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Question Number 113080 by bobhans last updated on 11/Sep/20

The perimeter of a triangle is 84 cm and it′s area is 336 square cm. If the length of one side of triangle is 30 cm, then what is the lengths of the remaining two sides of triangle ?

$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{is} \\ $$$$\mathrm{84}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{area}\:\mathrm{is}\:\mathrm{336}\:\mathrm{square}\:\mathrm{cm}.\:\mathrm{If}\:\mathrm{the}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{30}\:\mathrm{cm},\:\mathrm{then} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{two} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:? \\ $$

Answered by bemath last updated on 11/Sep/20

(i) perimeter = 84 = 30 +x+y lengths two sides is { (x),((54−x)) :} then the area of triangle is 336 = (√(42.(42−30).(42−x).(42−54+x))) 336 = (√(42.12.(42−x).(x−12))) 224 = (42−x)(x−12) 224 = −504+54x−x^2 x^2 −54x+728 = 0 x = ((54 ± (√4))/2) = ((54 ± 2)/2) = 27 ± 1 Hence the lengths of two sides of triangle is { ((x=28, y=26 or)),((x=26, y = 28)) :}

$$\left(\mathrm{i}\right)\:\mathrm{perimeter}\:=\:\mathrm{84}\:=\:\mathrm{30}\:+\mathrm{x}+\mathrm{y}\: \\ $$$$\:\:\:\mathrm{lengths}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{is}\:\begin{cases}{\mathrm{x}}\\{\mathrm{54}−\mathrm{x}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{is}\: \\ $$$$\mathrm{336}\:=\:\sqrt{\mathrm{42}.\left(\mathrm{42}−\mathrm{30}\right).\left(\mathrm{42}−\mathrm{x}\right).\left(\mathrm{42}−\mathrm{54}+\mathrm{x}\right)} \\ $$$$\mathrm{336}\:=\:\sqrt{\mathrm{42}.\mathrm{12}.\left(\mathrm{42}−\mathrm{x}\right).\left(\mathrm{x}−\mathrm{12}\right)} \\ $$$$\mathrm{224}\:=\:\left(\mathrm{42}−\mathrm{x}\right)\left(\mathrm{x}−\mathrm{12}\right) \\ $$$$\mathrm{224}\:=\:−\mathrm{504}+\mathrm{54x}−\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{54x}+\mathrm{728}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{54}\:\pm\:\sqrt{\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{54}\:\pm\:\mathrm{2}}{\mathrm{2}}\:=\:\mathrm{27}\:\pm\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{of} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\begin{cases}{\mathrm{x}=\mathrm{28},\:\mathrm{y}=\mathrm{26}\:\mathrm{or}}\\{\mathrm{x}=\mathrm{26},\:\mathrm{y}\:=\:\mathrm{28}}\end{cases} \\ $$

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