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.

Answer 1

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"`.