Question

9. Each of the two equal sides of a triangle
is 4 m less than three times the third
side. Find the sides of the triangle, if its
perimeter is 55 m.​

Answer 1

Answer:

9,23,23

Step-by-step explanation:

let the unequal side be x

then first equal side = 3x-4

second equal side = 3x-4

perimeter of triangle = side + side + side

so, x+3x-4+3x-4 = 55

7x-8 = 55

7x = 55+8= 63

x= 63/7 = 9

so, first side = x= 9 m

so, first side = x= 9 msecond side = 3x-4 = 3(9)-4 = 27-4 = 23

so, first side = x= 9 msecond side = 3x-4 = 3(9)-4 = 27-4 = 23third side = 3x -4 = 27-4 = 23

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Answer 2

$\bf{\underline{Given:-}}$

Perimeter of the triangle = 55 m

$\bf{\underline{To\:Find}}$

The sides of the triangle

$\bf{\underline{Assumption}}$

Let the third side be x

Two equal sides = 3x - 4

$\bf{\underline{Solution:-}}$

We know,

$\sf{Perimeter \:of \:the \:triangle = (side + side + side)\: units}$

Therefore,

ATQ,

$\sf{55 = x + (3x-4) + (3x-4)}$

$\sf{\implies 55 = x + 3x - 4 + 3x - 4}$

$\sf{\implies 55 = 7x - 8}$

$\sf{\implies 55 + 8 = 7x}$

$\sf{\implies 7x = 63}$

$\sf{\implies x = \dfrac{63}{7}}$

$\sf{\implies x = 9}$

Side of the Triangles :-

$\sf{1st\:side = 3x-4 = 3\times9-4 = 27 - 4 = 23\:cm}$

$\sf{2nd\:side = 3x - 4 = 3\times9-4 = 27 - 4 = 23\:cm}$

$\sf{3rd\:side = x = 9\:cm}$

$\bf{\underline{Verification}}$

Perimeter of the triangle = 55

Therefore,

$\sf{Side + side + side = perimeter}$

$\sf{23 + 23 + 9 = 55}$

$\sf{46 + 9 = 55}$

$\sf{55=55\:\:[Verified]}$