## RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

Other Exercises

- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.

In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.

Solution:

Given : In the figure, AB = CD, AD = BC

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 a 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

To prove : ∠PNM = ∠PLM

Proof : In ∆PQR,

∵ M and N are the mid points of sides PR and QR respectively

∴ MN || PQ and MN = \(\frac { 1 }{ 2 }\) 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

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 are helpful to complete your math homework.

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