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Published on: 29/08/2019
Areas of Parallelograms and Triangles
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Questions + Answers key
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1.
In the figure, BC = 2BE and area (\(\Delta\)ABC) = 60 cm2, then ar (\(\Delta\)AEC) is:
15 cm2
20 cm2
30 cm2
40 cm2
2.
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:
1:2
2:1
4:1
1:4
3.
In the given figure, AD is the median of \(\Delta\)ABC. The ratio of areas of \(\Delta\)ABD and \(\Delta\)ACD respectively is:
2:1
1:2
1:1
3:1
4.
In the figure, ABCD is a parallelogram of area 128 cm2 . If CF = 16 cm, the length of AD is
8 cm
4 cm
16 cm
10 cm
5.
In the figure, the area of parallelogram PQRS is:
\(PQ\times QB\)
\(QR\times QC\)
\(SR\times QC\)
\(PS\times SA\)
6.
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
4cm
1cm
2 cm
3 cm
7.
Area of a triangle is equal to
\(\frac { 1 }{ 2 } \times Base\times Corresponding\quad altitude\)
\(\frac { 1 }{ 4 } \times Base\times Corresponding\quad altitude\)
\(\frac { 1 }{ 3 } \times Base\times Corresponding\quad altitude\)
\( Base\times Corresponding\quad altitude\)
8.
Area of a parallelogram is equal to
\(\frac { 1 }{ 2 } \times Base\times Corresponding\quad altitude\)
\(\frac { 1 }{ 3 } \times Base\times Corresponding\quad altitude\)
\(\frac { 1 }{ 4 } \times Base\times Corresponding\quad altitude\)
\( Base\times Corresponding\quad altitude\)
9.
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 _______________
10.
Why we cannot construct a triangle of given sides as 5 cm, 5 cm and 10 cm?
1.
(c)
30 cm2
2.
A median of a triangle divides it into two triangles of equal area.
3.
A median of a triangle divides it with two triangles of equal areas.
4.
(a)
8 cm
5.
Area of parallelelogram=\(Base\times Corresponding\quad altitude\)
6.
(a)
4cm
7.
Theorem
8.
Theorem
9.
( )
The required ratio is 1:4.
10.
( )
As
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