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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
Let p be the length of AP.
It is given that $$\angle\ BAC\ =90\ $$ and $$\angle\ APC\ =90\ $$
Let $$\angle\ ABC\ =\theta\ $$, then $$\angle\ BAP\ =90-\theta\ $$ and $$\angle\ BCA\ =90-\theta\ $$
So $$\angle\ PAC\ =\theta\ $$
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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