Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 19150 by Tinkutara last updated on 06/Aug/17

$$\mathrm{A}\mathrm{semicircle}\mathrm{is}\mathrm{tangent}\mathrm{to}\mathrm{both}\mathrm{legs}\mathrm{of}\mathrm{a}$$$$\mathrm{right}\mathrm{triangle}\mathrm{and}\mathrm{has}\mathrm{its}\mathrm{centre}\mathrm{on}\mathrm{the}$$$$\mathrm{hypotenuse}.\mathrm{The}\mathrm{hypotenuse}\mathrm{is}$$$$\mathrm{partitioned}\mathrm{into}4\mathrm{segments},\mathrm{with}\mathrm{lengths}$$$$3,12,12,\mathrm{and}x,\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{figure}.$$$$\mathrm{Determine}\mathrm{the}\mathrm{value}\mathrm{of}{}^{\prime }{x}^{\prime }.$$

Commented by Tinkutara last updated on 06/Aug/17

Commented by ajfour last updated on 06/Aug/17

Commented by ajfour last updated on 06/Aug/17

$$\mathrm{AB}=\sqrt{{\left(15\right)}^2-{\left(12\right)}^2}=9$$$$\frac{12+\mathrm{x}}{12}=\frac{12+3}{9}$$$$\Rightarrow \mathrm{x}=8.$$

Commented by Tinkutara last updated on 06/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Answered by sandy_suhendra last updated on 06/Aug/17

Commented by allizzwell23 last updated on 06/Aug/17

$$\mathrm{FD}=\mathrm{BF}=12\left(\mathrm{radii}\right)$$$$\mathrm{AB}=\sqrt{{(3+12)}^2-{12}^2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)$$$$\mathrm{AB}=\sqrt{{15}^2-{12}^2}=9$$$$\therefore \mathrm{AB}=9$$$$\mathrm{BCDF}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{of}\mathrm{side}12$$$$\Rightarrow \frac{\mathrm{DE}}{\mathrm{DF}}=\frac{\mathrm{EC}}{\mathrm{AC}}\left(\mathrm{\Delta }\mathrm{DEF}\mathrm{is}\mathrm{similar}\mathrm{to}\mathrm{\Delta }\mathrm{ACE}\right)$$$$\Rightarrow \frac{\mathrm{DE}}{12}=\frac{\mathrm{DE}+12}{12+9}$$$$\Rightarrow \frac{\mathrm{DE}}{12}=\frac{\mathrm{DE}+12}{21}$$$$\Rightarrow 21\times \mathrm{DE}=12\times (\mathrm{DE}+12)$$$$\Rightarrow (21-12)\times \mathrm{DE}=12\times 12$$$$\Rightarrow 9\times \mathrm{DF}=12\times 12$$$$\therefore \mathrm{DE}=16$$$$\mathrm{Similarly},$$$${(12+x)}^2={12}^2+{16}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right)$$$$12+x=\sqrt{400}=20$$$$\therefore x=8$$

Commented by sandy_suhendra last updated on 06/Aug/17

Commented by Tinkutara last updated on 06/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com