If the dimensions of a 20 cm x 40 cm rectangle are each reduced 20%, by what percent is the area of the 20cm x 40 cm rectangle reduced? Please explain how you found your answer.
Answers (3)
a = 20cm
b = 40cm
The area p of this rectangle equals
p = a * b = 800cm²
Reducing the lengths of the sides by 20% yields:
a' = a - a * 0.2 = 20 - 4 = 16cm
b' = b - b * 0.2 = 40 - 8 = 32cm
The area p' of this reduced rectangle equals
p' = a' * b' = 16 * 32 = 512cm²
The area has been reduced by 800cm² - 512cm² = 288cm².
And 288 are x % of 800:
800 - 800 * x = 512 ;; TIMES (-1), PLUS(800)
800 * x = -512 + 800
x = 288 / 800
x = 0.36 = 36%.
The area of the rectangle with reduced sides is 36% less than the area of the original.
Oh wait now I understand your question. So 20 × 20% = 4 and 40 x 20% = 8 So the answer is 20 × 4 × 8 = 6.4.
So here are the steps that I did:
Step #1: 20 × 20% = 4
Step #2: 40 × 20% = 8
Step #3: 20 × 4 × 8 = 6.4