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In right triangle $PQR$, we have\($\angle Q = \angle R$ \)  and $PR = \(6\sqrt{2}\). What is the area of $\triangle PQR$?

 Mar 25, 2020
 #1
avatar+483 
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Because \(\angle Q = \angle R\), they must be acute angles(because if they were the right angle, then two right angles would already be 180 degrees, making the triangle degenerate). Name the angle measurement of one of these 2 angles x. We can write the expression:

90 + x + x = 180(the internal angle sum of a triangle).

90 + 2x = 180

2x = 180

x = 45

That means that we have a 45-45-90 triangle at hand, which happens to be a special right triangle. The ratio of a 45-45-90 triangle is:

\(1,1,\sqrt2\) where the legs are 1, and the hypotenuse is \(\sqrt2\)(Remember this! It will make math with certain right triangles a lot easier!)

We know that angle P must be the right angle, because angle Q and R are both 45 degrees.

We then have that the hypotenuse PR is equal to \(6\sqrt2\)

Because the ratios of a 45-45-90 triangle are 1,1,\(\sqrt2\), we know that right triangle PQR is scaled by a factor of 6 compared to a 45-45-90 of leg length 1.

That means the leg lengths of our triangle are:

 

6 * 1 = 6

Now that we have the length of the legs as 6, we just need to find the area.

The area is equivalent to the (height * base) /2. Since we have both legs, we can just set one of them as the height, and the other the base(since they share a 90 degree angle). We then have our area as:

 

(6*6)/2 = 18

 Mar 25, 2020
 #2
avatar+392 
0

the answer is wrong

 Mar 25, 2020
 #3
avatar+483 
0

 

Sorry the answer should be :

\((6\sqrt2 * 6\sqrt2)/2 = 36\)

since I mixed up the leg length for the hypotenuse length. The length of each leg is \(6\sqrt2\), not 6

jfan17  Mar 25, 2020

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