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In the figure, ABCD is a rectangle. The area of quadrilatera
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17 Mar 2019, 12:39
17 Mar 2019, 12:39
Expert Reply
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85% (02:07) correct
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In the figure, ABCD is a rectangle. The area of quadrilateral EBFD is one-half the area of the rectangle ABCD. Which one of the following is the value of AD ?
(A) 5
(B) 6
(C) 7
(D) 12
(E) 15
Re: In the figure, ABCD is a rectangle. The area of quadrilatera
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20 Mar 2019, 09:31
20 Mar 2019, 09:31
1
Expert Reply
In the triangle EBD, the base is 3 and the height is AB. It's area, then, is: \(\frac{3AB}{2}\)
Similarly, in the triangle BDF, the base is 4 and the height is AB. So it's area is \(\frac{4AB}{2}\)
Both of these together, \(\frac{3AB}{2}+\frac{4AB}{2}=\frac{7AB}{2}\) form the area of quadrilateral EBFD. This is one half of the area of the rectangle ABCD.
The area of ABCD is AB*AD. So:
\(\frac{7AB}{2} = \frac{AB*AD}{2}\)
Multiply both sides by 2:
\(7AB = AB*AD\)
Divide both sides by AB:
\(7 = AD\)
Similarly, in the triangle BDF, the base is 4 and the height is AB. So it's area is \(\frac{4AB}{2}\)
Both of these together, \(\frac{3AB}{2}+\frac{4AB}{2}=\frac{7AB}{2}\) form the area of quadrilateral EBFD. This is one half of the area of the rectangle ABCD.
The area of ABCD is AB*AD. So:
\(\frac{7AB}{2} = \frac{AB*AD}{2}\)
Multiply both sides by 2:
\(7AB = AB*AD\)
Divide both sides by AB:
\(7 = AD\)
Re: In the figure, ABCD is a rectangle. The area of quadrilatera
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13 Aug 2022, 05:16
13 Aug 2022, 05:16
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