a rectangular lot measure is (5w+3z)cm by (5w-3z)cm. If (2w)is added to both dimensions, how is the area affected? Explain.

Question

anonymous · Answer

using the formula (a+b)(a-b) = a² - b²
original area = (5w + 3z)(5w - 3z) = (5w)² - (3z)² = 25w² - 9z²

anonymous · Answer

where is the (2w)..?..and how is the area affected..?

anonymous · Answer

on adding 2w to both we get
(7w + 3z)(7w - 3z) = (7w)² - (3z)² = 49w² - 9z²

anonymous · Answer

how is the area affected?

id21 · Answer

The area of 1st lot is $$(5w+3z)(5w-3z)=25w^2-9z^2$$
if you add 2w for each dimension, you get:$$(5w+3z+2w)(5w-3z+2w)=(7w+3z)(7w-3z)=49w^2-9z^2$$

Are will be greater by this ratio: $$\frac{49w^2-9z^2}{25w^2-z^2}$$

anonymous · Answer

new area is more by (49w² - 9z²) - (25w² - 9z²) = 49w² - 25w² - 9z² + 9z² = 24w²

anonymous · Answer

new are will be 24w² more than the original area

id21 · Answer

Harkirat is right!

anonymous · Answer

@TheChariz
when we r answering, hv some patience.....☺

anonymous · Answer

ok..thank you very much..
i will try to be patience..next time..