Area of a triangle is equal to

1

(a)

\(\frac { 1 }{ 2 } \times Base\times Corresponding\quad altitude\)

(b)

\(\frac { 1 }{ 4 } \times Base\times Corresponding\quad altitude\)

(c)

\(\frac { 1 }{ 3 } \times Base\times Corresponding\quad altitude\)

(d)

\( Base\times Corresponding\quad altitude\)

In the following figure, find the area of quad. ABCD.

1

(a)

7 square units

(b)

12 square units

(c)

6 square units

(d)

24 square units

If length of the diagonal of a square is 8 cm, then its area will be

1

(a)

64cm²

(b)

32cm²

(c)

16cm²

(d)

48cm²

If P, Q, Rand S are the midpoints of a rectangle of area 36 sq. cm, then PQRS is a parallelogram of area

1

(a)

24 cm²

(b)

18cm²

(c)

12cm²

(d)

9cm²

Which of the following figures lie on the same base and between the same parallels?

1

(a)

(i) (iv)

(b)

(ii),(iii)

(c)

(ii),(iv)

(d)

(i),(iii)

If a parallelogram and a triangle are on the same base and between the same parallels, then

1

(a)

area of the triangle =\(\frac { 1 }{ 2 } \) area of the parallelogram

(b)

area of the triangle = area of the parallelogram

(c)

area of the triangle =\(\frac { 1 }{ 3 } \) area of the parallelogram

(d)

area of the triangle =\(\frac { 1 }{ 4 } \) area of the parallelogram

In the figure, ABCD is a parallelogram. If area (AOD) = 12 cm² then area (ABCD) is:

1

(a)

3 cm²

(b)

24 cm²

(c)

48 cm²

(d)

36 cm²

In the figure below, D is mid-point of side BC of \(\Delta\) ABC. O is mid-point of AD. If area of \(\Delta\)AOB 8 cm² , then area of \(\Delta\)ABC is:

1

(a)

16 cm²

(b)

24 cm²

(c)

32 cm²

(d)

4 cm²

In \(\Delta\)ABC, E is the mid-point of median AD. Then the ratio of areas of \(\Delta\)BED to area of \(\Delta\)ABC is:

1

(a)

1:2

(b)

2:1

(c)

4:1

(d)

1:4

In the figure, BC = 2BE and area (\(\Delta\)ABC) = 60 cm², then ar (\(\Delta\)AEC) is:

1

(a)

15 cm²

(b)

20 cm²

(c)

30 cm²

(d)

40 cm²

Why we cannot construct a triangle of given sides as 5 cm, 5 cm and 10 cm?

1

(a)

As 

In \(\Delta\)ABC, E is the mid-point of median AD, then the ratio of area of \(\Delta\)BED to the area \(\Delta\)ABC is _______________

1

(a)

The required ratio is 1:4.

In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 cm and 5 cm. Find measure of AD.

2

(a)

In the given figure, ABC and DBC are triangles on the same base and between parallel lines I and m. If AB = 3 cm, BC = 5 cm, \(\angle A=90°\), find area of \(\Delta \) DBC.

2

(a)

In the figure, ABCD is a parallelogram. P is a point on AB produced and \(DN\bot AB\) . If AB = 8 cm and DN = 3 cm. Find the area of \(\Delta \) CPD.

2

(a)

ABCD is a rectangle and BD is one of its diagonals. If ar(\(\Delta\)ABD) = 8cm², find ar(\(\Delta\) BCD).

2

(a)

In given figure, ABCD is a parallelogram and BE\(\bot \)AD. If BE=14 cm and AD=8 cm, find the area of \(\Delta \)DBC.

2

(a)

In a triangle ABC, E is the midpoint of median AD. Show that ar(\(\Delta \) BED) = \(\frac { 1 }{ 4 } \) ar(\(\Delta \) ABC).

3

(a)

ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

 

4

(a)

In the given figure, AB 11DC. Show that ar(BDE) = ar(ACED).
     
 

4

(a)

In the figure, diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O. Show that ar (\(\Delta \) AOD)= ar(\(\Delta \) BOC).

4

(a)

Areas of triangles on the same bases and between the same parallels are equal in. Prove it.

4

(a)