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?

- \(\frac{1}{5}\)
- \(\frac{2}{5}\)
- \(\frac{1}{2}\)
Key Concepts
Geometry
Square
Triangle
Check the Answer
Answer:\(\frac{1}{2}\)
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find the area of four triangles
Can you now finish the problem ……….
Four triangles are congruent
can you finish the problem……..

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 \(ab\)= \(\frac{1}{2}\)
Other useful links
- https://www.cheenta. https://www.youtube.com/watch?v=W9XdZd8zXPA
- https://www.cheenta.com/radius-of-a-semicircle-amc-8-2016-problem-25/