Two isosceles triangles have the same base length. The legs of one of the triangles are twice as long as the legs of the other. How do you find the lengths of the sides of the triangles if their perimeters are 23 cm and 41 cm?

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Answer 1 · 2017-05-01T20:46:11

Every step shown so a bit long. Skip over the bits you know.

Base is 5 for both
The smaller legs are 9 each
The longer legs are 18 each

Sometimes a quick sketch helps in spotting what to do

Tony B

For triangle 1 $->a+2b=23" "...............Equation(1)$
For triangle 2 $->a+4b=41" "...............Equation(2)$

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$color(blue)("Determine the value of "b)$

For equation(1) subtract $2b$ from both sides giving:

$a=23-2b" ".........................Equation(1_a)$

For equation(2) subtract $4b$ from both sides giving:

$a=41-4b" "......................Equation(2_a)$

Set $Equation(1_a)=Equation(2_a)$ through $a$

$23-2b=a=41-4b$

$23-2b=41-4b$

Ad $color(red)(4b)$ to both sides

$color(green)(23-2bcolor(red)(+4b)" "=" "41-4bcolor(red)(+4b))$

$23+2b" "=" "41+0$

Subtract $color(red)(23)$ from both sides

$color(green)(23color(red)(-23)+2b" "=" "41color(red)(-23))$

$0+2b" "=" "18$

Divide both sides by $color(red)(2)$

$color(green)(2/(color(red)(2)) xx b" "=" "18/(color(red)(2)))$

But $2/2=1$ giving $1xxb=b$

$b=9$
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$color(blue)("Determine the value of "a)$

Substitute for $b$ in $Equation(1)$

$a+2b=23" "->" "a+2(9)=23$

$" "a+18=23$

$" "a=5$
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Check using $Equation(2)$

$a+4b=41" "=>5+4(9)=41$
$" "5+36color(white)(.)=41 color(red)(larr" True")$

$a=5"; "b=9$
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