior angle of a triangle equals the sum of the opposite interior angles)

∠PQR +∠QPS > ∠PRQ +∠RPS

Therefore, from (i), (ii), (iii) and (iv), we get

∠PSR > ∠PSQ

Question 47: Show that of all the line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.

Solution 47:

First, we suppose “l” be a line segment, and “B” is a point lying on it. A line AB which is perpendicular to l, is drawn. Also, let C be any other point on l.

To prove:

AB < AC

Proof:

In ΔABC, ∠B = 90°

We know,

∠A+∠B+∠C = 180°

∴ ∠A +∠C = 90°

Thus, ∠C must be an acute angle implying ∠C < ∠B

So, AB < AC (As we know that the side opposite to the larger angle is always larger)

Question 48:  In the triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of Δ PQR must be equal to the side AB of Δ ABC so that the two given triangles are congruent? Give a reason for your answer.

Solution 48:

In triangles ABC and PQR, we have

∠A = ∠Q [Given]

∠B = ∠R [Given]

For the triangle to be congruent, AB must be equal to QR.

Thus, triangle ABC and PQR may be congruent by the ASA congruence rule.

Question 49:  In the triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of Δ PQR must be equal to side BC of Δ ABC so that the two triangles are congruent? Give a reason for your answer.

Solution 49:

In the triangles ABC and PQR, we have

∠A = ∠Q and ∠B = ∠R [Given]

For the triangles to be congruent, we should have

BC = RP

Thus, triangle ABC and PQR are congruent by following the AAS congruence rule.

Question 50:  “If the two sides and an angle of one of the triangles are equal to the two sides and an angle of another triangle, then the two triangles should be congruent.” Is this statement true? If so, why?

Solution 50:

No, the statement, “If the two sides and an angle of one of the triangles are equal to the two sides and an angle of another triangle, then the two triangles should be congruent.”, is false.

Justification:

Because by the congruency rule,

The two sides and the included angle of one triangle are equal to the two sides, and the included angle of the other triangle, i.e., the SAS rule.

Question 51: “If the two angles and a side of one of the triangles are equal to the two angles and a side of another triangle, then the two triangles must be congruent.” Is this statement true? Why?

Solution 51:

The statement, “If the two angles and a side of one of the triangles are equal to the two angles and a side of another triangle, then the two triangles must be congruent.” is true.

Justification:

The statement is true since the triangles would be congruent either by the ASA rule or the AAS rule. This is so because the two angles and one side are enough to construct two congruent triangles.

Question 52:  Is it possible to construct any triangle with lengths of sides 4 cm, 3 cm and 7 cm? Give a reason for your answer.

Solution 52:

No, it is not possible to construct any triangle with lengths of sides 4 cm, 3 cm and 7 cm.

Justification:

We know,

The sum of any two sides of a triangle is always greater than the third side.

But here, the sum of two sides whose lengths are 4 cm and 3 cm = 4 cm + 3 cm = 7 cm,

which is equal to the length of the third side, i.e., 7 cm.

Therefore, it is not possible to construct a triangle with lengths of sides 4 cm, 3 cm and 7 cm.

Question 53: It is given that Δ ABC ≅ Δ RPQ. Is it true to state that BC = QR? Why?

Solution 53:

It is False that BC = QR. This is because BC = PQ as ΔABC ≅ ΔRPQ.

Question 54: ABC is an isosceles triangle in which AB = AC and BD and CE are the two medians. Show that BD = CE.

Solution 54:

As per the question,

ΔABC is an isosceles triangle where AB = AC, BD and CE are its two medians

From ΔABD and ΔACE,

AB = AC (given in the question)

2 AE = 2 AD (since D and E are the midpoints)

∠A = ∠A (common between the triangles)

Thus, ΔABD ≅ ΔACE (using the SAS rule)

⇒ BD = CE (by CPCT rule)

Hence proved.

Question 55: In the figure given below, D and E are points on side BC of Δ ABC in such a way that BD = CE and AD = AE.

Show that Δ ABD ≅ Δ ACE.

Image source: NCERT textbook

Solution 55:

As per the question,

In ΔABC,

BD = CE and AD = AE.

As opposite angles to equal sides are equal,

Therefore we have,

Also, ∠AED + ∠AEC = 180° (since linear pair)

∠AEC = 180° – ∠AED

∠AEC = 180° – ∠ADE … eq.(3)

From equation (2) and (3)

BD = EC (given)

∠ADB = ∠AEC (from eq. (4)

Thus, ΔABD ≅ΔACE (by SAS rule)

Question 56: CDE is the equilateral triangle formed on the side CD of square ABCD in figure given below. Show that

Image source: NCERT textbook

Solution 56:

As per the question,

CDE is an equilateral triangle which is formed on a side CD of a square ABCD.

