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Question
Find the area of an isosceles triangle ABC in which AB = AC = 6 cm, ∠A = 90°. Also, find the length of perpendicular from A to BC.
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Solution
Area of ΔABC
= `(1)/(2) xx "AB" xx "AC"`
= `(1)/(2) xx 6 xx 6`
= 18cm2.
In right-angled ΔBAC,
BC2
= AB2 + AC2
= 62 + 62
= 36 + 36
= 72
⇒ BC = `6sqrt(2)"cm"`
Now,
area of ΔABC is also
= `(1)/(2) xx "BC" xx "AP"`
⇒ 18 = `(1)/(2) xx 6sqrt(2) xx "AP"`
⇒ 18 = `3sqrt(2) xx "AP"`
⇒ AP
= `(18)/(3sqrt(2)`
= `(6)/sqrt(2)`
= `(6)/sqrt(2) xx sqrt(2)/sqrt(2)`
= `(6sqrt(2))/(2)`
= `3sqrt(2)"cm"`.
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