If AD =10\sqrt{3} and ADC is a right angle, what is the area of triangle ABC?
(1) AC = 20
(2)angle BAD=30^{\circ}
This was asked in the Kaplan free test, but I think that the answer is wrong.
How I think the answer could be solved:
...
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(1) AC = 20
(2)angle BAD=30^{\circ}
This was asked in the Kaplan free test, but I think that the answer is wrong.
How I think the answer could be solved:
[Reveal] Spoiler:
Since\triangle ABD is similar to \triangleADC
\frac{AB}{AC}=\frac{AD}{DC}=\frac{BD}{AD} --------------(a)
1) We're given that AC=20
So by Pythagoras Theorem, CD = 10
and from (a), we can substitute values of 10 and10\sqrt{3} in\frac{AD}{DC}=\frac{BD}{AD} ,
thus getting the value of BD
\frac{AB}{AC}=\frac{AD}{DC}=\frac{BD}{AD} --------------(a)
1) We're given that AC=20
So by Pythagoras Theorem, CD = 10
and from (a), we can substitute values of 10 and10\sqrt{3} in\frac{AD}{DC}=\frac{BD}{AD} ,
thus getting the value of BD
...
Attachments
Untitled.png [ 30.27 KiB | Viewed 16 times ]
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