Question

John painted his most famous work, in his country, in 1930 on composition board with perimeter 104.99 in. If the rectangular painting is 5.01 in. taller than it is wide, find the dimensions of the painting.​

Answer 1

Answer:

The height of the painting is 28.7525 in  and the width is 23.7425 in

Step-by-step explanation:

Let

x -----> the height of the painting

y -----> the width of the painting

we know that

The perimeter of the painting is

$P=2(x+y)\\P=104.99\ in$

so

$104.99=2(x+y)$ ----> equation A

$x=y+5.01$ ----> equation B

substitute equation B in equation A and solve for y

$104.99=2(y+5.01+y)$

$52.495=2y+5.01$

$2y=52.495-5.01$

$y=23.7425\ in$

Find the value of x

$x=23.7425+5.01=28.7525\ in$

therefore

The height of the painting is 28.7525 in  and the width is 23.7425 in