CBSE Questions for Class 9 Maths Areas Of Parallelograms And Triangles Quiz 2 - MCQExams.com

For figures on same base and same parallels, the figures have to lie between same parallels
  • Yes
  • no
  • Only in exceptional cases
  • None of these
All triangles are equal in area.
  • True
  • False
All triangles have the same base and the same altitude. 
  • True
  • False
Triangles having the same base have equal area.
  • True
  • False
Name the common base on which both the figures, $$\triangle ABF$$ and parallelogram $$ABCD$$ lies (if any), on the basis of given diagram.
1949507_8898ae1c00ea4cdb86e22a27eb902d93.png
  • $$CD$$
  • $$AB$$
  • $$AD$$
  • $$BC$$
On the basis of given diagram and student's responses, choose the name of student whose answer is correct
1949494_e3584b2722484a2880ec2cd61a5b8efa.png
  • Student $$P$$
  • Student $$Q$$
  • Student $$R$$
  • Student $$S$$
In which of the following figures, you find two polygons on the same base and between the same parallels?
  • None of these
Area of any two parallelograms ia equal if they have common ____ and both are between same parallels.
  • diagonals
  • sides
  • base
  • none of these
On the basis of given figure, we can conclude that
1949485_45ff2f8cdfb541f9908953f80bb8325b.png
  • triangle $$DCF$$ and parallelogram $$ABCD$$ lies on same base
  • triangle $$DCF$$ and quadrilateral $$AECB$$ lies on same base
  • triangle $$ADE$$ and parallelogram $$ABCD$$ lies on same base
  • None of these
Observe the given diagram and write the common base on which figures lie and two parallel lines between which figures lie.
1949504_ae48ff53c1524837913bdf0cb7c70734.png
  • Base $$ QR$$ and Parallels $$= QR$$ & $$PS$$
  • Base $$ PA$$ and Parallels $$= QR$$ & $$AB$$
  • Base $$ BS$$ and Parallels $$= QR$$ & $$PA$$
  • Base $$ QR$$ and Parallels $$= AR$$ & $$PQ$$
Observe the given diagram and choose the correct answer:
1949521_6301590d5bf04e9db002a8680ea0de97.jpg
  • $$ar(\triangle ABC) = ar(\triangle ADB) + ar(\triangle ACD)$$
  • $$ar(\triangle ABC) = ar(\triangle ADB) - ar(\triangle ACD)$$
  • $$ar(\triangle ABC) = ar(\triangle ADB) + 2 ar(\triangle ACD)$$
  • $$ar(\triangle ABC) = ar(\triangle ADB) \times ar(\triangle ACD)$$
Name the two parallel lines in-between which both the figures, $$\triangle ABF$$ and parallelogram $$ABCD$$ lies (if any), on the basis of given diagram.
1949509_9d68b98beeee4398902f88e20db31425.png
  • $$AB$$ and $$CD$$
  • $$AF$$ and $$BC$$
  • $$AD$$ and $$BF$$
  • None of these
In given diagram, there are two different parallelograms with same base $$AB$$ and both are between same parallel lines. If the $$AB = 6$$ $$cm$$ and height of the parallelogram $$ABCD$$ is $$5$$ $$cm$$, then what will be the area of parallelogram $$ABEF$$
1949716_0daadfaa982244e5abeacb76a2a92139.jpg
  • $$25$$ $$cm^2$$
  • $$30$$ $$cm^2$$
  • $$36$$ $$cm^2$$
  • None of these
In adjoining figure, $$\triangle ABD$$ is a right angled at $$A$$ in which $$AB = 8$$ $$cm$$ and $$AD = 7$$ $$cm$$. If $$CD$$ is parallel to $$AB$$, then what will be the area of $$\triangle ABC$$?
1949739_30f89e46694b484781be9765aec1d0f6.png
  • $$24$$ $$cm^2$$
  • $$26$$ $$cm^2$$
  • $$28$$ $$cm^2$$
  • $$30$$ $$cm^2$$
It is given that $$E, F, G$$ and $$H$$ are the mid points of the sides of the parallelogram $$ABCD$$. Then can we conclude that $$ar(\triangle EFH) = ar(\triangle FGH)$$?
1949727_32a65837ab3d4b80b76639645dad3edb.png
  • Yes
  • No
  • Can't say anything
  • None of these
If two triangles are on the same base and are between same parallel lines they have double each other's area
  • True
  • False
Is $$\triangle ADB$$ and $$\triangle ABC$$ equal in area?
1949729_1fc0ece76af0421e84dab10e72da8ef4.png
  • True
  • False
If two triangles are on same ____ and are between same parallel lines they have equal area.
  • base
  • height
  • length
  • none of these
A parallelogram and a square are on equal base and between the same parallel lines. Then the ratio of their areas is ____.
  • $$1:1$$
  • $$2:1$$
  • $$3:1$$
  • $$2:3$$
In given figure, if $$P$$ be a mid point of $$AB$$ then
1949731_bfe0450819a641229f5cb75aeaebb569.png
  • $$ar(\triangle APD) = ar(\triangle BPC)$$
  • $$ar(\triangle APD) = 2 \times ar(\triangle BPC)$$
  • $$ar(\triangle APD) \neq ar(\triangle BPC)$$
  • None of these
0:0:1


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