Please help I cant figure this out.

Question

Rachel bought a picture of framed artwork. The diagonal of the frame is 20 inches. If the length of the frame is 4 inches greater than the width, find the dimensions of the frame

Answer 1

Hi Rachel--

 

Remember, it's a picture frame, so let's assume its rectangular.

 

Which means the triangle is a right triangle, so at some point the Pythagorean theorem is going to come into play. 

 

You know that your hypotenuse is 20, which squared is 400.

 

so a² + b²= 400, and...

 

a = b + 4 because the length is 4 more than the width.

 

If you sub in "b + 4" for "a" you can solve this:

 

(b + 4)² + b² = 400, so...

 

(b + 4)(b + 4) + b² = 400, so...

 

b² + 8b + 16 + b² = 400, so...

 

2b² + 8b = 384 so

 

b² + 4b = 192 so

 

b² + 4b - 192 = 0, so...

 

(b + 16)(b - 12) = 0

 

so b could either b - 16 or 12, and since a length can't be negative, it's got to be 12. The other leg is 12 + 4, which is 16.

 

12 squared (144) + 16 squared (256) = 20 squared (400)

 

Hope this helps!

 

Matt in New York

 

(PS: this is an extension of the 3-4-5 phenomenon of right triangles. Any RT where the legs are in a 3:4 ratio, the hypotenuse will the corresponding multiple of 5)