Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 125758 by ZiYangLee last updated on 13/Dec/20

$$\mathrm{\Delta }\mathrm{ABC}\mathrm{have}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{length}101313$$$$\mathrm{while}\mathrm{\Delta }\mathrm{PQR}\mathrm{have}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{length}$$$$131324.$$$$\mathrm{Find}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{Area}\mathrm{of}\mathrm{\Delta }\mathrm{ABC}:\mathrm{Area}$$$$\mathrm{of}\mathrm{\Delta }\mathrm{PQR}.$$

Answered by mr W last updated on 13/Dec/20

$${\mathrm{\Delta }}_{ABC}=\frac{1}{2}\times 10\times \sqrt{{13}^2-{\left(\frac{10}{2}\right)}^2}$$$${\mathrm{\Delta }}_{PQR}=\frac{1}{2}\times 24\times \sqrt{{13}^2-{\left(\frac{24}{2}\right)}^2}$$$$\frac{{\mathrm{\Delta }}_{ABC}}{{\mathrm{\Delta }}_{PQR}}=\frac{10}{24}\times \sqrt{\frac{{13}^2-5^2}{{13}^2-{12}^2}}=\frac{10\times 12}{24\times 5}=1$$

Answered by MJS_new last updated on 13/Dec/20

$$\mathrm{area}\mathrm{of}△aab=\frac{b}{4}\sqrt{4a^2-b^2}$$$$\Rightarrow \frac{\mathrm{area}△ABC}{\mathrm{area}△PQR}=\frac{1}{1}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com