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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 = 90o. Find the area of quadrilateral ABOC.
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Solution
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
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