Question 1164867
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Rectangle ABCD has sides AB = 21 and AD = 28. Let P be a point
inside the rectangle such that AP = 17 and BP = 10. 
Find the lengths PC and PD.
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1.  Make a sketch.

    In the sketch, draw the perpendicular from the point P to the side AB of the rectangle.
    Let E be the foot of this perpendicular at AB.

    In the sketch, draw the perpendicular from the point P to the side AD of the rectangle.
    Let F be the foot of this perpendicular at AD.



2.  Using the Heron's formula, find the area S of the triangle APB.  It is

        area S = {{{sqrt(24*(24-21)*(24-17)*(24-10))}}} = {{{sqrt(24*3*7*14)}}} = 84.

    Here in the formula  24 = {{{(21+17+28)/2}}}  is the semi-perimeter of the triangle APB.



3.  By knowing the area of the triangle APB (84 square units) and its base AB (21 units), 
    you can find its altitude, which is {{{(84*2)/21}}} = 8 units.



4.  Now in the right angled triangle APE, you know its hypotenuse AP = 17  and one of the legs PE = 8.
    Hence, the other leg is  {{{sqrt(17^2-8^2)}}} = 15 units.



5.  So, you know now the lengths of both perpendiculars  PE and PF, i.e. distances of the point P from 
    two sides of the rectangle.



6.  The rest of the solution is easy, and I leave it to you to complete it on your own.



<U>ANSWER</U>.  PD = 25; PC = {{{sqrt(436)}}} = {{{2*sqrt(109)}}}.
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