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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