Question

The perimeter of the triangle is 41 inches. The perimeter of the rectangle is 66 inches.

Answer 1

Answer:

Part A)  The system of equations is equal to

$41=4x+7y$

$33=(3x+7y)$

Part B)

The length is $24\ in$

The width is $9\ in$

Step-by-step explanation:

Part A)

we know that

The perimeter of the triangle is equal to the sum of the length of its three sides

In this problem

$41=x+3x+7y$

$41=4x+7y$ -----> equation A

The perimeter of rectangle is equal to

$P=2(L+W)$

In this problem

$66=2(3x+7y)$

$33=(3x+7y)$ ------> equation B

Part B)

Using a graphing tool

Solve the system of equations

We know that

The intersection point both graphs is the solution of the system of equations

$41=4x+7y$

$33=(3x+7y)$

The intersection point is $(8,9/7)$

see the attached figure

$x=8\ in$

$y=(9/7)\ in$

step 3

Find the length and width of the rectangle

The length is 3x

$3*8=24\ in$

The width is 7y

$7*(9/7)=9\ in$

Answer 2

PART A

Perimeter is the distance around the figure.

The perimeter of the triangle is 41 inches.

This implies that,

$3x + x + 7y = 41$

$4x + 7y = 41...(1)$

The perimeter of the rectangle is 66 inches.

This implies that,

$2(7y + 3x) = 66$
Or

$7y + 3x = 33...(2)$

PART B

The system of equations are

$4x + 7y = 41...(1)$

$7y + 3x = 33...(2)$

Subtract equation (2) from equation (1).

This gives us,

$4x - 3x = 41 - 33$

$x = 8$

Put this value into equation (2) to find y.

$7y + 3(8) =33$

$7y + 24 = 33$

$7y = 33- 24$

$7y =9$

$y = \frac{9}{7}$

The length of the rectangle is

$3(8)=24\:in$

The width of the rectangle is

$7(\frac{9}{7})=9\: in$