Question

.28
The sum of areas of two squares is 468cm? If the sum
of their perimeters is 120cm, then the difference of the
side is-





(1) 1.5 cm
(2) 2 cm
Q
(3) 4 cm
(4) 6 cm​

Answer 1

LET THE SIDE OF THE FIRST SQUARE = A

AND OF THE SECOND SQUARE = B

AREA OF SQUARE = $SIDE^{2}$

∴ AREA OF SQUARE 1 = $A^{2}$

AND AREA OF SQUARE 2 = $B^{2}$

SUM OF AREAS = $A^{2}$ + $B^{2}$

PERIMETER OF SQUARE = 4 X SIDE

∴ PERIMETER OF SQUARE 1 = 4A

AND PERIMETER OF SQUARE 2 = 4B

SUM OF PERIMETERS = 4A + 4B

GIVEN THAT

$A^{2}$ + $B^{2}$ = 468

AND

4A + 4B = 120

4 (A + B ) = 120

A + B = 30

USING THE IDENTITY

$(A + B)^{2}$ = $A^{2}$ + $B^{2}$ + 2AB

$30^{2}$ = 468 + 2AB

432 = 2AB

DIFFERENCE OF THE SIDES = A - B

USING THE IDENTITY

$(A - B)^{2}$ = $A^{2}$ + $B^{2}$ - 2AB

$(A - B)^{2}$ = 468 - 432

$(A - B)^{2}$ = 36

A - B = 6

∴ DIFFERENCE OF THE SIDES IS 6 CM