Question

How do you find the dimensions of a rectangle whose perimeter is 52 m and whose area is 160 m?

Answer 1

answer

perimeter

2(l+b) =52m  

l+b=26                            ...........................(i)[given]

area

lxb=160m^2                                        ..........................(ii)[given]

l=160/b                                                 ..........................(iii)

Substituting (iii) in (i),we get

(160/b)+b=26

(160+b^2)/b=26

b^2-26b+160=0

sum=-26

product=160

pair of numbers =-10,-16

by splitting the middle term

b^2-16b-10b+160=0

b[b-16]-10[b-16]=0

(b-10)(b-16)=0

b=10;b=16

Case 1:-If b=10

l=160/10                                                                                     [from(iii)]

l=16m

Case 2:-If b=16

l=160/16                                                                                     [from(iii)]

l=10m

but length is usually greater than breadth

THUS,

l=16m and b=10m