Question

In the following figure, find the sides marked x and y.

Answer 1

Given,

ΔABD is right angle triangle, right angle at B.

Her base (BD) = x + x = 2x = ?

Hypotenuse (AD) = 20

Perpendicular (AB) = 12

To find BD, apply Pythagoras theorem in ΔABD.

(AD)² = (AB)² + (BD)²

(20)² = (12)² + (2x)²

(2x)² = (20)² – (12)²

4x² = 400 – 144

4x² = 256

x² = `256/4`

x = `sqrt(64)`

x = 8

Thus, BD = 2x = 2 × 8 = 16

From the given figure, ΔBMC = ΔDMC

Now, ΔBMC is right angle triangle, right angle at M.

Her base (BM) = x = 8

Hypotenuse (CB) = y = ?

Perpendicular (CM) = 15

Apply Pythagoras theorem in ∆BMC.

CB² = CM² + MB²

y² = 15² + 8²

y² = 225 + 64

y = `sqrt(289)`

y = 17

Thus, CB = y = 17

Hence, x = 8, y = 17.