Area of Triangle and Square | 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?

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 \(ab\)= \(\frac{1}{2}\)

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