Asked by nisha
Found the area of TrapeziumABCD whose parallel side are 14cm and 24cm.its non parallel sides are 8cm and 6cm
Answers
Answered by
Reiny
Make a sketch, dropping perpendiculars from the shorter parallel line to the longer
Let that height be h
You now have a rectangle 24 by h, and two triangles.
One has hypotenuse 6, height of h, let its base be x
The other has hypotenuse 8, height of h, and base of 10-x
from the first triangle:
x^2 + h^2 = 36
from the 2nd:
(10-x)^2 + h^2 = 64
subtract them:
(10-x)^2 - x^2 = 28
100 - 20x + x^2 - x^2 = 28
20x = 72
x = 18/5
then (18/5)^2 + h^2 = 36
h^2 = 36-324/25 = 576/25
h = ± 24/5 , ignoring the negative height
so now you have x = 18/5 and y = 24/5
so find the area of the two triangles, and add that to the area of the rectangle
Let that height be h
You now have a rectangle 24 by h, and two triangles.
One has hypotenuse 6, height of h, let its base be x
The other has hypotenuse 8, height of h, and base of 10-x
from the first triangle:
x^2 + h^2 = 36
from the 2nd:
(10-x)^2 + h^2 = 64
subtract them:
(10-x)^2 - x^2 = 28
100 - 20x + x^2 - x^2 = 28
20x = 72
x = 18/5
then (18/5)^2 + h^2 = 36
h^2 = 36-324/25 = 576/25
h = ± 24/5 , ignoring the negative height
so now you have x = 18/5 and y = 24/5
so find the area of the two triangles, and add that to the area of the rectangle
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.