DE = CE (sides of an equilateral triangle)

Now,

Also, ∠EDC = ∠ECD = 60°

Thus, ∠ADE = ∠ADC + ∠CDE = 90° + 60° = 150°

And ∠BCE = ∠BCD + ∠ECD = 90° + 60° = 150°

AD = BC (sides of the square)

Therefore, ΔADE ≅ΔBCE (by SAS rule)

Question 57:  In figure given below, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that Δ ABC ≅ Δ DEF.

Image source: NCERT textbook

Solution 57:

As per the question,

BA ⊥ AC, DE ⊥ DF in such a way that BA = DE and BF = EC.

In ΔABC and ΔDEF

BA = DE (given)

BF = EC (given)

∠A = ∠D (both 90°)

BC = BF + FC

EF = EC + FC = BF + FC (∵ EC = BF)

⇒ EF = BC

Hence, ΔABC ≅ ΔDEF (by RHS)

Question 58:  Q is any point on the side SR of a Δ PSR so that PQ = PR. Prove that PS> PQ.

Solution 58:

Given: In ΔPSR, Q is any point on the side SR so that PQ = PR.

In ΔPRQ,

PR = PQ (given)

⇒ ∠PRQ = ∠PQR (angles opposite to equal sides are equal)

But ∠PQR> ∠PSR (i.e., the exterior angle of a triangle is greater than each of the opposite interior angles)

⇒ ∠PRQ > ∠PSR

⇒ PS> PR (sides opposite to the greater angle is greater)

⇒ PS > PQ (since PR = PQ)

Question 59: Find the angles of an equilateral triangle.

Solution 59:

We know in an equilateral triangle,

all sides are equal.

Thus, all angles are equal as well

Let x be the angles of an equilateral triangle

Now, following the  angle sum property,

We know the sum of the interior angles=180o.

x+x+x=180o

3x=180

x=60°

Hence, all angles of an equilateral triangle are 60°

Question 60: The image of an object which is placed at point A before a plane mirror LM such that it is seen at point B by an observer standing at D, as shown in figure given below. Prove that the image is as far behind the mirror as the object is in front of the mirror.

[Note: CN is normal to mirror. Also, angle of incidence = angle of reflection].

Image source: NCERT textbook

Solution 60:

We suppose that AB intersects LM at O. We have to prove that AO = BO.

Now, ∠i = ∠r …(1)

[∵Angle of incidence = Angle of reflection]

∠B = ∠i [Corresponding angles] …(2)

And ∠A = ∠r [Alternate interior angles] …(3)

From (1), (2) and (3), we get

∠B = ∠A

⇒ ∠BCO = ∠ACO

In ΔBOC and ΔAOC, we have

∠1 = ∠2 [Each = 90o]

OC = OC [i.e., Common side]

And ∠BCO = ∠ACO [Proved earlier]

ΔBOC ≅ ΔAOC [ASA congruency rule]

Therefore, AO = BO [CPCT]

Question 61: ABC is an isosceles triangle in which AB = AC, and D is the point on BC in such a way that AD ⊥ BC (in figure given below). To prove that ∠BAD = ∠CAD, a student started as follows:

In Δ ABD and Δ ACD,

AB = AC (Given)

∠B = ∠C (since AB = AC)

Thus Δ ABD Δ Δ ACD (AAS)

f

Image source: NCERT textbook

What is the defect/problem in the above arguments?

[Hint: Recall how ∠B = ∠C is proved when AB = AC].

Solution 61:

Image source: NCERT textbook

In Δ ABD and Δ ADC, we have

As per the question,

AB = BC

By the RHS criterion of congruence rule,

We have,

Δ ABD ≅ Δ ACD

Hence Proved.

Question 62: P is the point on the bisector of ∠ABC. If the line passing through P, parallel to BA, meets BC at Q, we are to prove that BPQ is an isosceles triangle.

Solution 62:

Image source: NCERT textbook

To prove: BPQ is an isosceles triangle.

As per the question,

As BP is the bisector of ∠ABC,

∠1 = ∠2 … eq. (1)

Here, PQ is parallel to BA, and BP cuts them

∠1 = ∠3 [Since alternate angles] … eq.(2)

Now from equations (1) and (2),

We have

∠2 = ∠3

In Δ BPQ,

We have

∠2 = ∠3

PQ = BQ

Therefore, BPQ is an isosceles triangle.

Question 63: ABCD is a quadrilateral with AB = BC and AD = CD. Show that BD is bisecting both the angles ABC and ADC.

Solution 63:

As per the question,

In ΔABC and ΔCBD,

We have

AB = BC

BD = BD [As common side]

ΔABC ≅ ΔCBD [By SSS congruence rule]

⇒ ∠1 = ∠2 [CPCT rule]

And ∠3 = ∠4

Therefore, BD bisects both ∠ABC and ∠ADC.

Question 64:  ABC is a right triangle with AB = AC. The bisector of ∠A meets BC at D. Prove that BC = 2 AD.

