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Question Number 170912 by MathsFan last updated on 03/Jun/22

$$\mathbf{In}\mathbf{a}\mathbf{triangle}\mathit{Q}\mathit{R}\mathit{S},\mathit{Q}\mathit{R}=14\mathit{c}\mathit{m},\mathit{R}\mathit{S}=12\mathit{c}\mathit{m}$$$$\mathbf{and}\mathit{S}\mathit{Q}=9\mathit{c}\mathit{m}.\mathbf{An}\mathbf{incircle}\mathbf{is}\mathbf{drawn}\mathbf{to}$$$$\mathbf{touch}\mathbf{the}\mathbf{three}\mathbf{sides}\mathbf{of}\mathbf{the}\mathbf{triangle}$$$$\mathbf{at}\mathit{X},\mathit{Y},\mathit{Z}\mathbf{respectively}.\mathbf{Find}\mathbf{the}\mathbf{lenghts}$$$$\mathit{Q}\mathit{X},\mathit{R}\mathit{Y}\mathbf{and}\mathit{S}\mathit{Z}.$$

Answered by som(math1967) last updated on 03/Jun/22

$$letQX=QZ=a$$$$RX=RY=b$$$$SY=SZ=c$$$$QR=14$$$$QX+XR=14$$$$a+b=14....i$$$$RS=12$$$$RY+SY=12$$$$b+c=12....ii$$$$QS=9$$$$SZ+ZQ=9$$$$c+a=9...iii$$$$i)+ii)+iii)$$$$2(a+b+c)=14+12+9$$$$a+b+c=17.5$$$$\Rightarrow (a+b+c)-(b+c)=17.5-12$$$$\therefore a=5.5$$$$b=14-a=14-5.5=8.5$$$$c=12-8.5=3.5$$$$\therefore QX=5.5cm$$$$RY=8.5cm$$$$SZ=3.5cm$$

Commented by som(math1967) last updated on 03/Jun/22

Commented by MathsFan last updated on 03/Jun/22

$$thankyousir$$$$iamverygreatful$$

Commented by otchereabdullai@gmail.com last updated on 03/Jun/22

$$\mathrm{nice}!$$

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