Question

A poster is 8 in taller than it is wide. It is mounted on a backing board that provides a 2 in border on each side of the poster. If the area of the backing board is 308 in2 what are the dimensions of the poster

Answer 1

Answer:

The dimensions of poster are 32.5 in wide and 40.5 in tall.

Step-by-step explanation:

Given:

A poster is 8 in taller than it is wide.

It is mounted on a backing board that provides a 2 in border on each side of the poster.

The area of the backing board is 308 in².

Now, to find the dimensions of the poster.

Let $x$ be the length of the poster.

And $y$ be the width of the poster.

As given, poster is 8 in taller than it is wide.

So,

$x=y+8$   ......(1)

Area = 308 in².

So, it is mounted on a backing board that provides a 2 in border on each side of the poster.

According to question:

$2\times (2(y+4))+2\times (2\times x)=308$

Now. substituting the value from equation (1) in the place of $x$ we get:

$2\times (2y+8)+2(2\times (y+8))=308$

$2\times (2y+8)+2(2y+16)=308$

$4y+16+4y+32=308$

$8y+48=308$

Subtracting both sides by 48 we get:

$8y=260$

Dividing both sides by 8 we get:

$y=32.5\ in.$

The width of the poster = 32.5 in.

Now, substituting the value of $y$ in equation (1):

$x=y+8$

$x=32.5+8$

$x=40.5\ in.$

Length of the poster = 40.5 in.

Therefore, the dimensions of poster are 32.5 in wide and 40.5 in tall.