Solution 64:

Given: A right-angled triangle with AB = AC bisector of ∠A meets BC at D.

Proof:

According to the question,

In the right Δ ABC,

AB = AC

Since hypotenuse is the longest side,

BC is hypotenuse

∠BAC = 90o

Now,

We have,

AC = AB

Since AD is the bisector of ∠A,

∠1 = ∠2

Now,

By SAS criterion of congruence,

We get,

CD = BD [CPCT]

Since the mid-point of the hypotenuse of the right triangle is equidistant from the three vertices of the triangle.

AB = BD = CD …eq.(1)

Here, BC = BD + CD

Hence, proved.

Question 65:  O is the point in the interior of the square ABCD in such that OAB is an equilateral triangle. Show that Δ OCD is an isosceles triangle.

Solution 65:

Given in the question, A square ABCD and OA = OB = AB.

To prove: Δ OCD is an isosceles triangle.

Proof:

In square ABCD,

As ∠1 and ∠2 is equal to 90o

∠1 = ∠2 …eq.(1)

Now, in Δ OAB, we have

As ∠3 and ∠4 is equal to 60o

∠3 = ∠4 …(2)

Subtracting equations (2) from (1),

We get

∠1−∠3 = ∠2 −∠4

⇒ ∠5 = ∠6

Again,

In Δ DAO and Δ CBO,

∠5 = ∠6 [Proved earlier]

OA = OB [Given]

By the SAS criterion of congruence rule,

We have

Δ DAO ≅ Δ CBO

OD = OC

⇒ Δ OCD is an isosceles triangle.

Hence, proved.

Question 66: ABC and DBC are the two triangles on the same base BC such that A and D lie on opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Solution 66:

Given: Δ ABC and Δ DBC lie on the same base BC. Also, AB = AC and BD = DC.

To prove: AD is the perpendicular bisector of BC, i.e., OB = OC

AB = AC [Given]

BD = CD [Given]

So, by the SSS criterion of congruence, we have

∠1 = ∠2 [CPCT]

Now, in Δ BAO and Δ CAO, we have

AB = AC [Given]

∠1 = ∠2 [Proved above]

AO = AO [Common side]

So, by the SAS criterion of congruence, we have

Δ BAO ≅ Δ CAO

BO = CO [CPCT]

And, ∠3 = ∠4 [CPCT]

But, ∠3+∠4 =180o [Linear pair axiom]

⇒ ∠3+∠3 =180

⇒ 2∠3 =180

⇒ ∠3 =180/2

⇒ ∠3 = 90o

Since BO = CO and ∠3 = 90o,

AD is a perpendicular bisector of BC.

Hence proved.

Question 67: ABC is an isosceles triangle with AC = BC. AD and BE are two altitudes to sides BC and AC, respectively. Prove that AE = BD.

Solution 67:

Image source: NCERT textbook

As per the question,

In Δ ADC and Δ BEC,

We have

AC = BC [Given] …eq(1)

As ∠ADC and ∠BEC = 90o

∠ACD = ∠BCE [Since common angle]

Δ ADC ≅ Δ BEC [By ASA congruence rule]

CE = CD … (2) [CPCT rule]

Subtracting equation (2) from (1),

We have

AC – CE = BC – CD

⇒ AE = BD

Hence proved.

Question 68: Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Solution 68:

As per the question,

We have Δ ABC with AD as its median.

To prove:

Construction:

Join EC.

Proof:

In Δ ADB and Δ EDC,

∠1 = ∠2 [Vertically opposite angles are equal]

DB = DC [Given]

So, by the SAS criterion of congruence, we have

AB = EC [CPCT]

And ∠3 = ∠4 [CPCT]

Now, in Δ AEC,

Since the sum of the lengths of any two sides of a triangle must be greater than the third side,

We have

AC + CE > AE

⇒ AC + CE > AD + DE

⇒ AC + CE > 2AD

⇒ AC + AB > 2AD [∵AB = CE]

Similarly,

We get,

Hence, proved.

Question 69: ABCD is a quadrilateral where AD = BC and ∠DAB = ∠CBA. Prove that

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Solution 69:

As per the question,

∠DAB = ∠CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by following SAS rule congruency as

AB = BA (the common arm)

∠DAB = ∠CBA and AD = BC (given)

Thus, triangles ABD and BAC are similar

i.e. ΔABD ≅ ΔBAC. (Hence proved).

(ii) Since it is already proven,

ΔABD ≅ ΔBAC

Thus,

BD = AC (by CPCT)

(iii) As ΔABD ≅ ΔBAC

So, the angles are equal,

∠ABD = ∠BAC (by CPCT).

Question 70: AD and BC are equal and perpendicular to the line segment AB. Show that CD bisects AB.

Solution 70:

Given AD and BC are two equal perpendiculars to AB.

