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Suppose P, Q, and R are points in the plane such that PQ=1.8,PR=8.2, and angle PQR=\(90^\circ\). What is the perimeter of triangle PQR, and area.

Guest Aug 23, 2017
edited by Guest  Aug 23, 2017

Best Answer 

 #1
avatar+554 
+1

Using the Pythagorean theorem...

1.8^2 (That is PQ) + (QR)^2 = 8.2^2 (That is PR)

Subtract 1.8^2 to both sides to get QR^2 = 8.2^2-1.8^2

QR^2 = 64

QR=8, and QR is the base. PQ is the height.

Get the triangle's area.

8*1.8*1/2=7.2

 

Is this correct???

Gh0sty15  Aug 23, 2017
 #1
avatar+554 
+1
Best Answer

Using the Pythagorean theorem...

1.8^2 (That is PQ) + (QR)^2 = 8.2^2 (That is PR)

Subtract 1.8^2 to both sides to get QR^2 = 8.2^2-1.8^2

QR^2 = 64

QR=8, and QR is the base. PQ is the height.

Get the triangle's area.

8*1.8*1/2=7.2

 

Is this correct???

Gh0sty15  Aug 23, 2017

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