A-semicircle-is-tangent-to-both-legs-of-a-right-triangle-and-has-its-centre-on-the-hypotenuse-The-hypotenuse-is-partitioned-into-4-segments-with-lengths-3-12-12-and-x-as-shown-in-the-figure-Det

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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

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

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