trapezoid area - increase b1 and b2 - how effect area?

[jsvlad]

q24 answer 1.
im not sure how this is? should i plug in numbers?


thanks!

 

[soroban]

Hello jsvlad!

24. \(\displaystyle x\) is the long base of an isosceles trapezoid.
\(\displaystyle y\) is the short base, and \(\displaystyle h\) is the altitude.
If \(\displaystyle x\) is multiplied by 2 and \(\displaystyle y\) is multiplied by \(\displaystyle \frac{1}{2}\)
by how much will the trapezoid's area change?

. . \(\displaystyle \text{(1) It will increase by }\frac{h}{2}\left(x - \frac{y}{2}\right)\)

. . \(\displaystyle \text{(2) It will decrease by }\frac{h}{2}(x+y)\)

. . \(\displaystyle \text{(3) It will not change.}\)

. . \(\displaystyle \text{(4) Impossible to determine from the data.}\)

The original trapezoid has bases \(\displaystyle x\) and \(\displaystyle y\), and altitude \(\displaystyle h.\)
. . Its area is: .\(\displaystyle A_1 \;=\;\frac{h}{2}(x+y)\)

The new trapezoid has bases \(\displaystyle 2x\) and \(\displaystyle \frac{y}{2}\), and altitude \(\displaystyle h.\)
. Its area is: .\(\displaystyle A_2 \;=\;\frac{h}{2}\left(2x + \frac{y}{2}\right) \)


The change of area is:

. . \(\displaystyle \Delta A \;=\;A_2 - A_1\)

. . . .. . \(\displaystyle =\;\frac{h}{2}\left(2x+\frac{1}{2}y\right) - \frac{h}{2}(x + y)\)

.. . . . . \(\displaystyle =\;\frac{h}{2}\left(2x + \frac{y}{2} - x - y\right)\)

.. . . . . \(\displaystyle =\;\frac{h}{2}\left(x - \frac{y}{2}\right) \:\hdots\:\text{answer (1)}\)