Question 11
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is
Solution
Let p be the length of AP.
It is given thatĀ $$\angle\ BAC\ =90\ $$ andĀ $$\angle\ APC\ =90\ $$
LetĀ $$\angle\ ABC\ =\theta\ $$, thenĀ $$\angle\ BAP\ =90-\theta\ $$ andĀ $$\angle\ BCA\ =90-\theta\ $$
SoĀ $$\angle\ PAC\ =\theta\ $$
Triangles BPA and APC are similar
$$p^2=x\left(20-x\right)$$
We have to maximize the value of p, which will be maximum when x=20-x
x=10
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