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19 Aug 2017, 10:06
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87% (01:01) correct
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If the area of the triangle in the figure above is 100, what is the length of side AB?
(A) 10√3
(B) 10√5
(C) 20
(D) 24
(E) 25
Attachment:
2017-08-19_2104.png [ 11.52 KiB | Viewed 4000 times ]
19 Aug 2017, 10:14
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Since we have BC = 20 and we know that the triangle ABC is right angled at B,
We are also given area of the triangle = 100
Area = \(\frac{1}{2} * Base * Height = 100\)
BC is the Height and let x be the Base(AC)
\(\frac{1}{2}*x*20 = 100\)
\(x = \frac{200}{20}\) => \(x = 10\)
In a right angled triangle, using Pythagoras theorem
\(AB^2 = AC^2 + BC^2\)
\(AB^2 = 400 + 100 = 500\)
\(AB = \sqrt{500} = \sqrt{5*10*10}\)
Therefore the length of AB = \(10\sqrt{5}\) (Option B)
19 Aug 2017, 10:18
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Ans is B
0.5 AC x BC = 100 , where AC =20
BC= 10
using pythogoras theorm AC^2+BC^2 =AB^2
AB= √500
= 10√5
hence B is correct
0.5 AC x BC = 100 , where AC =20
BC= 10
using pythogoras theorm AC^2+BC^2 =AB^2
AB= √500
= 10√5
hence B is correct
