Question

An isosceles triangle has its base on the side of a square. The areas of the triangle and the square are equal. What is the length of one of the equal sides of the isosceles triangle in terms of the length of one side of the square? Give an exact answer. ​

Answer 1

√(17) * s/2 = one of equal sides of triangle

Explanation:

Sorry for my handwriting, but the figure will look somewhat like it.

Given, area of triangle = area of square.

We know ar(tri) = 1/2 * b* h

Also, ar(sq) = side²

• Given ar (tri) = ar(sq)

It is also given,base of triangle = side of square.

Let us be familiar with the properties:

• All of the sides of square are equal.
• Pythagoras theorem: a² +b² = c²(Only for 90° triangle)

Hence,

• 1/2 * b* h = s²
• [ b = s, given]
• 1/2 * b* h = b²
• 1/2 * h = b
• h = 2b.
• h = 2s.

Draw perpendicular from top vertex to the base (height of triangle)

We get a right angled triangle.

• Property Of Isoceles Triangle: Perpendicular from top vertex to the base divides the triangle into congruent parts.(Well, there's a prove on congruency, do let me know in comments if you need!)

Base of the right angled triangle= side of square / 2 (Reason: c.p.c.t)

• b1 = b/2 = s/2

By Pythagorean theorem,

(leg1)² +(leg2)² = Hypotenuse ²

Here hyp = one of equal side of triangle = a

leg1 = b1 = b/2 = s/2 (b = s)

leg2 = 2s

On putting values,

• (s/2)² + (2s)² = a²
• s²/4 + 4s² = a²
• (Do LCM and take square root on both sides)
• √(17)* s/2 = a

Abbreviations:

• a =One of Equal side of triangle
• s = Side of Square
• h = height of triangle
• b = base of triangle.