In the given figure, angle $A C P = ∠ B D P = 90^{o}, A C = 12 m, B D = 9 m$ and $P A = P B = 15 m$ . Find
(i) $C P$
(ii) $P D$
(iii) $C D$

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Solution

It is given that
$A C P = ∠ B D P = 90 °$
$A C = 12 m$
$B D = 9 m$
$P A = P B = 15 m$
(i) In the right angled triangle ACP
$A P^{2} = A C^{2} + C P^{2}$
Substituting the values
$15^{2} = 12^{2} + C P^{2}$
By further calculation
$225 = 144 + C P^{2}$
$C P^{2} = 225 - 144 = 81$
So we get
$C P = \sqrt 81 = 9 m$
(ii) In the right angled triangle BPD
$P B^{2} = B D^{2} + P D^{2}$
Substituting the values
$152 = 92 + P D^{2}$
By further calculation
$225 = 81 + P D^{2}$
$P D^{2} = 225 - 81 = 144$
So we get
$P D = \sqrt 144 = 12 m$
(iii) We know that
$C P = 9 m$
$P D = 12 m$
So we get
$C D = C P + P D$
Substituting the values
$C D = 9 + 12 = 21 m$

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