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
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
∠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.
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.
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.
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.
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.
Answer :
∠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
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.
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.
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 :
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 :
Question 14. In each of the following diagrams, find the values of x and y.
Answer :
— : 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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