Find the area of an isosceles triangle the measure of one of its equal side being b and the third sides a
Answers
Length of sides=a,a,b
Heron's formula
=√s(s-a)(s-b)(s-c)
where s=(a+b+c)/2
Substituting values
s=(a+a+b)/2
=(2a+b)/2
√(2a+b)(2a+b-a)(2a+b-a)(2a+b-b)
=√(2a+b)(a+b)²(2a)
=(a+b)√4a²+b
Area of triangle = b/2 √( a²- (b²/4)
Given:
In an isosceles triangle, the measure of one of its equal side is b and the third sides a
To find:
The area of an isosceles triangle
Solution:
Given triangle is an isosceles triangle
Given length of equal side = a and third side = b
As we know Area of a triangle = √s(s-a)(s-b)(s-c)
Here a, b and c are sides and s is semi perimeter of triangle
From above data a = a, b = a and c = b
Semi perimeter of triangle = (a+a+b)/2 = 2a+b/2 = a+b/2
⇒ s = a+b/2
Area of a triangle = √s(s-a)(s-a)(s-b)
= √s(s-a)²(s-b)
= (s-a) √s(s-b)
Substitute s value
= (a + b/2 -a) √ [ (a+b/2) (a+b/2 - b) ]
= b/2 √( a+b/2) (a -b/2)
= b/2 √( (a)²- (b/2)²)
= b/2 √( a²- (b²/4)
Area of triangle = b/2 √( a²- (b²/4)
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