Q). The sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m, find the sides of the two squares.
(Q 35 B – 30/1/3 – CBSE 2026 Question Paper)
Ans:
Let’s consider that the side of a square is X m and the side of the other square is Y m.
Step 1: By 1st given condition, A1 + A2 = 640
∴ X2 + Y2 = 640 ……….. (i)
Step 2: By 2nd given condition, P1 – P2 = 64
∴ 4 X – 4 Y = 64
∴ X – Y = 16
∴ X = 16 + Y ………….(ii)
Step 3: By substituting value of X from equation (ii) in equation (i), we get:
X2 + Y2 = 640
∴ (16 + Y)2 +Y2 = 640
∴ 256 + Y2 + 32 Y + Y2 = 640
∴ 2 Y2 + 32 Y + 256 – 640 = 0
∴ 2 Y2 + 32 Y – 384 = 0
∴ Y2 + 16 Y -192 = 0
Step 4: We solve the above quadratic equation by mid term splitting:
∴ Y2 +24 Y – 8 Y + 66 = 0
∴ Y (Y +24) – 8 (Y + 24) = 0
∴ (Y + 24) (Y – 8) = 0
∴ Y = – 24 and Y = 8
Here, we reject Y = – 24 because side can not have negative value.
∴ Y = 8
Step 5: By substituting value of Y in equation (i), we get:
X = 16 + Y
∴ X = 16 + 8 = 24
Therefore, the sides of one square is 24 m and the side of the other square is 8 m.
Check:
Sum of areas = 24 2 + 8 2 = 576 + 64 = 640 m 2 ……. it meets the given condition.
Difference of perimeters = 4 x 24 – 4 x 8 = 96 – 32 = 64 m ………it also meets the given condition.
Since both the given conditions in the question are met, hence our solution is correct.
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