Question

in given figure, angle D=90° , AB =8cm,BC=6cm and CA=3cm. Find CD.
​

Answer 1

Given:

∠D = 90°

AB = 8cm

BC = 6cm

CA = 3cm

To find:

The value of CD

Solution:

In Δ ACD, using the Pythagoras theorem, we get

$AC^2 = AD^2 + CD^2$

substituting the value of AC

$\implies (3)^2 = AD^2 + CD^2$

$\implies AD^2 = 9 - CD^2$ ....... (i)

In Δ ABD, using the Pythagoras theorem, we get

$AB^2 = AD^2 + BD^2$

$\implies AB^2 = AD^2 + (BC + CD)^2$

substituting the value of AB & BC

$\implies (8)^2 = AD^2 + (6 + CD)^2$

$\implies 64 = AD^2 + (6^2 + 12CD + CD^2)$

$\implies AD^2 = 64 - (6^2 + 12CD + CD^2)$

$\implies AD^2 = 64 - 36 - 12CD - CD^2$

$\implies AD^2 = 28 - 12CD - CD^2$ ..... (ii)

Now, from (i) & (ii), we get

$9 - CD^2 = 28 - 12CD - CD^2$

$\implies 9 - 28 = - 12CD - CD^2 + CD^2$

$\implies -19 = - 12CD$

$\implies 19 = 12CD$

$\implies CD = \frac{19}{12}$

$\implies\bold{ CD = 1.58\:cm}$

Thus, the value of CD is 1.58 cm.

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