When the sides of a square are increased by 4 cm, the area of the square increases by 21 cm.?

Question

Find the length of a side of the original square.

Thanks

Answer 1

The original side length is s and the original area is s^2. The new side length is s + 4 and the new area is s^2 + 21. However, the new area is also (s + 4)^2. So set (s + 4)^2 = s^2 + 21 and solve.

Answer 2

Let x = the length of the side of the original square.
A = area of original square.

This means that the area of the square is
A = x^2

We also know that when the sides of a square are increased by 4cm, then x+4 is the length of the new side. This means (x+4)(x+4) is the area of the new square. Let's call the area A2.

A2 = (x+4)^2
But we were given that A2 = A + 21, so now we have

A + 21 = (x+4)^2

And now, we solve for x

A + 21 = x^2 + 8x + 16
A = x^2 + 8x - 5
0 = x^2 + 8x - 5 - A
0 = x^2 + 8x - (5+A)

We have to use the quadratic formula to solve this.

x = [-8 +/- sqrt(8^2 - 4(5+A))]/2
x = [-8 +/- sqrt(64 - 20 - 4A)]/2
x = [-8 +/- sqrt(44 - 4A)]/2
x = [-8 +/- 2 sqrt(11 - A)] / 2
x = -4 +/- sqrt(11 - A)

x = -4 + sqrt(11 - A) or
x = -4 - sqrt(11 - A) [we discard this solution

x = -4 + sqrt(11 - A)

Since x cannot be negative, then

-4 + sqrt(11 - A) > 0
sqrt(11 - A) > 4
11 - A > 4
-A > -7
A < 7

The length of a side of the original square is
x = -4 + sqrt(11 - A)
for all A < 7

Answer 3

easy (I assume you mean "by 21 cm square")

original area, A:
A = c^2 (1)

modified area, A' = A+21 = (c+4)^2
which you can develop
A + 21 = (c+4)^2 = c^2+8c+16 (2)

now you can substract (1) from (2):
A + 21 - A = c^2 +8c + 16 -c^2

21 = 8c +16
c = (5/8)

so your original area, A, is c^2 = 0.3906...

and the new area, A', is (c+4)^2 = 21.3906...

so basically, you had a very small square to start with, and end up with a much larger one


hope this helps

Answer 4

(s+4)^2 = s^2 + 21
s^2 + 8s + 16 = s^2 + 21
8s = 5
s = 5/8

Answer 5

4 times 21 will equal 84 dived by 2 and that is your answer 42.

Answer 6

x^2+8a+16=x^2+21
8a=5
a=5/8cm.