ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 1/4 ar ( ABC).
If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP).
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is
(A) 1 : 3 (B) 1 : 2 (C) 3 : 1 (D) 1 : 4
What do you think the famous cartoonist R.K.
Laxman is trying to convey in this cartoon?
How does it relate to the 2016 law that you
read about on page 103?
ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
(A) is a rectangle (B) is always a rhombus
(C) is a parallelogram (D) need not be any of (A), (B) or (C)
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(A) 1 : 2 (B) 1 : 1 (C) 2 : 1 (D) 3 : 1
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