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Question Number 95464 by i jagooll last updated on 25/May/20

Commented by i jagooll last updated on 25/May/20

$$\mathrm{my}{\mathrm{son}}^{\prime }\mathrm{s}\mathrm{exam}\mathrm{questions}$$

Commented by john santu last updated on 25/May/20

Commented by john santu last updated on 25/May/20

$$\mathrm{area}\mathrm{of}\mathrm{square}=16{\mathrm{cm}}^2$$$$\mathrm{area}\mathrm{of}\mathrm{\Delta }\mathrm{PBQ}=\frac{1}{4}\times 16{\mathrm{cm}}^2=4{\mathrm{cm}}^2$$$$\mathrm{BT}:\mathrm{PS}=1:2\Rightarrow \mathrm{BT}=2\mathrm{cm}$$$$\frac{\mathrm{area}\mathrm{of}\mathrm{\Delta }\mathrm{CBT}}{\mathrm{area}\mathrm{of}\mathrm{\Delta }\mathrm{PCS}}=\frac{\mathrm{a}}{4\mathrm{a}}=\frac{1}{4}$$$$\mathrm{in}\mathrm{\Delta }\mathrm{PBT};\mathrm{PC}:\mathrm{CB}=1:2$$$$\frac{\mathrm{area}\mathrm{of}\mathrm{\Delta }\mathrm{CBT}}{\mathrm{area}\mathrm{of}\mathrm{\Delta }\mathrm{PCT}}=\frac{\mathrm{a}}{2\mathrm{a}}$$$$\Rightarrow 6\mathrm{a}=4\Rightarrow \mathrm{a}=\frac{2}{3}.$$$$\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{region}=2\mathrm{a}=\frac{4}{3}$$$$$$

Commented by mr W last updated on 25/May/20

$$A_{shade}=\mathrm{\Delta }BPQ-2\times \mathrm{\Delta }ATQ$$$$A_{shade}=\frac{4\times 2}{2}-2\times \frac{2\times \frac{4}{3}}{2}=\frac{4}{3}$$

Commented by john santu last updated on 25/May/20

$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{height}\mathrm{of}\mathrm{\Delta }\mathrm{ATQ}=\frac{4}{3}$$

Commented by i jagooll last updated on 25/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Commented by mr W last updated on 25/May/20

$$\mathrm{\Delta }ATQ\sim \mathrm{\Delta }ARS$$$$\frac{heightof\mathrm{\Delta }ATQ}{heightof\mathrm{\Delta }ARS}=\frac{TQ}{RS}=\frac{1}{2}$$$$\Rightarrow heightof\mathrm{\Delta }ATQ=\frac{1}{1+2}\times 4=\frac{4}{3}$$

Commented by john santu last updated on 25/May/20

$$\mathrm{oo}\mathrm{i}\mathrm{know}.\mathrm{by}\mathrm{using}\mathrm{Ladder}\mathrm{theorem}$$$$\mathrm{let}\mathrm{height}\mathrm{of}\mathrm{\Delta }\mathrm{ATQ}=\mathrm{t}$$$$\frac{1}{\mathrm{t}}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\Rightarrow \mathrm{t}=\frac{4}{3}$$

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