-4x^2+20-4 in standard form
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<h3>Given</h3>
trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8
<h3>Find</h3>the area of PSTK
<h3>Solution</h3>It helps to draw a diagram.
∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as
... KS² = PS² + KP²
... 13² = PS² + 12²
... PS = √(169 -144) = 5
This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is
... A = (1/2)(b1 +b2)h
... A = (1/2)(12 +8)·5
... A = 50
The area of trapezoid PSTK is 50 square units.