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-

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

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

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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