To prove: CD is the bisector of AB

Proof:

The triangles ΔAOD and ΔBOC are similar by AAS rule congruency

Since:

(i) ∠A = ∠B (perpendicular angles)

(ii) AD = BC (given in the question)

(iii) ∠AOD = ∠BOC (i.e., vertically opposite angles)

∴ ΔAOD ≅ ΔBOC.

Thus, AO = OB ( by CPCT).

Hence, CD bisects AB (Hence proved).

Question 71: Line l is the bisector of the angle ∠A, and B is any point on the line l. BP and BQ are the perpendiculars from B to the arms of ∠A. Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Image source: NCERT textbook

Solution 71:

It is given in the question that line “l” is the bisector of angle ∠A, and the line segments BP and BQ are perpendiculars drawn from line l.

(i) ΔAPB and ΔAQB are similar by following the AAS congruency rule because;

∠P = ∠Q (right angles)

AB = AB (common arm)

∠BAP = ∠BAQ (l is the bisector of angle A)

Thus, ΔAPB ≅ ΔAQB.

(ii) By CPCT rule, BP = BQ. Thus, we can say that point B is equidistant from the arms of ∠A.

Question 72: AB is the line segment, and P is the mid-point. D and E are the points on the same side of AB, so  ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that

(i) ΔDAP ≅ ΔEBP

Solution 72:

Given that P is the mid-point of line segment AB.

And, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) Given, ∠EPA = ∠DPB

Now, adding ∠DPE on both sides,

∠EPA + ∠DPE = ∠DPB + ∠DPE

Thus it can be said that angles DPA and EPB are equal, i.e. ∠DPA = ∠EPB

Now, considering the triangles DAP and EBP.

∠DPA = ∠EPB

AP = BP (As P is the mid-point of line segment AB)

Thus, by the ASA congruency criterion,

ΔDAP ≅ ΔEBP.

(ii) By CPCT rule,

Question 73: In the right triangle, ABC, which is right-angled at C and M, is the mid-point of hypotenuse AB. C is joined to M and is produced to a point D so that DM = CM. Point D is joined to point B. Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = 1/2 AB

Solution 73:

We know M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM

(i) We consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given)

∠CMA = ∠DMB (Vertically opposite angles)

So, by SAS congruency criterion, ΔAMC ≅ ΔBMD.

(ii) ∠ACM = ∠BDM (by CPCT)

∴ AC ∥ BD as alternate interior angles is equal.

Now, ∠ACB + ∠DBC = 180° (Since they are co-interiors angles)

⇒ 90° + ∠B = 180°

∴ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Since common side)

∠ACB = ∠DBC (Both are right angles)

DB = AC (by CPCT rule)

Thus, ΔDBC ≅ ΔACB by SAS congruency rule.

(iv) DC = AB (Since ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (Since M is the mid-point)

Thus, DM + CM = BM + AM

Therefore, CM + CM = AB

⇒ CM = (½) AB

Question 74: ABC is an isosceles triangle with altitudes BE and CF, which are drawn to equal sides, AC and AB, respectively. Show that the altitudes are equal.

Image source: NCERT textbook

Solution 74:

Given:

(i) BE and CF are the altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar by the AAS congruency rule since;

∠A = ∠A (Since common arm)

∠AEB = ∠AFC (As both are right angles)

AB = AC (Given in the question)

Therefore, ΔAEB ≅ ΔAFC

and BE = CF (by CPCT rule).

Question 75: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Image source: NCERT textbook

Solution 75:

Given, AB = AC and AD = AB

To prove: ∠BCD is a right angle.

Proof:

Consider ΔABC,

AB = AC (Given)

Also, ∠ACB = ∠ABC (Angles opposite to equal sides)

Now, consider ΔACD,

Also, ∠ADC = ∠ACD (Angles opposite to equal sides)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

Therefore, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — eq.(i)

In a similar manner in ΔADC,

∠CAD = 180° – 2∠ACD — eq.(ii)

Again,

∠CAB + ∠CAD = 180° (BD is the straight line.)

Adding equations (i) and (ii), we have,

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

Question 76: ΔABC and ΔDBC are the two isosceles triangles on the same base BC, with vertices A and D on the same side of BC. If AD is extended so that it intersects BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A and ∠D.

(iv) AP is perpendicular bisector of BC.

Solution 76:

We have ΔABC and ΔDBC are the two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruency rule because:

AB = AC (ΔABC is isosceles)

BD = CD (ΔDBC is isosceles)

∴ ΔABD ≅ ΔACD.

(ii) ΔABP and ΔACP are similar because:

AP = AP (common side)

∠PAB = ∠PAC ( by CPCT rule since ΔABD ≅ ΔACD)

AB = AC (As ΔABC is isosceles)

Thus, ΔABP ≅ ΔACP by SAS congruency rule.

(iii) ∠PAB = ∠PAC by CPCT rule as ΔABD ≅ ΔACD.

