Question

What must be the side of an equilateral triangle so that its area may be equal to the area of an issosceles triabgle whose base and equal sides are 12m and 10m respectively?

Answer 1

Answer:32root3.

Step-by-step explanation:

Here is answer....

Answer 2

Given,

The base of an isosceles triangle = 12m

The equal sides of an isosceles triangle = 10m

To Find,

The side of an equilateral triangle so that its area may be equal to the area of an isosceles triangle=?

Solution,

The base of an isosceles triangle(b)  = 12m

The equal sides of an isosceles triangle (a) = 10m

The height of an isosceles triangle = $\sqrt{a^2 - \frac{b^2}{4} }$

$h = \sqrt{10^2 - \frac{12^2}{4} } \\h = \sqrt{100 - \frac{144}{4} } \\h = \sqrt{100 - 36 } \\h = \sqrt{64}\\ h = 8$

Height of the isosceles triangle = 8m

The area of the isosceles triangle = 1 / 2(b * h)

The area of the isosceles triangle = 1 / 2(12 * 8)

The area of the isosceles triangle = 48 m²

The area of the equilateral triangle = The area of the isosceles triangle

The area of the equilateral triangle = 48m²

$\frac{\sqrt{3} }{4} a^2 = 48 \\\\ a^2 = 48 * 4 / \sqrt{3} \\ a^2 = 192 * \sqrt{3} / 3\\ a^2 = 64\sqrt{3}\\a = 8 * \sqrt[4]{3} \\a = 8 * 1.31\\a = 10.5$

Hence, the side of an equilateral triangle should be 10.5m.