Mensuration Practice Test 4

Other side $\frac{2}{6}\times 24$ = 8 cm
Hypotenuse = $\sqrt{8^2+6^2}$
= $\sqrt{64+36}$ = 10 cm

Let altitude to hypotenuse = h
h = $\sqrt{36-{\mathit{x}}^2}$ ... (1)
and h = $\sqrt{64-{\left(10-\mathit{x}\right)}^2}$ ... (2)
Equating (1) and (2) we get:
$\sqrt{36-{\mathit{x}}^2}$ = $\sqrt{64-{\left(10-\mathit{x}\right)}^2}$
$36-{\mathit{x}}^2$ = 64 - 100 - ${\mathit{x}}^2$ + 20x
$36-{\mathit{x}}^2$ = $-36-{\mathit{x}}^2+20\mathit{x}$
72 = 20x
⇒ x = 3.6
b = $\sqrt{36-{\left(3.6\right)}^2}$
= $\sqrt{36-12.96}$
= $\sqrt{23.04}$
= 4.8 cm