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Question Number 200325 by cherokeesay last updated on 17/Nov/23

$$\mathrm{A}\mathrm{point}\mathrm{P}\mathrm{is}\mathrm{taken}\mathrm{inside}\mathrm{the}\mathrm{rectangleC}$$$$\mathrm{ABD}.\mathrm{This}\mathrm{point}\mathrm{joins}\mathrm{the}\mathrm{four}\mathrm{vertices}\mathrm{ofh}$$$$\mathrm{te}\mathrm{rectangle}.\mathrm{Knowing}\mathrm{that}\mathrm{PA}\mathrm{is}15\mathrm{cm}.\mathrm{B}$$$$\mathrm{P}24\mathrm{cm}\mathrm{and}\mathrm{PC}20\mathrm{cm}\mathrm{find}\mathrm{the}\mathrm{distancef}$$$$\mathrm{o}\mathrm{point}\mathrm{P}\mathrm{from}\mathrm{point}\mathrm{D}.$$

Answered by som(math1967) last updated on 17/Nov/23

$${PB}^2+{PC}^2={PA}^2+{PD}^2$$$${24}^2+{20}^2={15}^2+{PD}^2$$$$\Rightarrow PD=\sqrt{{24}^2+{20}^2-{15}^2}$$$$=\sqrt{576+175}$$$$=\sqrt{751}=27.04cm\left(approx\right)$$

Commented by cherokeesay last updated on 18/Nov/23

$$thankyou!$$

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