AP bisects ∠A. ………… eq. (1)

Again, ΔBPD and ΔCPD are similar by SSS congruency rule as

PD = PD (Since common side)

BD = CD (Since ΔDBC is an isosceles triangle.)

BP = CP (by CPCT rule as ΔABP ≅ ΔACP)

Thus, ΔBPD ≅ ΔCPD.

Thus, ∠BDP = ∠CDP by CPCT. ……………. eq.(2)

Now we compare equations (1) and (2), and it can be interpreted that AP is bisecting ∠A and ∠D.

(iv) ∠BPD = ∠CPD (by CPCT rule as ΔBPD ≅ ΔCPD)

and BP = CP — eq.(1)

also,

∠BPD + ∠CPD = 180° (Since BC is a straight line.)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90° —eq.(2)

Now, from equations (1) and (2), it can be said that

AP is the perpendicular bisector of BC.

Question 77: Two sides AB and BC and median AM of one triangle ABC are equal to sides PQ and QR, respectively, and the median PN of ΔPQR. Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR

Solution 77:

Given in the question;

AB = PQ,

BC = QR and

AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)

And BC = QR

Thus, 1/2 BC = 1/2QR

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN and AB = PQ (Given)

BM = QN (proved earlier)

∴ ΔABM ≅ ΔPQN by SSS congruency rule.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (Given)

∠ABC = ∠PQR (by CPCT rule)

So, ΔABC ≅ ΔPQR by SAS congruency rule.

Question 78: In the Figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR> ∠PSQ.

Image source: NCERT textbook

Solution 78:

Given in the question, PR> PQ and PS bisects ∠QPR

To prove: ∠PSR> ∠PSQ

Proof:

∠QPS = ∠RPS — eq.(1) (PS bisects ∠QPR)

∠PQR > ∠PRQ — eq.(2) (Since PR > PQ as angle opposite to the larger side is always larger)

∠PSR = ∠PQR + ∠QPS — eq.(3) (Since the exterior angle of a triangle equals the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — eq.(4) (As the exterior angle of a triangle equals the sum of opposite interior angles)

By adding equations (1) and (2), we get

∠PQR + ∠QPS > ∠PRQ + ∠RPS

Now, from (1), (2), (3) and (4), we have

∠PSR > ∠PSQ

Question 79:  AD and BC are equal perpendiculars to line segment AB. Show that CD bisects AB.

Solution 79:

In △BOC and △AOD

BOC = AOD [Since Vertically Opposite angles]

BOC = AOD [By ASA congruency rule]

OB  =  OA and OC  =  OD [By C.P.C.T. rule]

Question 80: In quadrilateral ABCD. AC = AD and AB are bisecting ∠A. Show that △ABC ≅ △ABD. What can you depict about BC and BD?

Solution 80: Given: In quadrilateral ABCD

AC  = AD and AB bisect ∠A.

To prove △ABC ≅△ABD.

Proof: In △ABC and △ABD,

∠BAC   = ∠BAD [ AB bisects ∠A]

AB  =  AB [Common side]

△ABC≅ △ABD [By SAS congruency rule]

Hence, BC  =  BD [By C.P.C.T. rule]

Question 81:  ABCD is a quadrilateral where AD = BC and ∠ DAB = ∠CBA. Prove that:

(i) △ABD ≅△BAC

Solution 81: In ABC and ABD,

DAB  = CBA [Given]

AB  =  AB [Common side]

ABC≅ABD[By SAS congruency rule]

Thus AC  =  BD [By C.P.C.T.]

(ii) BD = AC

Solution: Since △ABC ≅△ABD

AC  =  BD [By C.P.C.T.]

(iii) ∠ ABD = ∠ BAC

Solution: Since △ABC≅ △ABD

∠ABD  = ∠BAC [By C.P.C.T.]

Question 82:  l and m are two parallel lines which are intersected by another pair of parallel lines, p and q. Show that △ABC ≅ △CDA.

Solution 82: AC is a transversal.

Given ∠DAC  = ∠ACB [Alternate angles]

Now p∥q [Given]

AC is a transversal. [Given]

Thus ∠BAC = ∠ACD [Since Alternate angles]

∠ACB  = ∠DAC [Proved above]

∠BAC  = ∠ACD [Proved above]

AC  =  AC [Common]

△ABC≅ △CDA [By ASA congruency]

Question 83: Line l is the bisector of the angle A. B is any point on BP. BP and BQ are perpendicular from B to the arms of A. Show that:

(i). △APB ≅ △AQB

Solution 83:

Given: Line l bisects ∠A

∠BAP  = ∠BAQ

In △ABP and △ABQ

∠BAP  = ∠BAQ[Given]

∠BPA  = ∠BQA  = [Given]

AB  =  AB [Common]

△APB≅ △AQB [By ASA congruency]

(ii). BP = BQ or P is equidistant from the arms of ∠A

Solution: As △APB≅△AQB

BP  =  BQ [By C.P.C.T.]

