A median of a triangle divides it into two triangles of
The area of quadrilateral ABCD is:
Area of an equilateral triangle is equal to
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. If area (ΔAOD) = 37sq cm, then area (ΔBOC) =
ABCD is a parallelogram whose diagonals intersect at O. A line through O intersects AB at P and DC at Q. Then
The area of a right triangle is 30 sq cm. If the base is 5 cm, then the hypotenuse must be
AE is a median to side BC of triangle ABC. If area(ΔABC) = 24 cm, then area(ΔABE) =
D is the mid point of side BC of ABC and E is the mid point of BD. If O is the mid point of AE, then
In the figure, ∠PQR = 90°, PS = RS, QP = 12 cm and QS = 6.5 cm. The area of ΔPQR is
In the given figure, if parallelogram ABCD and rectangle ABEF are of equal area then:
In ΔABC, if L and M are points on AB and AC respectively such that LM BC then ar (LMC) is equal to
In a parallelogram ABCD, P is a point in interior of parallelogram ABCD. If ar (||gm ABCD) = 18 cm2 then [arΔAPD) + ar(ΔCPB)] is :
Diagonals AC and BD of a trapezium ABCD with AB ll DC intersect each other at O. If area (ΔAOD)=37sq cm, then area (ΔBOC) = ?
In ΔPQR, if D and E are points on PQ and PR respectively such that DE || QR, then ar (PQE) is equal to
ABC is a triangle in which D is the mid point of BC and E is the mid point of AD. Then area (ΔBED) is equal to
In a parallelogram ABCD, E and F are any two points on the sides AB and BC respectively. If ar (ΔDCE) is 12 cm2, then ar(ΔADF) is
AD is one of the medians of a Δ ABC. X is any point on AD. Then, the area of ΔABX is equal to
ABC and BDE are two equilateral triangles such that D is the mid point of BC. AE intersects BC in F. Then ar (BDE) is equal to
In figure P, Q are points on sides AB, AC respectively of ΔABC such that ar (BCQ) = ar (BCP). Then,
In ΔABC, E is the mid-point of median AD. Area(ΔBED) =
AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal area.
∴ Area (ΔABD) = Area (ΔACD)
⇒Area (ΔABD ) = (1/2) area (Δ ABC) ------------(1)
In ΔABD, E is the mid-point of AD.
Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
Area (ΔBED) = (1/2)Area (ΔABD)
Area (ΔBED) = (1/2 ) x(1/2) Area (ΔABC) [From (1)]
∴ Area (ΔBED) = (1/4)Area (ΔABC).