This page 10th CBSE math solution for exercise 6.2 part 1 is going to provide you solution for every problems that you find in the exercise no 6.2

(1) In fig 6.17 (i) and (ii) DE ∥ BC. Find EC in (i) and AD in (ii)

Solution:

Let EC = x

In triangle ABC,the side DE is parallel to BC

by BPT theorem

(AD/DB) = (AE/EC)

AD = 1.5 cm DB = 3 cm AE = 1 cm

(1.5/3) = (1/x)

1.5 x = 3

x = 3/1.5

x = 2 cm

Solution:

Let AD = x

In triangle ABC,the side DE is parallel to BC

by BPT theorem

(AD/DB) = (AE/EC)

AD = x DB = 7.2 cm AE = 1.8 cm EC = 5.4 cm

(x/7.2) = (1.8/5.4)

5.4 x = 1.8 x 7.2

x = (1.8 x 7.2)/5.4

x = 2.4 cm

(2) E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases,state whether EF

∥ QR.

(i) PE = 3.9 cm , EQ = 3 cm,PF = 3.6 cm and FR = 2.4 cm

(PE/EQ) = (PF/FR)

(3.9/3) = (3.6/2.4)

1.3 = 1.5

So EF is not parallel to QR.

(ii) PE = 4 cm , EQ = 4.5 cm,PF = 8 cm and FR = 9 cm

(PE/EQ) = (PF/FR)

(4/4.5) = (8/9)

0.88 = 0.88

So EF is parallel to QR.

(iii) PQ = 1.28 cm PR = 2.56 cm PE = 0.18 cm and PF = 0.36 cm

PQ = PE + EQ

1.28 = 0.18 + EQ

EQ = 1.28 - 0.18

= 1.1

PR = PF + FR

2.56 = 0.36 + FR

2.56 - 0.36 = FR

FR = 2.2

(PE/EQ) = (PF/FR)

(0.18/1.1) = (0.36/2.2)

0.1636 = 0.1636

So EF is parallel to QR.

(3) In fig 6.18,if LM ∥ CB and LN ∥ CD prove that (AM/AB) = (AN/AD)

Solution:

In triangle ABC

LM is parallel to BC

(AM/MB) = (AL/LC) ------- (1)

NL is parallel to DC

(AN/ND) = (AL/LC) ------(2)

(1) = (2)

(AM/MB) = (AN/ND)

Hence proved.

In the page 10th CBSE math solution for exercise 6.2 part 1 we are going to see the solution of next problem

(4) In fig 6.19,DE ∥ AC and DF ∥ AE. prove that (BF/FE) = (BE/EC)

Solution :

In triangle ABC

DE ∥ AC

(BD/DA) = (BE/EC) -----(1)

In triangle AEB

DF ∥ AE

(BD/DA) = (BF/FE) -----(2)

(1) = (2)

(BE/EC)= (BF/FE)

Hence proved

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