ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions

ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions Ch-10. Step by Step Answer of Exercise-10.2 of Triangles of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-10 Triangles
Topics Solution of Exe-10.2 Questions
Academic Session 2024-2025

Solution of Exe-10.2 Questions

ML Aggarwal Triangles Class 9 ICSE Maths Solutions Ch-10

Question 1. In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.

Answer: In triangle ABC and triangle PQR

∠A = ∠Q

∠B = ∠R

1. In triangles ABC and PQR, and . Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent Give reason for your answer

AB = QP

Because triangles are congruent of their corresponding two angles and included sides are equal

Question 2. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.

Answer: In ∆ABC and ∆PQR

1. In triangles ABC and PQR, and . Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent Give reason for your answer

∠A = ∠Q

∠B = ∠R

Their included sides AB and QR will be equal for their congruency.

Therefore, BC = PR by corresponding parts of congruent triangles.

Question 3. “If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?

Answer:  The given statement can be true only if the corresponding (included) sides are equal otherwise not.

Question 4. In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.

Answer: Given in ∆ABC, AD is median BM and CN are perpendicular to AD form B and C respectively.

4. In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN

To prove:

BM = CN

Proof:

In ∆BMD and ∆CND

BD = CD (because AD is median)

∠M = ∠N

∠BDM = ∠CDN (vertically opposite angles)

∆BMD ≅ ∆CND (AAS axiom)

Therefore, BM = CN.

Question 5. In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

5. In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD

Answer:  Given in figure BM and DN are perpendicular to AC

BM = DN

To prove:

AC bisects BD that is BE = ED

Construction:

Join BD which intersects Ac at E

Proof:

In ∆BEM and ∆DEN

BM = DN

∠M = ∠N (given)

∠DEN = ∠BEM (vertically opposite angles)

∆BEM ≅ ∆DEN

BE = ED

Which implies AC bisects BD

Question 6. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

6. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA

Answer: In the given figure, two lines l and m are parallel to each other and lines p and q are also a pair of parallel lines intersecting each other at A, B, C and D. AC is joined.

To prove: ∆ABC ≅ ∆CDA

Proof:

In ∆ABC and ∆CDA

AC = AC (common)

∠ACB = ∠CAD (alternate angles)

∠BAC = ∠ACD (alternate angles)

∆ABC ≅ ∆ DCA (ASA axiom)

Question 7. In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.

7. In the given figure, two lines AB and CD intersect each other at the point O such that and Show that O is the mid-point of both the line segments AB and CD

Answer: In the given figure, lines AB and CD intersect each other at O such that BC ∥ AD and BC = DA

To prove:

O is the midpoint of Ab and Cd

Proof:

Consider ∆AOD and ∆BOC

AD = BC (given)

∠OAD = ∠OBC (alternate angles)

∠ODA = ∠OCB (alternate angles)

∆AOD ≅ ∆BOC (SAS axiom)

Therefore, OA = OB and OD = OC

Therefore O is the midpoint of AB and CD.

Question 8. In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that

(i) ∆ACD ≅ ∆BDC
(ii) BC = AD
(iii) ∠A = ∠B.

In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that (i) ∆ACD ≅ ∆BDC (ii) BC = AD (iii) ∠A = ∠B.

Answer : 

In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that (i) ∆ACD ≅ ∆BDC (ii) BC = AD (iii) ∠A = ∠B.

∠1= ∠2 and ∠3 = ∠4

∠1+∠3 = ∠2+∠4

∠ACD = ∠BDC

In ∆ACD and ∆BDC

∠ADC = ∠BCD (given)

CD = CD (COMMON)

∠ACD = ∠BDC [from (i)]

∆ACD  ≅  ∆BDC (ASA rule)

AD = BC and ∠A = ∠B (CPCT)

Question 9.  In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that

(i) ∆EBC ≅ ∆DCB
(ii) ∆OEB ≅ ∆ODC
(iii) OB = OC.

Answer :  given:-. ∠ABC=∠ACB. , BE=CD

In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that (i) ∆EBC ≅ ∆DCB (ii) ∆OEB ≅ ∆ODC (iii) OB = OC.

TO prove:-

(i) triangle EBC  triangle DCB

(ii) triangle OEB  triangle ODC

(iii) OB=OC

proof :-

in tri(EBC) and tri(DCB)

∠EBC=∠DCB. ( given ,∠ABC=∠ACB)

BE = CD. (given)

BC= BC. (common)

so , △EBC≡ △DCB. ( by SSS property).

now subtract common (OBC) from both sides

△EBC – (OBC)≡ △DCB -(OBC)

△OEB  △ ODC.

OB = OC ( by C.P.C.T. )

Hence proved

Question 10. ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.

ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Answer:  ABC is an isosceles triangle with
AB = AC.
AP ⊥ BC
To Prove: ∠B = ∠C

Question 11. In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.

triangle ml class 9 chapter 10 img 20

Answer : AC = EF, we have to prove ΔABC = ΔDEF

Here BA = DE (given) …(i)

BA ⊥ AC and DE ⊥ FE

∠BAC = ∠DEF = 90° (each) …(ii)

Also BF = CD (given) …(iii)

Adding FC in equation (iii),

BF + FC = CD + FC

⇒ BC = FD …(iv)

From (i), (ii) and (iv), we get

ΔABC = ΔDEF (by RHS congruency property)

⇒ AC = EF (by c.p.c.t)

Question 12. ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.

Answer :

ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX. Read more on Sarthaks.com - https://www.sarthaks.com/1072281/abcd-is-square-and-are-points-on-sides-ad-and-respectively-such-that-prove-that-and-bay-abx

Given that ABCD is a square, X and Y are points on the sides AD and BC respectively.

AY = BX

We have to prove:

BY = AX and ∠BAY = ∠ABX

Join B and X, A and Y

Since, ABCD is a square

∠DAB = ∠CBA = 90°

∠XAB = ∠YBA = 90°………………(i)

Now, consider ΔXAB and ΔYBA

∠XAB = ∠YBA = 90°[From (i)]

BX = AY (Given)

AB = BA (Common side)

So, by RHS congruence rule, we have

ΔXAB ≅ ΔYBA

BY = AX and ∠BAY = ∠ABX(By c.p.c.t)

Hence, proved

Question 13. (a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that

(i) ∆XTQ ≅ ∆XSQ
(ii) PX bisects the angle P.

(b) In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that

(i) AD = BC
(ii) AC = BD.

(c) In the figure (3) given below, BA || DF and CA II EG and BD = EC . Prove that, .

(i) BG = D
(ii) EG = CF.

Answer :

(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that (i) ∆XTQ ≅ ∆XSQ (ii) PX bisects the angle P.

(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that (i) ∆XTQ ≅ ∆XSQ (ii) PX bisects the angle P.

Chapter triangle ml aggrawal _3

Question 14. In each of the following diagrams, find the values of x and y.

Answer :

triangle ml class 9 chapter 10 img 22

triangle ml class 9 chapter 10 img 23

—  : End of ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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