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Puzzle #21

A square with a green isosceles triangle inside. The triangle has one vertex at the top left vertex of the square and the other two vertices in the bottom and right side of the square such that they are equidistant from the first vertex. A vertical double arrow with the length of the square's side is labeled/labelled six.

The two equal sides of an isosceles triangle trisects the top left vertex of the square with side length 6.

What is the area of the triangle?

Solution

As the two equal sides of the isosceles triangle trisects the top left vertex of the square, the size of its unequal angle is 90° ÷ 3 = 30°.

The isosceles triangle is between two congruent 30-60-90 triangles.

The side lengths of the top triangle are:

A square with a green isosceles triangle inside. The triangle has one vertex at the top left vertex of the square and the other two vertices in the bottom and right side of the square such that they are equidistant from the first vertex. A right(-angled) triangle is formed by the top side of the square and part of the right side of the square and the equal side of the isosceles triangle as the hypotenuse. The sides of the right(-angled) triangle are labeled/labelled six, two square root of three and four square root of three, respectively.

Then, the two equal sides of the isosceles triangle measure

4\sqrt{3}

So, the area of the isosceles triangle is

\frac{1}{2} \times \left( 4\sqrt{3} \right)^2 \times \sin 30 ^{\circ}

\frac{1}{2} \times 16 \times 3 \times \frac{1}{2}

12

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