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11 Jul 2019, 00:29
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D
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90% (00:34) correct
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In the figure to the right, what is the area of the triangle?
(1) c^2 = 6^2 + 8^2
(2) x = 90
Attachment:
2019-07-11_1113.png [ 14.76 KiB | Viewed 1443 times ]
11 Jul 2019, 00:35
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Bunuel
S1:
Since for right-angled triangles,
\(c^2 = 6^2 + 8^2\)
=> x =90
\(Area = \frac{6*8}{2} = 24\)
S2:
x=90
\(Area = \frac{6*8}{2} = 24\)
IMO D
firas92
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11 Jul 2019, 00:38
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In a triangle with side lenghts a, b & c
If c^2<a^2+b^2, the triangle is an acute triangle
If c^2=a^2+b^2, the triangle is a right angled triangle
If c^2>a^2+b^2, the triangle is an obtuse triangle
From (1) we can deduce that the given triangle is a right angled
In a right angled triangle, one of the two perpendicular sides can be considered the base while the other can be considered the height and so
Area = 1/2*6*8 = 24
Sufficient
From (2), we are directly given that the triangle is a right angled triangle
Again Area=1/2*6*8=24
Sufficient
Answer is (D)
If c^2<a^2+b^2, the triangle is an acute triangle
If c^2=a^2+b^2, the triangle is a right angled triangle
If c^2>a^2+b^2, the triangle is an obtuse triangle
From (1) we can deduce that the given triangle is a right angled
In a right angled triangle, one of the two perpendicular sides can be considered the base while the other can be considered the height and so
Area = 1/2*6*8 = 24
Sufficient
From (2), we are directly given that the triangle is a right angled triangle
Again Area=1/2*6*8=24
Sufficient
Answer is (D)
