Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.

Area of a Triangle- AMC 8, 2012 – Problem 25

A square with area 4 is inscribed in a square with area 5,  with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?

Area of triangle and square

  • \(\frac{1}{5}\)
  • \(\frac{2}{5}\)
  • \(\frac{1}{2}\)

Key Concepts




Check the Answer

But try the problem first…


Suggested Reading

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints

First hint

Find the area of four triangles

Can you now finish the problem ……….

Second Hint

Four triangles are congruent

can you finish the problem……..

Final Step

Area of Triangle and Square 2

Total area of the big square i.e ABCD is 5 sq.unit

and total area of the small square i.e EFGH is 4 sq.unit

So Toal area of the \((\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)=(5-4)=1\) sq.unit

Now clearly Four triangles\(( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )\) are congruent.

Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit

So area of the one triangle is \(\frac{1}{4}\) sq.unit

Now “a” be the height and “b” be the base of one triangle

The area of one triangle be \((\frac{1}{2} \times base \times height )\)=\(\frac{1}{4}\)

i.e \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)

i.e \(ab\)= \(\frac{1}{2}\)

Subscribe to Cheenta at Youtube