Question

What is the area of to the nearest tenth of a square centimeter? Use special right triangles to help find the height. Show your work.

Answer 1

That is a 30 60 90 triangle.
In such a triangle the "middle" size = short side * sq root (3)
height = 5 * sq root (3) =  8.6602540378
Area = (1/2) * 5 * 8.6602540378
Area = 2.5 * 8.6602540378
Area = 21.6506350946 square centimeters

Answer 2

Answer:

$21.7\text{ cm}^2$

Step-by-step explanation:

We have been given an image of a 30-60-90 triangle. We are asked to find the area of our given triangle.

$\text{Area of triangle}=\frac{1}{2}\times (\text{Base}\times\text{Height})$

First of all, we need to find the height of our given triangle.

In a 30-60-90 triangle, the shorter leg that corresponds to 30 degree angle is $x$ units long. The sides corresponding to 60 and 90 degree angles are $x\sqrt{3}$ and $2x$ respectively.

Since the value of $x$ is 5 cm, so the height of our given triangle (corresponding to 60 degree angle) will be:

$x\sqrt{3}=5\sqrt{3}$

Upon substituting our given values in area of triangle formula we will get,

$\text{Area of triangle}=\frac{1}{2}\times (5\times 5\sqrt{3})$

$\text{Area of triangle}=\frac{1}{2}\times (25\sqrt{3})$

$\text{Area of triangle}=12.5\sqrt{3}$

$\text{Area of triangle}=21.65063509\approx 21.7$

Therefore, the area of our given triangle is 21.7 square centimeters.