AMC 10

Area of polygon – AMC 10B, 2019 Problem 8

Try this beautiful problem of geometry in which we have to find area of triangle and square mixed. You may use sequential hints to help you solve the problem.

Area of polygon

Polygon means shape composed of multiple sides , for example square , triangle, trapezium pentagon etc. A regular polygon means all of its sides have same length. Square ,equilateral triangle is a regular polygon. Let us learn to find the area of polygon.

Try the problem

The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

area of polygon- knowledge graph
$\textbf{(A) } 4 \qquad \textbf{(B) } 12 - 4\sqrt{3} \qquad \textbf{(C) } 3\sqrt{3}\qquad \textbf{(D) } 4\sqrt{3} \qquad \textbf{(E) } 16 - 4\sqrt{3}$

AMC 10B, 2019 Problem 8

Area of polygon -square and triangle

6 out of 10

challenges and thrills of pre college mathematics

Knowledge Graph

Use some hints

Split the square in 4 identical parts by drawing two perpendicular (horizontal and vertical )lines from the center. And then we will get smaller square with two similar right angle triangles in it and one fourth part of the shaded region.

When we split an equilateral triangle in half, we get two triangles with angles 30,60 and 90 degrees. Therefore, the altitude, which is also the side length of one of the smaller squares, is \(\sqrt{3}\).

the area of the two congruent triangles will be $2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}$.


The area of the each small squares is the square of the side length, i.e. $\left(\sqrt{3}\right)^2 = 3$. Therefore, the area of the shaded region in each of the four squares is $3 – \sqrt{3}$. Since there are $4$ of these squares, we multiply this by $4$ to get $4\left(3 – \sqrt{3}\right) = {\textbf{(B) } 12 – 4\sqrt{3}}$.

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