B is equidistant from the arms of ∠A.

Question 84:  In the given triangle, AC = AB, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Solution 84:

we get,∠BAD  + ∠DAC  = ∠EAC  + ∠DAC

and AC  =  AE [Given]

∠BAC  = ∠DAE [From eq. (i)]

△ABC≅ △ADE [By SAS rule congruency]

BC  =  DE [By C.P.C.T. rule]

Question 85: AB is a line segment, and P is the mid-point. D and E are the points on the same side of AB, so BAD = ABE and EPA = DPB. Show that: (i) DAF FBPE D (ii) AD = BE

Solution 85:

Given that ∠EPA = ∠DPB

Adding ∠EPD on both sides, we get,

∠EPA  + ∠EPD  = ∠DPB  + ∠EPD

∠APD  = ∠BPE ………….eq.(i)

Here, in △APD and △BPE,

AP  = PB [P is the mid-point of AB]

∠APD  = ∠BPE [From eq. (i)]

∠DPA = ∠EBP [By ASA congruency]

AD  =  BE [ By C.P.C.T. rule]

Question 86:  In an isosceles triangle ABC, where AB = AC, the bisectors of B and C are intersecting each other at O. Join A to O. Show that:

(i) OB = OC

Solution 86:

ABC is an isosceles triangle where AB = AC

⇒ C  =  B [Angles opposite to equal sides]

△ OCA  + △OCB  = △OBA  + △OBC

OB bisects ∠B, and OC bisects ∠C

∠OBA  = ∠OBC and ∠OCA  = ∠OCB

∠OCB  + ∠OCB  = ∠OBC  + ∠OBC

2∠OCB  =  2∠OBC

∠OCB  = ∠OBC

Now in △OBC,

∠OCB  = ∠OBC [Proved above]

OB  =  OC [Sides opposite to equal sides]

(ii) AO bisects A

Solution: In △AOB and △AOC,

AB  =  AC [Given]

∠OBA  = ∠OCA[Given]

And ∠B  = ∠C

12∠B = 12∠C

∠OBA  = ∠OCA

OB  =  OC [Proved above]

△AOB ≅△AOC [By SAS congruency]

Question 87: In ABC, AD is a perpendicular bisector of BC. Show that ABC is an isosceles triangle where AB = AC.

Solution 87:

In △AOB and △AOC,

BD  =  CD [AD bisects BC]

∴△ABD ≅ △ACD [By SAS congruency]

therefore, AB  =  AC [By C.P.C.T.]

Thus, ABC is an isosceles triangle.

Question 88: ABC is an isosceles triangle whose altitudes BE and CF are drawn to sides AC and AB, respectively. Show that these altitudes are equal.

Solution 88: In △ABE and △ACF,

∠A = ∠A [Common]

∠AEB  = ∠AFC[ Since   AD⊥BC] [Given]

AB  =  AC [Given]

△ABE ≅ △ACF [By ASA congruency]

BE  =  CF [By C.P.C.T.]

Therefore, the altitudes are equal.

Question 89: ABC is a triangle whose altitudes BE and CF to sides AC and AB are equal. Show that:

(i). In △ABE and △ACF,

Solution 89: In △ABE and △ACF,

∠A = ∠A [Common]

[Given]

∠AEB  = ∠AFC  = 900[Given]

BE  =  CF [Given]

△ABE ≅ △ACF [By ASA congruency]

(ii) AB = AC or △ABC is an isosceles triangle.

Solution: Since, △ABE ≅ △ACF

BE  =  CF [By C.P.C.T.]

Thus, △ABC is an isosceles triangle.

Question 90: ABC and DBC are the two isosceles triangles on the same base BC. Show that

∠ABD = ∠ACD.

Solution 90: In the isosceles triangle ABC, AB  =  AC [Given]

∠ACB  = ∠ABC ……….(i) [i.e. Angles opposite to equal sides]

Also, in the isosceles triangle BCD.

BD  =  DC

∠BCD  = ∠CBD ………..(ii)[i.e., Angles opposite to equal sides]

Adding eq. (i) and (ii), we get

∠ACB  + ∠BCD  = ∠ABC  + ∠CBD

∠ACD  = ∠ABD or

∠ABD  = ∠ACD

Question 91: ABC is a right-angled triangle where ∠A = 900 and AB = AC. Find ∠B and ∠C.

Solution 91: △ABC is a right triangle where,

∠A  =  900 And AB  =  AC

In △ABC,

AB  =  AC ⇒∠C  = ∠B …………eq(i)

We know,in △ABC, ∠A  + ∠B  + ∠C  =   1800 [Angle sum property]

900 +  B  +  B  = 1800 [ ∠A  =  900(given) and ∠B  = ∠C (from eq. (i)]

2 ∠B  =   900

∠B  =   900

Also ∠C  = 450 [ ∠B  = ∠C]

Question 92: AD is the altitude of an isosceles triangle ABC w AB = AC. Show that:

Solution 92: In △ABD and △ACD,

AB  =  AC [Given]

△ABD ≅ △ACD [RHS rule of congruency]

BD  =  DC [By C.P.C.T.]

