Question

12E
30. a) The sum of the areas of two squares is 274 sq.m. If the difference of
their perimeters is 32 m, find the sides of the two squares.
(Or)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

let the side of one square = x m

perimeter of the square is =4x m

given that perimeters are differ by 32 m

means another square perimeter is 4x-32m.

the side of the second square =4x-32/4

=4(x-8)/4

=x-8 m

sum of there areas are 274 m2

area of square is s^2

then,

(x)^2+(x-8)^2=274

x^2+x^2-16x+64=274

2x^2-16x-274+64=0

2x^2-16x-210=0

x^2-8x-105=0

x^2-15x+7x-105=0

x(x-15)+7(x-15)=0

(x+7)(x-15)=0

x= -7 or 15(discard -7)

another square side=(x-8)m

=(15-8)m

=7m

hope this AP maths pre final question answer helps you.if you want ask any question on maths paper 2 ask I will answer it.