The area of a square is 16x² + 40x + 25 square units. Find the perimeter of the square.

- (a)

- (b)

- (c)

- (d)

Answer
Why
Correct — D. The area of a square is side², so the expression must be a perfect square.
16x² + 40x + 25
= (4x)² + 2·(4x)·5 + 5²
= (4x + 5)²
So the side is 4x + 5.
Perimeter = 4 × side = 4(4x + 5) = 16x + 20
Now match the printed forms: 2(8x + 10) = 16x + 20 → option (d).
Why the others are wrong
- (a)Expands to 24x + 30, which is 6(4x + 5) — six sides, not four. The multiplier for a square's perimeter is 4.
- (b)Expands to 30x + 24, the 10 and the 8 swapped. It is not a multiple of the side 4x + 5, so it cannot be this square's perimeter.
- (c)Expands to 20x + 16, which is 4(5x + 4) — the perimeter of a square of side 5x + 4. But (5x + 4)² = 25x² + 40x + 16, not the given area.
Concept
Two ideas meet here: area = side² for a square, and the perfect square trinomial a² + 2ab + b² = (a + b)².
Recognise the pattern from the ends. 16x² is (4x)² and 25 is 5², so if this is a perfect square its middle term must be 2 × 4x × 5 = 40x. It is, so the side is 4x + 5.
The perimeter follows as 4(4x + 5) = 16x + 20. SSC then prints that in a factored form you have to recognise.
Every option is written factored, and none of them reads 16x + 20 on its face.
Expand before you compare — 2(8x + 10) is the same expression. Matching an answer to a differently factored option is half the work in this item.
Key facts
- 16x² + 40x + 25 = (4x + 5)², because 2 × 4x × 5 = 40x matches the middle term.
- A square of side s has area s² and perimeter 4s.
- The perimeter here is 4(4x + 5) = 16x + 20, printed as 2(8x + 10).
Study next
Common traps
- Square-rooting only the first and last terms without checking the middle term.
- Stopping at the side 4x + 5 when the question asks for the perimeter.
- Not expanding the options and missing that 2(8x + 10) equals 16x + 20.
SSC wraps identity recognition inside a mensuration sentence, so the geometry is one line and the algebra is the real question.
The same shift tests the square identities in pure algebra at Quant Q.18, where x² + y² = 280 and xy = 120 have to become (x ± y)².
Related PYQs
No directly related past PYQ was found.