Question

the length of the sides of triangle are 3x+2y,4x+4/3y and 3(x+1)+3/2(y-1). if the triangle is equilateral then length of its sides is​
a 8 unit
b 10 unit
c 12 unit
d 16 unit
guys in solution please

Answer 1

Answer:

Step-by-step explanation:

Since its an equilateral triangle, all the sides are equal

3x + 2y = 4x +4/3y = 3(x+1) + 3/2(y-1)

3x + 2y = 4x + 4/3y

3x-4x = 4/3y-2y

-x = 4y-6y/3

-x = -2y/3

x-2y/3 = 0

3x-2y = 0 _____ (i)

4x+4/3y = 3(x+1) +3/2(y-1)

4x + 4/3y = 3x +3 +3/2y - 3/2

4x-3x +4/3y + 3/2y = 6-3/2

x + 8y+9y/6 = 3/2

x+ 17y/6 = 3/2

6x + 17y = 9 _____(ii)

equations we have;

3x - 2y = 0    _______(i)

6x + 17y = 9  _______(ii)

solving (i) and (ii), we get,

x = 2/7 and y = 3/7

∴ length of triangle is 12 units

Answer 2

Answer:

Hence the correct answer is option (c) 12 units

Step-by-step explanation:

Given,

The length of the sides of an equilateral triangle are 3x+2y,4x+$\frac{4}{3}$y, and 3(x+1)+$\frac{3}{2}$(y-1)

To find,

The length of its sides

Recall the concept

All the three sides of an equilateral triangle are of equal length.

Solution:

Since all the three sides of an equilateral triangle we have

3x+2y = 4x+$\frac{4}{3}$y =  3(x+1)+$\frac{3}{2}$(y-1)

3x+2y = 4x+$\frac{4}{3}$y

3x -4x =$\frac{4}{3}$y - 2y

-x = $\frac{4y -6y}{3}$ = $\frac{-2y}{3}$

x = $\frac{2y}{3}$

3x -2y = 0 ----------------(1)

Again,

3x+2y  = 3(x+1)+$\frac{3}{2}$(y-1)

3x+2y  = 3x +3 +$\frac{3}{2}$y- $\frac{3}{2}$

2y - $\frac{3}{2}$y = 3 - $\frac{3}{2}$

$\frac{4y-3y}{2}$ = $\frac{6-3}{2}$

$\frac{y}{2} = \frac{3}{2}$

y = 3

Substituting the value of y in equation(1) we get

3x - 2 ×3 = 0

3x = 6

x = 2

Hence, the length of the side of the triangle = 3x+2y = 3×2+2×3 =6+6 = 12

∴Length of the side  = 12 units

Hence the correct answer is option (c) 12 units

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