In-a-square-ABCD-a-triangle-APQ-inscribed-in-it-AP-4-cm-PQ-3-cm-and-AQ-5-cm-Point-P-is-on-the-side-BC-and-point-Q-is-on-side-CD-Find-the-area-of-the-square-ABCD-

Question Number 134097 by bobhans last updated on 27/Feb/21

In a square ABCD, a triangle

APQ inscribed in it . AP = 4 cm,

PQ = 3 cm and AQ = 5 cm . Point

P is on the side BC and point Q

is on side CD . Find the area of the

square ABCD .

Answered by mr W last updated on 27/Feb/21

Commented by mr W last updated on 27/Feb/21

B P = \sqrt{4^{2} - a^{2}}

\frac{P C}{3} = \frac{a}{4}

\Rightarrow P C = \frac{3 a}{4}

\sqrt{4^{2} - a^{2}} + \frac{3 a}{4} = a

4^{2} - a^{2} = \frac{a^{2}}{16}

a^{2} = \frac{16^{2}}{17}

\Rightarrow a = \frac{16}{\sqrt{17}}

a r e a o f A B C D = a^{2} = \frac{16^{2}}{17} = \frac{256}{17} \approx 15.05

Commented by bobhans last updated on 28/Feb/21

thanks

Commented by otchereabdullai@gmail.com last updated on 12/Mar/21

Thanks prof w

Answered by liberty last updated on 01/Mar/21

Commented by liberty last updated on 01/Mar/21

{(4 u)}^{2} + u^{2} = 16

\Leftrightarrow {16 u}^{2} + u^{2} = 16; u^{2} = \frac{16}{17}

so area of square is {16 u}^{2} = \frac{256}{17} \approx 15.0588

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