Question

In the given figure, ∠A = ∠CED, prove that ∆CAB ∼ ∆CED. Also, find the value of x.

Answer 1

SOLUTION :

GIVEN :∠A = ∠CED. In the figure :  

AB = 9 cm , CE= 10 cm , AD = 7cm , DC= 8cm ,BE= 2 cm , DE = x cm

In ∆CAB and ∆CED

∠C = ∠C

[Common]

∠A = ∠CED

[Given]

∴ ∆CAB ~ ∆CED [By AA similarity criterion ]

CA/CE = AB /DE = CB /CD

[Since corresponding sides of two similar triangles are proportional]

AB/DE = CB/CD

9/x = (CE+EB)/CD

9/x = (10+2)/8

9/x = 12/8

12x = 8×9

x = (8×9)/12

x = (2×9)/3

x = 2× 3

x = 6 cm

Hence , the value of x= 6 cm.

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Answer 2

in △abc and △ced

<c=<c  [common]

<a=<ced [given]

therefore△cab ∼ced    [by AAsimilarity]

the value of X=6

becoause in △DEC

RIGHT ANGLED at D

by P.G.T

H square=B square + P square

10sq=DEsq + 8 sq

100=DEsq+64

100-64=X            36=Xsq

as 36 is square root of 6

therefore value  of X=6