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27 May 2021, 22:20
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B
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Difficulty:
55%
(hard)
Question Stats:
47% (01:50) correct
53%
(02:12)
wrong
based on 34
sessions
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In ABC triangle, two sides, AB and BC, are in the ratio 2:3. A bisector is drawn from the vertex B, crossing AC at point D. What is the ratio of the areas of ABD and BDC?
A. 4:9
B. 2:3
C. 15:25
D. 3:10
E. 8:25
A. 4:9
B. 2:3
C. 15:25
D. 3:10
E. 8:25
27 May 2021, 23:29
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Sherbeck
Not seen a question on GMAT testing Angle bisector theorem, but the theorem is:
Angle bisector theorem: If the angle bisector of angle B bisects side AC at a point D, the angle bisector theorem states that the ratio of the length of the line segment AD to the length of segment CD is equal to the ratio of the length of side AB to the length of side BC
So the ratio of AD:CD = AB:BC=2:3
The height of the triangles ABD and ACD is the same and the ratio of their areas = \(\frac{1}{2}*h*AD:\frac{1}{2}*h*CD=AD:CD=2:3\)
B
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