Where are the triangles so I can compare to find the triangle congruence?
<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.
Answer: No. If two lines have the same slope they are parallel lines (or they are the same line), so they cannot intersect at one point. Because the triangles are congruent, the angles where the lines meet the x-axis are congruent. Because the lines form the same angle with the x-axis, they are parallel.
Step-by-step explanation:
what is your clear question? it lacks some details
Answer:
r≈4.75
Step-by-step explanation: