Question

P is a point inthe interiour of a rectangle ABCD.if P is joined to each of the vertices of the rectangle and the lengths of PA ,PB and PC are 3cm,4cm and 5cm respectively prove that PA2+PC2=PB2+PD2 and find the length of CE

Answer 1

Answer on Question #45234 – Math - Geometry

P is a point in the interior of a rectangle ABCD. if P is joined to each of the vertices of the rectangle and the lengths of PA, PB and PC are 3cm, 4cm and 5cm respectively prove that $PA^2 + PC^2 = PB^2 + PD^2$ and find the length of PD

Solution:

$$\left\{ \begin{array}{l} a ^ {2} + d ^ {2} = P A ^ {2} = 3 ^ {2} = 9 \\ a ^ {2} + b ^ {2} = P B ^ {2} = 4 ^ {2} = 1 6 \\ b ^ {2} + c ^ {2} = P C ^ {2} = 5 ^ {2} = 2 5 \\ c ^ {2} + d ^ {2} = P D ^ {2} \end{array} \right.$$

From (1) + (3) and (2) + (4) we will obtain:

$$\left\{ \begin{array}{l} P A ^ {2} + P C ^ {2} = a ^ {2} + d ^ {2} + b ^ {2} + c ^ {2} \\ P B ^ {2} + P D ^ {2} = a ^ {2} + b ^ {2} + c ^ {2} + d ^ {2} \end{array} \right.$$

From this, we can see:

$$P A ^ {2} + P C ^ {2} = P B ^ {2} + P D ^ {2}$$

So we can find PD

$$P D ^ {2} = P A ^ {2} + P C ^ {2} - P B ^ {2} = 9 + 2 5 - 1 6 = 1 8$$ $$P D = \sqrt {1 8} = 3 \sqrt {2} c m$$

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