Question

I need help with the area only for 7 8 and 9

Answer 1

We will get the areas of each one of the given triangles:

7) A = 12.69 mm²

8) A = 19.21 in²

9) A = 16.81 yd²

How to get the area of the given triangles?

7) First, we have an equilateral triangle, where all the sides measure 5.4mm

For an equilateral triangle of side length S, the area is:

$A = \frac{\sqrt{3} }{4}S^2$

In this case, S = 5.4 mm, replacing we get:

$A = \frac{\sqrt{3} }{4}(5.4mm)^2 = 12.62 mm^2$

8) Now we have two equal sides and one different. The general area of a triangle of base B and height H is:

A = B*H/2.

In this case, the base measures 3.4 in.

To get the height, let's divide the triangle into two right triangles, such that one cathetus is 3.4in/2 = 1.7 in.

The hypotenuse measures 5.9 in

And the other cathetus is the height of the triangle.

Then, by using the Pythagorean theorem, we see that the height is:

$H = \sqrt{(5.9 in)^2 - (1.7in)^2} = 5.65 in$

Then the area of this triangle is:

$A = (3.4 in)*(5.65 in)/2 = 19.21 in^2$

9) Here the base measures 8.2 yds, and the height 4.1 yds, so the area is just:

$A = (4.1 yd)*(8.2 yd)/2 = 16.81 yd^2$

If you want to learn more about triangles:

brainly.com/question/2217700

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