Solution:

Question 93: Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution 93:  We suppose that ABC be a right-angled triangle, ∠ B =900.

To prove: Hypotenuse AC is the longest side.

∠A  + ∠B  + ∠C  = ∠A  +  900 + ∠C  =   [∠ B  =  900

∠A  + ∠C  = 1800 –  900

And ∠B  =  900

∠B  > ∠C and ∠B  > ∠A

Since the greater angle has a side longer as opposed to it.

AC  >  AB and AC  >  AB

Therefore, ∠B being the greatest angle, has the longest opposite side AC, i.e. hypotenuse.

Question 94: In ABC, the sides AB and AC are extended to points P and Q, respectively. Also PBC < QCB. Show that AC> AB.

Solution 94: Given: In △ABC, ∠PBC < ∠QCB

To prove: AC  >  AB

Proof: In △ABC, ∠4  > ∠2 [Given]

Now ∠1  + ∠2  = ∠3  + ∠4  = 1800 [Linear pair]

∠1  > ∠3 [ ∵∠4  > ∠ 2]

AC> AB [Side opposite to the greater angle is longer]

Question 95: In the given triangle, B < A and C < D. Show that AD < BC.

Solution 95: In △AOB

∠B  < ∠A [Given]

OA  < OB………..(i)[Side opposite to the greater angle is longer]

In △COD, ∠C  < ∠ D [Given]

OD  <  OC ……….(ii) [opposite side to greater angle is longer]

Adding eq. (i) and (ii), we get

OA  +  OD  <  OB  +  OC

Question 96:  In a triangle, locate any point in its interior that is equidistant from all sides of a triangle.

Solution 96: Let △ABC be a triangle.

We draw the bisectors of ∠B and ∠C.

We suppose that these angle bisectors intersect each other at a point I.

Draw IK⊥BC

Also, draw IJ⊥AB and IL⊥AC.

Join AI.

In △BIK and △BIJ,

∠ IKB  = ∠IJB  = 900[By construction]

∠ IBK  = ∠IBJ

BI is the bisector of ∠B (By construction)]

BI  =  BI [Common side]

△BIK ≅ △BIJ [ASA criteria of congruency rule]

IK  =  IJ [By C.P.C.T.rule]……….eq.(i)

In a similar way, △CIK ≅ △CIL

∴IK  =  IL [By C.P.C.T.]……….eq.(ii)

From eq (i) and (ii),

IK  =  IJ  =  IL

Thus, I is a point of intersection of the angle bisectors of any two angles of △ABC which are equidistant from its sides.

Question 97: In quadrilateral ACBD, AB=AD and AC are bisecting at A. show ABC ≅ ACD?

Solution 97: In  △ABC and △ACD,

∠BAC = ∠CAD ……(AC bisects A)

And AC = AC……………… (Common side)

△ABC ≅ △ACD………….. (By SAS axiom)

Question 98: If DA and CB are equal perpendiculars to line segment AB. Show that CD is bisecting AB.

Solution 98:  In

△AOD and △BOC,

∠A = ∠B and

∠AOD  = ∠BOC(vertically opp. Angles)

∴∠AOD  = ∠BOC(AAS rule)

∴OA = OB(CPCT)

Hence, CD bisects AB.

Question 99: l and m are two parallel lines that are intersected by another pair of parallel lines, p and q. show that △ABC ≅ △CDA.

Solution 99:

L∥M and AC cuts them (given)

P ∥ Q and AC cuts them (Given)

∴∠CAB=∠ACD(Alternate angles)

AC = CA(common)

∴ΔABC≅ΔCDA(ASA rule)

Benefits of Solving Important Questions Class 9 Mathematics Chapter 7

Triangles are a crucial topic since the underlying ideas discussed there are applied at higher educational levels. Students are advised to complete the additional textbook questions that the CBSE Board has approved for this purpose after finishing the questions from the NCERT Textbook. Our Mathematics specialists make sure to give all the questions, which are based on the NCERT textbook and the CBSE curriculum. The only method to ace the examinations is to correctly answer these Mathematics Class 9 Chapter 7 Important Questions. Thus, the exams can be taken without any fear by the students.

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Be in touch with us at Extramarks to get the important exam questions for CBSE Class 9 Mathematics, all chapters, with solutions. Use the additional questions offered here by clicking on the links below to fasten your preparation.

