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Question Number 96032 by i jagooll last updated on 29/May/20

Commented by i jagooll last updated on 29/May/20

$$\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{with}\mathrm{an}\mathrm{area}240{\mathrm{cm}}^2$$$$\mathrm{if}\mathrm{DP}=3\mathrm{PC},\mathrm{then}\mathrm{what}\mathrm{is}\mathrm{the}$$$$\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}?$$

Commented by mr W last updated on 29/May/20

$$saysidelengthofsquare=a$$$$a^2=240$$$$$$$$DP/PC=3/1$$$$\Rightarrow DP=\frac{3}{4}DC=\frac{3a}{4}$$$$\Rightarrow PC=\frac{a}{4}=SD$$$$A_{\mathrm{\Delta }DSP}=\frac{1}{2}\times \frac{3a}{4}\times \frac{a}{4}=\frac{3a^2}{32}$$$$A_{\mathrm{\Delta }DSP}+A_{\mathrm{\Delta }OSP}=\frac{square}{4}=\frac{a^2}{4}$$$$A_{\mathrm{\Delta }OSP}=\frac{a^2}{4}-\frac{3a^2}{32}=\frac{5a^2}{32}$$$$A_{shade}=2\times A_{\mathrm{\Delta }OSP}=\frac{5a^2}{16}=\frac{5}{16}\times square$$$$=\frac{5}{16}\times 240=75$$

Commented by i jagooll last updated on 29/May/20

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

Answered by john santu last updated on 29/May/20

Commented by john santu last updated on 29/May/20

$$\mathrm{area}\mathrm{ABCD}={16\mathrm{a}}^2=240;{\mathrm{a}}^2=\frac{240}{16}=15$$$$\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{region}=\frac{1}{2}{\left(\mathrm{a}\sqrt{10}\right)}^2$$$$={5\mathrm{a}}^2=5\times 15{\mathrm{cm}}^2=75{\mathrm{cm}}^2$$

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