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03 Apr 2020, 08:33
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:57) correct
39%
(02:53)
wrong
based on 76
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In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is
A. 12√3
B. 15√3
C. 18√3
D. 24√3
E. 32√3
Are You Up For the Challenge: 700 Level Questions
A. 12√3
B. 15√3
C. 18√3
D. 24√3
E. 32√3
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 08:50
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Bunuel
Area of parallelogram = \(72\)
--> \(9*h = 72\)
--> \(h = 8\)
Triangle ADP is right angled
--> \(AD^2 = AP^2 + PD^2\)
--> \(16^2 = 8^2 + PD^2\)
--> \(PD^2 = 16^2 - 8^2 = (16 - 8)(16 + 8) = 8*24 = 8*8*3\)
--> \(PD = \sqrt{8*8*3} = 8\sqrt{3}\)
Area of triangle APD = \(\frac{1}{2}*PD*AP = \frac{1}{2}*8\sqrt{3}*8 = 32\sqrt{3}\)
Option E
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4.JPG [ 15.14 KiB | Viewed 3651 times ]
03 Apr 2020, 14:16
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Area of the parallelogram ABCD = (base)(height) = (CD)(AP) = 72 sq.cm.
CD)(AP) = 72
or,9(AP) = 72
=> AP = 8
DP =√(AD^2-AP^2) = √(16^2-8^2) = 8√3
Area of triangle APD =1/2 * (AP) * (AD) =32√3
correct answer is E
CD)(AP) = 72
or,9(AP) = 72
=> AP = 8
DP =√(AD^2-AP^2) = √(16^2-8^2) = 8√3
Area of triangle APD =1/2 * (AP) * (AD) =32√3
correct answer is E
Attachments
area.jpg [ 1.3 MiB | Viewed 3596 times ]
