Area of a parallelogram is equal to

1

(a)

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

(b)

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

(c)

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

(d)

\( Base\times Corresponding\quad altitude\)

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)

\( Base\times Corresponding\quad altitude\)

The areas of a parallelogram and a triangle are equal and they lie on the same base. If the altitude of the parallelogram is 2 cm, then the altitude of triangle is

1

(a)

4cm

(b)

1cm

(c)

2 cm

(d)

3 cm

In the figure, the area of parallelogram PQRS is:

1

(a)

\(PQ\times QB\)

(b)

\(QR\times QC\)

(c)

\(SR\times QC\)

(d)

\(PS\times SA\)

In the figure, ABCD is a parallelogram of area 128 cm² . If CF = 16 cm, the length of AD is

1

(a)

8 cm

(b)

4 cm

(c)

16 cm

(d)

10 cm

In the given figure, AD is the median of \(\Delta\)ABC. The ratio of areas of \(\Delta\)ABD and \(\Delta\)ACD respectively is:

1

(a)

2:1

(b)

1:2

(c)

1:1

(d)

3:1

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.