RD Sharma Class 9 Solutions Chapter 12 Exercise 12.4: Download the Free PDF of RD Sharma Class 9 Solutions Maths from the link given in this blog. You can prepare for your Maths exams with this ultimate guide. All the solutions of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.4 are prepared by subject matter experts. To know more, read the whole blog.
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RD Sharma Class 9 Solutions Chapter 12 Exercise 12.4
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Question 1.
In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 – 1
Solution:
Given : In the figure, AB = CD, AD = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 – 2
To prove: ∆ADC = ∆CBA
Proof: In ∆ADC and ∆CBA
CD = AB (Given)
AD = BC (Given)
CA = CA (Common)
∴ ∆ADC ≅ ∆CBA (SSS axiom)
Question 2.
In an APQR, if PQ = QR and L, M, and N are the mid-points of the sides PQ, QR, and RP respectively. Prove that LN = MN.
Solution:
Given: In ∆PQR, PQ = QR
L, M, and N are the mid-points of sides PQ, QR, and RP respectively. Join LM, MN, and LN
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 – 2A
To prove: ∠PNM = ∠PLM
Proof: In ∆PQR,
∵ M and N are the midpoints of sides PR and QR respectively
∴ MN || PQ and MN = 12 PQ …(i)
∴ MN = PL
Similarly, we can prove that
LM = PN
Now in ∆NML and ∆LPN
MN = PL (Proved)
LM = PN (Proved)
LN = LN (Common)
∴ ∆NML = ∆LPN (SSS axiom)
∴ ∠MNL = ∠PLN (c.p.c.t.)
and ∠MLN = ∠LNP (c.p.c.t.)
⇒ ∠MNL = ∠LNP = ∠PLM = ∠MLN
⇒ ∠PNM = ∠PLM
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