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Question-66851




Question Number 66851 by John Kaloki Musau last updated on 20/Aug/19
Commented by John Kaloki Musau last updated on 20/Aug/19
A piece of wire, pcm long, is bent to form the shape shown in the figure above. The figure consists of a semicircular arc of radius rcm and two perpendicular sides of length xcm each. Express x in terms of p and r, hence show that the area of the figure is given by A=1/2πr^2+1/8(p-πr)^2
Answered by John Kaloki Musau last updated on 20/Aug/19
The answers are,  (i) x=((p−πr)/2)cm  (ii)A=(1/2)πr^2 +(1/2)×x×x=(1/2)πr+(1/2)(((p−πr)/3))^2   Those are the answers but I   don′t know how to calculate  them.Please show me.
$${The}\:{answers}\:{are}, \\ $$$$\left(\boldsymbol{{i}}\right)\:\boldsymbol{{x}}=\frac{\boldsymbol{{p}}−\pi{r}}{\mathrm{2}}{cm} \\ $$$$\left({ii}\right)\boldsymbol{{A}}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×{x}×{x}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{p}−\pi{r}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{T}}{ho}\boldsymbol{{se}}\:\boldsymbol{{are}}\:\boldsymbol{{the}}\:\boldsymbol{{answers}}\:\boldsymbol{{but}}\:\boldsymbol{{I}}\: \\ $$$$\boldsymbol{{don}}'\boldsymbol{{t}}\:\boldsymbol{{know}}\:\boldsymbol{{how}}\:\boldsymbol{{to}}\:\boldsymbol{{calculate}} \\ $$$$\boldsymbol{{them}}.\boldsymbol{{Please}}\:\boldsymbol{{show}}\:\boldsymbol{{me}}. \\ $$
Answered by $@ty@m123 last updated on 20/Aug/19
Length of half circumference  =πr (formula)  ∴ Length of wire=x+x+πr  ⇒p=2x+πr  ⇒2x=p−πr  ⇒x=((p−πr)/2)  −−−−−−−−−−−−−−−−−  Area of semi circle=((πr^2 )/2)   ....(1)   (formula)  Area of triangle=(1/2)×base×height   (formula)  Area of triangle=(1/2)×(((p−πr)/2))^2 .... (2)  ∴ Area of shape=(1)+(2)
$${Length}\:{of}\:{half}\:{circumference} \\ $$$$=\pi{r}\:\left({formula}\right) \\ $$$$\therefore\:{Length}\:{of}\:{wire}={x}+{x}+\pi{r} \\ $$$$\Rightarrow{p}=\mathrm{2}{x}+\pi{r} \\ $$$$\Rightarrow\mathrm{2}{x}={p}−\pi{r} \\ $$$$\Rightarrow{x}=\frac{{p}−\pi{r}}{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$${Area}\:{of}\:{semi}\:{circle}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}\:\:\:….\left(\mathrm{1}\right)\:\:\:\left({formula}\right) \\ $$$${Area}\:{of}\:{triangle}=\frac{\mathrm{1}}{\mathrm{2}}×{base}×{height}\:\:\:\left({formula}\right) \\ $$$${Area}\:{of}\:{triangle}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{{p}−\pi{r}}{\mathrm{2}}\right)^{\mathrm{2}} ….\:\left(\mathrm{2}\right) \\ $$$$\therefore\:{Area}\:{of}\:{shape}=\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$

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