Question

One side of a triangle twice as long as the shortest side. The third side is 4 times longer than the shortest side. The perimeter of the triangle is 35. Find all three sides.



A.
5, 10, 20

B.
10, 10, 15

C.
5, 10, 15

D.
4, 8, 16

Answer 1

Answer:

option d

Step-by-step explanation:

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Step-by-step explanation:

THE SHORTEST SIDE = X

THE LONGEST SIDE = 2X

THE THIRD SIDE = X+4

PERIMETER = X+2X+X+4

35 = 4X+4

35-4=4X

31=4X

X=31/4

Answer 2

Given :-

• One side = Twice as long as the shortest side.
• Third side = Four times longer than the shortest side.
• The perimeter of the triangle = 35 units.

To find :-

• All the three sides = ?

Solution :-

Let the sides of triangle :-

• Shortest side = x
• Second side = 2x
• Third side = 4x

As we know,

• $\boxed{Perimeter~=~Sum~of~all~the~sides.}$

$\implies$ 35 = x + 2x + 4x

$\implies$35 = 7x

$\implies$$\sf\dfrac{35}{7}$

$\implies$x = 5

• The value of x = 5

_______________________

The measure of the three sides :-

• Shortest side = x = 5 units.
• Second side = 2x = 2 × 5 = 10 units.
• Third side = 4x = 4 × 5 = 20 units.

Hence, Option A is correct.