Question

the sides of a triangle are 3x+7,4x-3 and 5x+3 Find the sides of the triangle if it's perimeter is 55 units​

Answer 1

Given :-

Sides of triangle = 3x +7 , 4x -3 and 5x +3

Perimeter= 55 units

Solution :-

We know that perimeter of triangle is sum of all sides

3x +7 + 4x -3 + 5x +3 = 55

12x + 7 = 55

12x = 55-7

12x = 48

x = 48/12

x = 4.

So sides will be

3(4) +7 = 19

4(4) -3 = 13

5(4)+3 = 23

Answer 2

Answer :-

The sides of triangle are :-

• 19 units
• 13 units
• 23 units

Step-by-step explanation :-

Given that,

• The perimeter of triangle = 55 units
• The sides are 3x + 7, 4x - 3 and 5x + 3

According the question,

As we know that,

$\dag \: \: \underline{\boxed{ \red{ \sf{Perimeter \: of \: triangle = Sum \: of \: all \: sides \: of \: \triangle}}}}$

Therefore,

$\longrightarrow \sf{(3x + 7) + (4x - 3)+ (5x + 3) = 55} \\ \\ \\ \longrightarrow \sf{3x + 7 + 4x - 3+ 5x + 3= 55} \\ \\ \\ \longrightarrow \sf{3x + 4x + 5x + 7 - 3 + 3= 55} \\ \\ \\ \longrightarrow \sf{12x + 7= 55} \\ \\ \\ \longrightarrow \sf{12x = 55 - 7} \\ \\ \\ \longrightarrow \sf{12x = 48} \\ \\ \\ \longrightarrow \sf{x = \dfrac{48}{12}} \\ \\ \\ \longrightarrow \sf{x = \cancel{ \dfrac{48}{12}}} \\ \\ \\ \longrightarrow \frak{ \pink{4 \: units}} \: \: \: \: \bigstar$

Now, the sides are :-

First side [ 3x + 7 units ]

$\longrightarrow \: \sf{3x + 7} \\ \\ \longrightarrow \: \sf{3 \times 4 + 7} \\ \\ \longrightarrow \: \sf{12 + 7} \\ \\ \longrightarrow \: \frak{ \pink{ 19 \: units}}$

Second side [ 4x - 3 units ]

$\longrightarrow \: \sf{4x - 3} \\ \\ \longrightarrow \: \sf{4 \times 4 - 3} \\ \\ \longrightarrow \: \sf{16 - 3} \\ \\ \longrightarrow \: \frak{ \pink{ 13 \: units}}$

Third side [ 5x + 7 units ]

$\longrightarrow \: \sf{5x + 7} \\ \\ \longrightarrow \: \sf{5 \times 4 + 3} \\ \\ \longrightarrow \: \sf{20+ 3} \\ \\ \longrightarrow \: \frak{ \pink{ 23 \: units}}$

The sides of triangle are :- 19 units, 13 units and 23 units.