Question

AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠BOC = 90^o. Find the area of quadrilateral ABOC.

Answer 1

Area of ΔABC =`"b"/4 sqrt(4"a"^2 - "b"^2)`

`= 36/4 xx sqrt( 4 xx 30^2 - 36^2 )`

= `9 xx sqrt2304`

= 9 × 48

= 432 cm²

∠ BOD = ∠ COD = 45°

OB = OC = x

In Δ BOC,

`"H"^2 = "P"^2 + "B"^2`

`(36)^2 = x^2 + x^2`

`36 xx 36 = 2x^2`

`sqrt(36 xx 18)` = x

`sqrt(6 xx 6 xx 3 xx 3 xx 2)` = x

∴ x = `18sqrt2`

Now,

Area of ΔBOC = `1/2 xx "base" xx "height"`

`= 1/2 xx 18sqrt2 xx 18sqrt2`

`= 162 xx 2`

= 324 cm²

Area of ABOC = Area of ΔABC - Area of ΔBOC

= 432 - 324

= 108 cm²