Asked by Anonymous...please help!
ABCD is a square with AB=25. P is a point within ABCD such that PA=24 and PB=7. What is the value of PD^2?
Answers
Answered by
Reiny
Did you notice that
25^2 = 24^2+7^2 ?
so angle APB = 90°
sin(angle BAP) = 7/25
So cos(angle DAP) = 7/25 , properties of complementary angles, since angle A = 90°
by the cosine law:
PD^2 = 25^2 + 24^ - 2(25)(24)cos DAP
= 625 + 576 - 2(25)(24)(7/25)
= 1201 - 336 = 865
25^2 = 24^2+7^2 ?
so angle APB = 90°
sin(angle BAP) = 7/25
So cos(angle DAP) = 7/25 , properties of complementary angles, since angle A = 90°
by the cosine law:
PD^2 = 25^2 + 24^ - 2(25)(24)cos DAP
= 625 + 576 - 2(25)(24)(7/25)
= 1201 - 336 = 865
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