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Question Number 181783 by mr W last updated on 30/Nov/22
one side of a triangle is 20 cm. the  other two sides are in ratio 1:3.  1) what is the maximum area of the  triangle, if exists?  2) what is the maximun perimeter  of the triangle, if exists?
$${one}\:{side}\:{of}\:{a}\:{triangle}\:{is}\:\mathrm{20}\:{cm}.\:{the} \\ $$$${other}\:{two}\:{sides}\:{are}\:{in}\:{ratio}\:\mathrm{1}:\mathrm{3}. \\ $$$$\left.\mathrm{1}\right)\:{what}\:{is}\:{the}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${triangle},\:{if}\:{exists}? \\ $$$$\left.\mathrm{2}\right)\:{what}\:{is}\:{the}\:{maximun}\:{perimeter} \\ $$$${of}\:{the}\:{triangle},\:{if}\:{exists}? \\ $$
Answered by Frix last updated on 30/Nov/22
Let the sides x, 3x  5<x<10  The collapsed triangle 10,20,30 has the  maximum perimeter 60  The area is A(x)=2(√(−x^4 +125x^2 −2500))  A′(x)=((−4x(2x^2 −125))/(A(x)))=0∧5<x<10 ⇒ x=((5(√(10)))/2)  ⇒ maximum area is 75
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{sides}\:{x},\:\mathrm{3}{x} \\ $$$$\mathrm{5}<{x}<\mathrm{10} \\ $$$$\mathrm{The}\:\mathrm{collapsed}\:\mathrm{triangle}\:\mathrm{10},\mathrm{20},\mathrm{30}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{perimeter}\:\mathrm{60} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{is}\:{A}\left({x}\right)=\mathrm{2}\sqrt{−{x}^{\mathrm{4}} +\mathrm{125}{x}^{\mathrm{2}} −\mathrm{2500}} \\ $$$${A}'\left({x}\right)=\frac{−\mathrm{4}{x}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{125}\right)}{{A}\left({x}\right)}=\mathrm{0}\wedge\mathrm{5}<{x}<\mathrm{10}\:\Rightarrow\:{x}=\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{is}\:\mathrm{75} \\ $$
Commented by mr W last updated on 30/Nov/22
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by MJS_new last updated on 30/Nov/22
generally for a given side a and a ratio  b:c=1:v with v>1 we get the maximum  area  ((a^2 v)/(2(v^2 −1))) with b=((√(v^2 +1))/(v^2 −1))a and c=vb
$$\mathrm{generally}\:\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:\mathrm{side}\:{a}\:\mathrm{and}\:\mathrm{a}\:\mathrm{ratio} \\ $$$${b}:{c}=\mathrm{1}:{v}\:\mathrm{with}\:{v}>\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{area} \\ $$$$\frac{{a}^{\mathrm{2}} {v}}{\mathrm{2}\left({v}^{\mathrm{2}} −\mathrm{1}\right)}\:\mathrm{with}\:{b}=\frac{\sqrt{{v}^{\mathrm{2}} +\mathrm{1}}}{{v}^{\mathrm{2}} −\mathrm{1}}{a}\:\mathrm{and}\:{c}={vb} \\ $$

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