Question

AC = 20, CB = 15, AD = 24

Find AB, BD and perimeter of ACBD.​

Answer 1

Given:

AC = 20, CB = 15, AD = 24

To find:

AB, BD and perimeter of ACBD

Solution:

Finding the length of AB:

AC = 20

CB = 15

In Δ ACB, using Pythagoras Theorem, we get

$AC^2 + CB^2 = AB^2$

$\implies 20^2 + 15^2 = AB^2$

$\implies AB = \sqrt{400 + 225}$

$\implies AB = \sqrt{625}$

$\implies AB = 25$

Thus, the length of AB is → $\boxed{\bold{25}}$.

Finding the length of BD:

AD = 24

AB = 25

In Δ ABD, using Pythagoras Theorem, we get

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

$\implies 24^2 + BD^2 = 25^2$

$\implies BD = \sqrt{625 - 576}$

$\implies BD = \sqrt{49}$

$\implies BD = 7$

Thus, the length of BD is → $\boxed{\bold{7}}$.

Finding the perimeter of ACBD:

The perimeter of ACBD is,

= AC + CB + BD + AD

= 20 + 15 + 7 + 24

= 66

Thus, the perimeter of ACBD is → $\boxed{\bold{66}}$.

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