RD Sharma Solutions Class 9 Congruent Triangles Exercise 10.4

RD Sharma Solutions Class 9 Chapter 10 Exercise 10.4

RD Sharma Class 9 Solutions Chapter 10 Ex 10.4 Free Download

RD Sharma Solutions Class 9 Chapter 10 Ex 10.4

(1) In fig (10).9(2) It is given that AB = CD and AD = BC. Prove that \(\Delta ADC \cong \Delta CBA\).

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Solution:

Given that in the figure AB = CD  and AD = BC.

We have to prove \(\Delta ADC \cong \Delta CBA\)

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Now,

Consider \(\Delta\) ADC and \(\Delta\) CBA.

We have

AB = CD                     [Given]

BC = AD                     [Given]

And AC = AC             [Common side]

So, by SSS congruence criterion, we have

\(\Delta ADC \cong \Delta CBA\)

Hence proved

(2) In a \(\Delta\) PQR. IF PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.

Sol: Given that in \(\Delta\) PQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively

We have to prove LN = MN.

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Join L and M, M and N, N and L

We have PL = LQ, QM = MR and RN = NP

[Since, L, M and N are mid-points of Pp. QR and RP respectively]

And also PQ = QR

  • PL = LQ = QM = MR = \(\frac{PQ}{2}\) = \(\frac{QR}{2}\) ……(i) Using mid-point theorem,

We have

MN \(\parallel\) PQ and MN = \(\frac{PQ}{2}\)

  • MN = PL = LQ ……(ii)

Similarly, we have

LN \(\parallel\)  QR and LN = (1/2)QR

  • LN = QM = MR ……(iii)

From equation (i), (ii) and (iii), we have

PL = LQ = QM = MR = MN = LN

LN = MN

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