Question

in the figure triangleABC similar to triangle EDC,if we have AB = 4cm,ED=3cm,CE=4.2cm and CD=4.8cm,then the value of CA and CB are

Answer 1

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Answer 2

The value of CA is 5.6 cm and the value of CB is 6.4 cm

Step-by-step explanation:

Given:

ΔABC $\sim$ ΔEDC

AB = 4 cm

ED = 3 cm

CE = 4.2 cm

CD = 4.8 cm

We need to find the value of CA and CB

Now We know that,

ΔABC $\sim$ ΔEDC

Hence by Properties of Similar Triangles we get;

$\frac{ED}{AB}=\frac{CE}{CA} = \frac{CD}{CB}$

Now Substituting the values we get;

$\frac{3}{4}=\frac{4.2}{CA} = \frac{4.8}{CB}$

Now first we will solve for CA.

$\frac{3}{4}=\frac{4.2}{CA}$

$CA = \frac{4\times4.2}{3} = 5.6\ cm$

Hence the value of CA is 5.6 cm.

Also

$\frac{3}{4}=\frac{4.8}{CB}$

$CB = \frac{4\times4.8}{3} = 6.4\ cm$

Hence the value of CB is 6.4 cm.