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If AB=6, DC=8, what is the area of the trapezoid ABCD? A. â

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If AB=6, DC=8, what is the area of the trapezoid ABCD? A. â

by Max@Math Revolution » Mon Mar 07, 2016 3:56 pm
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If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7


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by Dario@VinciaPrep » Tue Mar 08, 2016 1:06 am
Look at triangle ABO.
AB is 6, AO and BO are 4. It is an isosceles triangle.
We can find the height by using Pythagoras on half of it.
4^2 - 3^2 = 16 - 9 = 7 = BH^2
Thus BH is the square root of 7.
To find the area, we have to calculate
(AB+CD)*BH/2=(6+8)*sqrt(7)/2=7*sqrt(7)

The answer is (C)
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by Max@Math Revolution » Fri Mar 11, 2016 4:53 am
If AB=6, DC=8, what is the area of the trapezoid ABCD?

A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7

->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.
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by jain2016 » Fri Mar 11, 2016 8:01 am
HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.
[/quote]

Hi,

How come HC= 1 and OH= 3?

Please explain.

Thanks,

SJ
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by Shazi1711 » Mon Mar 14, 2016 6:51 am
Hello,

Can anyone please provide a detailed step by step solution to this problem ?

Thanks

SD
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by Max@Math Revolution » Mon Mar 14, 2016 5:11 pm
Shazi1711 wrote:Hello,

Can anyone please provide a detailed step by step solution to this problem ?

Thanks

SD

-> (8-6)/2=1
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