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Q.1 Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Marks:4
Ans

$\begin{array}{l}\text{GivenatriangleABCinwhichAD}\text{BC,BE}\text{ACandCF}\text{AB.}\\ \mathrm{To}\text{}\mathrm{Prove}\text{}:\text{}\mathrm{AD}+\mathrm{BE}+\mathrm{CF}\text{}<\text{}\mathrm{AB}+\mathrm{BC}+\mathrm{AC}\\ \mathrm{AB}>\mathrm{AD}\text{}\mathrm{and}\text{}\mathrm{AC}>\mathrm{AD}\text{}\left(\mathrm{AD}\mathrm{BC}\right)\\ \mathrm{AB}\text{}+\text{}\mathrm{AC}\text{}>\text{}\mathrm{Ad}\text{}+\text{}\mathrm{AD}\\ \mathrm{AB}\text{}+\text{}\mathrm{AC}\text{}>\text{}2\mathrm{AD}\text{}...\left(1\right)\\ \mathrm{BC}\text{}>\text{}\mathrm{BE}\text{}\mathrm{and}\text{}\mathrm{BA}\text{}>\text{}\mathrm{BE}\text{}\left(\mathrm{BE}\mathrm{AC}\right)\\ \mathrm{BC}\text{}+\text{}\mathrm{BA}\text{}>\text{}\mathrm{BE}+\mathrm{BE}\\ \mathrm{BC}\text{}+\text{}\mathrm{BA}\text{}>\text{}2\mathrm{BE}\text{}...\left(2\right)\text{}\mathrm{and},\\ \mathrm{AC}>\mathrm{CF}\text{}\mathrm{and}\text{}\mathrm{BC}\text{}>\text{}\mathrm{CF}\text{}\left(\mathrm{CF}\mathrm{AB}\right)\\ \mathrm{AC}\text{}+\text{}\mathrm{BC}\text{}>\text{}2\mathrm{CF}\text{}...\left(3\right)\\ \text{Adding}\left(\text{1}\right)\text{,}\left(\text{2}\right)\text{and}\left(\text{3}\right)\text{,weget}\\ \left(\mathrm{AB}+\mathrm{AC}\right)\text{}+\text{}\left(\mathrm{AB}+\mathrm{BC}\right)+\left(\mathrm{AC}+\mathrm{BC}\right)>2\mathrm{AD}+2\mathrm{BE}+2\mathrm{CF}\\ \text{}2\left(\mathrm{AB}+\mathrm{BC}+\mathrm{AC}\right)\text{}>\text{}2\mathrm{AD}+2\mathrm{BE}+2\mathrm{CF}\\ \text{}\mathrm{AB}+\mathrm{BC}+\mathrm{AC}>\mathrm{AD}+\mathrm{BE}+\mathrm{CF}\end{array}$

Q.2 ABCD is a parallelogram and BEFC is a square. Show that triangles ABE and DCF are congruent.

Marks:2
Ans

In the parallelogram ABCD,
BA = CD.
In the square BEFC,
EB = FC.
Since EB is parallel to FC and BA is parallel to CD then,
EBA = FCD
Now, in ABE and DCF, we have
EBA = FCD
BA = CD
EB = FC
Therefore, ABE … DCF (By SAS congruence criterion)

Q.3 ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD is a right angle.

Marks:4
Ans

In ABC,
AB = AC (Given)
ACB = ABC (Angles opposite to equal sides of a triangle are also equal)
In ACD,
ADC = ACD (Angles opposite to equal sides of a triangle are also equal)
In BCD,
ABC + BCD + ADC = 180 (Angle sum property of a triangle)
ACB + ACB + ACD + ACD = 180
2( ACB + ACD) = 180
2( BCD) = 180
BCD = 90

Q.4 D is a point on side BC of ABC such that AD = AC. Show that AB > AD.

Marks:3
Ans

In DAC,
So, ADC = ACD (Angles opposite to equal sides)
Now, ADC is an exterior angle for ABD.
or, ACD > ABD
or, ACB > ABC
So, AB > AC (Side opposite to larger angle in ABC)

Q.5 In Figure, B < A and C < D. Show that AD < BC.

Marks:4
Ans

$\begin{array}{l}\text{Given:-Infigure}\text{B<}\text{Aand}\text{C <}\text{D}\\ \text{Toprove:-}\mathrm{AD}\text{}<\text{}\mathrm{BC}.\\ \text{Proof:-In}\mathrm{AOB},\text{}\mathrm{B}<\mathrm{A}\\ \text{}\mathrm{so},\text{}\mathrm{AO}<\mathrm{BO}\text{}...\left(1\right)\text{}\left[\text{Oppositesideofsmallerangleissmaller}\right]\\ \mathrm{in}\text{}\mathrm{COD},\text{}\mathrm{C}<\mathrm{D}\\ \mathrm{So},\text{}\mathrm{OD}<\mathrm{OC}\text{}...\left(2\right)\text{}\left[\text{Oppositesideofsmallerangleissmaller}\right]\\ \text{Addingrelation}\left(\text{1}\right)\text{and}\left(\text{2}\right)\text{,weâ€‹get}\\ \text{AO+OD