Question Video: Using Trigonometric Formulas for Area of Triangles to Find the Area of a Shape Composed of a Right-Angled Triangle and an Isosceles Triangles Mathematics

Find the area of the figure below, giving the answer to three decimal places.


Video Transcript

Find the area of the figure below, giving the answer to three decimal places.

In this video, we have an equilateral triangle, which we can see from these markings, so each side must be 34 meters. And then we also have this right triangle. And since a triangle adds up to 180 degrees and we have a 90-degree angle and a 60-degree angle, that means our last angle must be 30.

And that’s a special right triangle; it’s a 30-60-90 triangle. In a triangle that has 30 degrees, 60 degrees, and 90 degrees, there’s a relationship between their sides. The side across from the 30-degree angle is 𝑥; the side across from the 60-degree angle would be 𝑥 times square root two; and then the longest side, the hypotenuse, the side across from the 90-degree angle is two times 𝑥.

And in our right triangle, we know that the hypotenuse is equal to 34. So we can set 34 equal to 2𝑥 and then solve for 𝑥. So we would solve by dividing both sides by two and find that 𝑥 is 17. So if 𝑥 is 17, this side must be 17, the one across from the 30-degree angle. So we can label that on our figure. And then the other side that’s missing, we plug 17 in for 𝑥 and multiply it by square root two, which is about 29.4449.

So now that we know all of the lengths of the sides of the triangles, we can begin finding the area. So we can find the area of this figure by finding the area of each triangle, the two of them, and then adding them together. So let’s begin with the right triangle. The area of a triangle is equal to one-half times the base times the height.

And luckily, with a right triangle, we can use either of the two legs, which are the nonhypotenuse sides, as the base or the height. So we can take one-half times 17 as the base and the height of that triangle being 29.4449. So the area of the right triangle would be be 250.28165 meters squared.

Okay, now that we have that triangle, now I need to find the area of the other triangle. And the problem is we don’t know the height, because if we would look at any of the sides as a base, the height would have to be perpendicular to it and we don’t know that value.

However, there’s something called Heron’s formula that we can use. It’s the square root of 𝑝 times 𝑝 minus 𝑎 times 𝑝 minus 𝑏 times 𝑝 minus 𝑐 where 𝑎, 𝑏, and 𝑐 are sides and 𝑝 is equal to half of the perimeter. First let’s find 𝑝. It’s half of the perimeter, so the perimeter would be 34 plus 34 plus 34 because the perimeter is the length around the triangle. So our perimeter would be 102.

So to find 𝑝, we need to take half of that. So the perimeter will be 51 meters, so we can replace 𝑝 with 51. Now our side lengths are all the same — they’re all 34 — so we can replace 𝑎 with 34, 𝑏 with 34, and 𝑐 with 34. So to simplify what we have, 51 minus 34 is seventeen. So we have 51 times 17 times 17 times 17, which would be equal to the square root of 250563, which is equal to 500.5627 meters squared.

So the area of the equilateral triangle is 500.5627 meters squared, and the area of the other triangle, the right triangle, was 250.28165 meters squared. So our total area would be equal to the equilateral triangle plus the right triangle, so 500.5627 meters squared plus 250.28165 meters squared, which, after running three decimal places, would equal 750.844 meters squared.

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