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

In a triangle QRS, QR=14cm, RS=12cm and SQ=9cm. An incircle is drawn to touch the three sides of the triangle at X,Y,Z respectively. Find the lenghts QX, RY and SZ.

$$\boldsymbol{\mathrm{In}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{{QRS}},\:\boldsymbol{{QR}}=\mathrm{14}\boldsymbol{{cm}},\:\boldsymbol{{RS}}=\mathrm{12}\boldsymbol{{cm}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{{SQ}}=\mathrm{9}\boldsymbol{{cm}}.\:\boldsymbol{\mathrm{An}}\:\boldsymbol{\mathrm{incircle}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{drawn}}\:\boldsymbol{\mathrm{to}} \\ $$$$\boldsymbol{\mathrm{touch}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{three}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{{X}},\boldsymbol{{Y}},\boldsymbol{{Z}}\:\boldsymbol{\mathrm{respectively}}.\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{lenghts}} \\ $$$$\boldsymbol{{QX}},\:\boldsymbol{{RY}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{SZ}}. \\ $$

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

let QX=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 ⇒(a+b+c)−(b+c)=17.5−12 ∴a=5.5 b=14−a=14−5.5=8.5 c=12−8.5=3.5 ∴QX=5.5cm RY=8.5cm SZ=3.5cm

$${let}\:{QX}={QZ}={a} \\ $$$${RX}={RY}={b} \\ $$$${SY}={SZ}={c} \\ $$$${QR}=\mathrm{14} \\ $$$${QX}+{XR}=\mathrm{14} \\ $$$$\:{a}+{b}=\mathrm{14}\:....{i} \\ $$$${RS}=\mathrm{12} \\ $$$${RY}+{SY}=\mathrm{12} \\ $$$$\:{b}+{c}=\mathrm{12}\:....{ii} \\ $$$${QS}=\mathrm{9} \\ $$$${SZ}+{ZQ}=\mathrm{9} \\ $$$${c}+{a}=\mathrm{9}\:...{iii} \\ $$$$\left.{i}\left.\right)\left.\:+{ii}\right)+{iii}\right) \\ $$$$\:\mathrm{2}\left({a}+{b}+{c}\right)=\mathrm{14}+\mathrm{12}+\mathrm{9} \\ $$$${a}+{b}+{c}=\mathrm{17}.\mathrm{5} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)−\left({b}+{c}\right)=\mathrm{17}.\mathrm{5}−\mathrm{12} \\ $$$$\therefore{a}=\mathrm{5}.\mathrm{5} \\ $$$$\:{b}=\mathrm{14}−{a}=\mathrm{14}−\mathrm{5}.\mathrm{5}=\mathrm{8}.\mathrm{5} \\ $$$${c}=\mathrm{12}−\mathrm{8}.\mathrm{5}=\mathrm{3}.\mathrm{5} \\ $$$$\therefore{QX}=\mathrm{5}.\mathrm{5}{cm} \\ $$$${RY}=\mathrm{8}.\mathrm{5}{cm} \\ $$$${SZ}=\mathrm{3}.\mathrm{5}{cm} \\ $$

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

Commented by MathsFan last updated on 03/Jun/22

thank you sir i am very greatful

$${thank}\:{you}\:{sir} \\ $$$${i}\:{am}\:{very}\:{greatful} \\ $$

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

nice!

$$\mathrm{nice}! \\ $$

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