Question

prove that ( b/4√(4a2-b2) ) is the equation to find the area of an isosceles triangle

Answer 1

We know,

• Isosceles triangle has two equal and an unequal side.

Let,

• ΔABC  be an isosceles triangle with sides  AB = AC = a  and  BC = b.

• Altitude be drawn from A to BC. Then altitude  AD = h.

Then,

• Area of  ΔABC is

$\huge\boxed{\sf S=\dfrac{b\times h}{2}}$

But,

• We don't have value for  h .

• The altitude AD will be at 90° with BC.

• Hence it will form 2 right triangles.

In ΔABD,

$\huge\boxed{\sf AD^2=AB^2-(\dfrac{BC}{2})^2}$

$\implies\sf { h^2=a^2-\dfrac{b^2}{4}}$

Therefore,

$\huge\boxed{\bold{\sf{h=\dfrac{\sqrt{4a^2-b^2}}{2}}}}$

Hence Proved !!!!

Answer 2

Step by step explaination :

area of triangle: 1/2×b×h, sq.units

Proof:- isosceles triangle = b/4(4a²-b²), sq.units

For proving the area of isosceles triangle, first of all we have to find height.

Here, base = b, units

In isosceles triangle, 90° angle formed.

so, h² = a² - (b/2)²

h = √a² - b²/4

= √( 4a²- b²)/4

= 1/2√4a²-b²

Now, height = 1/2√4a²-b², units

Area = 1/2 × b × 1/2√4a²-b², sq.units

= b/4 (4a²-b²), sq.units

Proved.