Question

A poster of the OU gymnastics team has a width that is 6 cm less than 3 times the length of the poster. The frame around the poster is 44 cm. What is the width and length of the poster? How much glass is needed to fit the frame?

Answer 1

Given that width is 6 cm less than 3times the length. If we take length as L and width as B, then we have

B=3L-6 ............(1)

Also given that frame around the poster is 44cm i.e. the perimeter of the poster is 44. But the perimeter is given by 2(L+B). So we have

L+B=44/2 = 22 ...........(2)

Now we solve above two equations. From (2) we get B=22-L. Putting this value of B in (1) we get

22-L=3L-6 which gives 28 = 4L i.e. L=7. Then Putiing L=7 in (2) gives B=15.

So length = 7cm and width = 15cm.

Also, the glass needed is actually the area of the frame i.e. L*B = 7*15= 105 square cm.