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22 Aug 2019, 04:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
55% (01:47) correct
45%
(01:42)
wrong
based on 184
sessions
History
Date
Time
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Not Attempted Yet
If AB is a diameter of the circle above, what is the area of the circle?
(1) BD = 3
(2) \(AC = \frac{2\sqrt{3}}{3}*BD\)
Attachment:
image486.png [ 14.57 KiB | Viewed 4643 times ]
If AB is a diameter of circle,
—> then triangle ABD and triangle ACB are right-angled triangles.
What is the area of circle?—(\(П*r^{2}\))
Statement1: BD = 3
AD= 2
-–> \(AB^{2}\)= \(BD^{2}+AD^{2}\)=9+4=13
—> AB= \(13^{1/2}\)
The radius of the circle =\(\frac{AB}{2}\)= r
Now, We can find the area of the circle.
Sufficient
Statement2 AC=(\(\frac{2√3}{3}\))∗BD
Create the two equations by using Pythagorean theorem :
\(AD^{2} + BD^{2}\)= \(AB^{2}\)
\(AC^{2} + BC^{2}\) = \(AB^{2}\)
4+\(BD^{2}\)=1 +\(( \frac{4}{3})*BD^{2}\)
\((\frac{1}{3})*BD^{2}\)= 3
—> BD= 3 —> the same thing as Statement1. Now we can find the radius of the circle —> the area of the circle
Sufficient
The answer is D.
Posted from my mobile device
—> then triangle ABD and triangle ACB are right-angled triangles.
What is the area of circle?—(\(П*r^{2}\))
Statement1: BD = 3
AD= 2
-–> \(AB^{2}\)= \(BD^{2}+AD^{2}\)=9+4=13
—> AB= \(13^{1/2}\)
The radius of the circle =\(\frac{AB}{2}\)= r
Now, We can find the area of the circle.
Sufficient
Statement2 AC=(\(\frac{2√3}{3}\))∗BD
Create the two equations by using Pythagorean theorem :
\(AD^{2} + BD^{2}\)= \(AB^{2}\)
\(AC^{2} + BC^{2}\) = \(AB^{2}\)
4+\(BD^{2}\)=1 +\(( \frac{4}{3})*BD^{2}\)
\((\frac{1}{3})*BD^{2}\)= 3
—> BD= 3 —> the same thing as Statement1. Now we can find the radius of the circle —> the area of the circle
Sufficient
The answer is D.
Posted from my mobile device
firas92
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22 Aug 2019, 04:54
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Triangles ADB and ABC have vertices on the circumference and diameter as base.
Therefore ADB and ABC are right triangles
(1) BD = 3
From right triangle ABD,
\(AD^2+BD^2=AB^2\)
\(AB^2=2^2+3^2\)
From this we can find \(AB\) and thus the area
(1) is sufficient
(2) \(AC=\frac{2√3}{3}∗BD\)
From right triangle ABC,
\(AB^2=AC^2+BC^2=\frac{4}{3}BD^2+BC^2\)
From right triangle ADB,
\(AB^2=AD^2+BD^2\)
So, \(\frac{4}{3}BD^2+BC^2 = AD^2+BD^2\)
\(BD^2 = 3*(2+1)\)
\(BD=3\)
Once again we can solve as we did in Statement (1)
(2) is sufficient
Answer is (D)
Therefore ADB and ABC are right triangles
(1) BD = 3
From right triangle ABD,
\(AD^2+BD^2=AB^2\)
\(AB^2=2^2+3^2\)
From this we can find \(AB\) and thus the area
(1) is sufficient
(2) \(AC=\frac{2√3}{3}∗BD\)
From right triangle ABC,
\(AB^2=AC^2+BC^2=\frac{4}{3}BD^2+BC^2\)
From right triangle ADB,
\(AB^2=AD^2+BD^2\)
So, \(\frac{4}{3}BD^2+BC^2 = AD^2+BD^2\)
\(BD^2 = 3*(2+1)\)
\(BD=3\)
Once again we can solve as we did in Statement (1)
(2) is sufficient
Answer is (D)
