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 77164 by Maclaurin Stickker last updated on 03/Jan/20

Commented by Maclaurin Stickker last updated on 03/Jan/20

$$Igotx=5.Ismyanswercorrect?$$

Commented by mr W last updated on 03/Jan/20

$$correct!$$

Answered by mr W last updated on 03/Jan/20

$$b=widthofrectangle$$$$\sqrt{{10}^2-x^2}+\frac{bx}{10}=10$$$$\Rightarrow b=\frac{10}{x}(10-\sqrt{{10}^2-x^2})$$$$b+x+\frac{b}{10}\sqrt{{10}^2-x^2}=10$$$$x+\frac{10}{x}(10-\sqrt{{10}^2-x^2})(1+\frac{\sqrt{{10}^2-x^2}}{10})=10$$$$10x^2+10(10-\sqrt{{10}^2-x^2})(10+\sqrt{{10}^2-x^2})=100x$$$$20x^2=100x$$$$\Rightarrow x=5$$

Commented by Maclaurin Stickker last updated on 03/Jan/20

$$ohyeah!$$

Commented by Tawa11 last updated on 29/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com