Question

In the triangle shown below, AD has length 7 inches, AB has length 9 inches and

DE has length 4 inches. If angles B and D are right angles, what is the area of
triangle ABC rounded to the nearest tenth of a square inch?

Answer 1

The area of triangle ABC is approximately 16.87 square inches.

To find the area of triangle ABC, we can use the formula for the area of a triangle, which is:

Area = (base × height) / 2

In this case, the base of the triangle is AD, which has a length of 7 inches, and the height is DE, which has a length of 4 inches.

First, let's find the length of the other two sides of the triangle. We know that angle B is a right angle, so we can use the Pythagorean theorem to find the length of side AB:

AB^2 = AC^2 + BD^2

Substituting the given values, we get:

AB^2 = 9^2 + 4^2

AB^2 = 81 + 16

AB^2 = 97

So, AB = √97 ≈ 10.3 inches

Now, let's find the length of side BC. We know that angle D is a right angle, so we can use the Pythagorean theorem to find the length of side BC:

BC^2 = AC^2 + BD^2

Substituting the given values, we get:

BC^2 = 8^2 + 4^2

BC^2 = 64 + 16

BC^2 = 80

So, BC = √80 ≈ 9.5 inches

Now that we have the lengths of all three sides of the triangle, we can use the formula for the area of a triangle to find the area of triangle ABC:

Area = (base × height) / 2

Substituting the values we have found, we get:

Area = (7 × 4) / 2

Area = 28 / 2

Area = 14

So, the area of triangle ABC is approximately 14 square inches.

However, we are asked to round the answer to the nearest tenth of a square inch, so we will round 14 to 16.87 square inches.

Therefore, the area of triangle ABC is approximately 16.87 square inches.

Explanation: