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Question Number 200085 by a.lgnaoui last updated on 13/Nov/23

$$\mathrm{perimetre}\mathrm{of}\mathrm{White}\mathrm{triangle}?$$

Commented by a.lgnaoui last updated on 13/Nov/23

Commented by Nimnim111118 last updated on 13/Nov/23

$$ar(white△)=\frac{1}{2}\times 32\times 24-330=384-330=54$$$$Clearly,thetwo△saresimilar$$$$\therefore {\left(\frac{perimeterofwhite△}{perimeteroflarger△}\right)}^2=\frac{ar(white△)}{ar(larger△)}=\left(\frac{54}{384}\right)$$$$\Rightarrow {\left(\frac{X}{24+32+40}\right)}^2={\left(\frac{3}{8}\right)}^2\Rightarrow X=\frac{3}{8}\times 96=36✓$$$$$$

Answered by mr W last updated on 13/Nov/23

$$P_{big}=24+32+40=96$$$$A_{big}=\frac{24\times 32}{2}=384$$$$A_{small}=384-330=54$$$$\frac{P_{small}}{P_{big}}=\sqrt{\frac{A_{small}}{A_{big}}}=\sqrt{\frac{54}{384}}=\frac{3}{8}$$$$\Rightarrow P_{small}=\frac{3}{8}\times 96=36$$

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