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The median of a triangle divides it into two

(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles

In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels?

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is :

(A) a rectangle of area 24 cm2 
(B) a square of area 25 cm2
(C) a trapezium of area 24 cm2
(D) a rhombus of area 24 cm2

In Fig. 9.4, the area of parallelogram ABCD is :


 

(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL

In Fig. 9.5, if parallelogram ABCD and rectangle ABEM are of equal area, then :

(A) Perimeter of ABCD = Perimeter of ABEM
(B) Perimeter of ABCD < Perimeter of ABEM
(C) Perimeter of ABCD > Perimeter of ABEM
(D) Perimeter of ABCD = (Perimeter of ABEM)

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

(A)  \frac{1}{2}ar(ABC)
(B)  \frac{1}{3}ar(ABC)
(C)  \frac{1}{4}ar(ABC)

(D)  ar(ABC)

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

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)

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

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is


 

(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)

Write true or false and justify your answer.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.

Write true or false and justify your answer.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ.
If PS = 5 cm,then ar (PAS) = 30 cm2.

Write true or false and justify your answer.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of DASR = 90 cm2.

Write true or false and justify your answer.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = \frac{1}{4} ar ( ABC).

Write true or false and justify your answer.
In Fig., ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = \frac{1}{2} ar ( EFGD).

In Fig., PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA \parallel QB \parallel RC. Prove that ar (PQE) = ar (CFD).

X and Y are points on the side LN of the triangle LMN such that LX=XY=YN. Through X, a line is drawn parallel to LM to meet MN at Z (see Fig.). Prove that ar (LZY) = ar (MZYX)

 

The area of the parallelogram ABCD is 90 cm^{2} (see Fig.9.13). Find

(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)

In \triangle ABC, D is the mid-point of AB and P is any point on BC. If CQ\parallel PD meets AB in Q (Fig.), then prove that ar (BPQ)=\frac{1}{2}ar(ABC).

ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig.), prove that ar (AER) = ar(AFR)

O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO).

ABCD  is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F.
If ar (DFB) = 3cm^{2}, find the area of the parallelogram ABCD.

In trapezium ABCD, AB\parallel DC and L is the mid-point of BC. Through L, a line PQ\parallel AD has been drawn which meets AB in P and DC produced in Q  (Fig.). Prove that ar (ABCD) = ar (APQD)  
 


 

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

 

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