Areas of Parallelogram and Triangles Class 9 MCQs Questions with Answers

1. If ABCD and EFGH are two parallelograms between same parallel lines and on the same base, then:

2. A median of a triangle divides it into two

3. In a triangle ABC, E is the mid-point of median AD. Then

4. The median of a triangle divides it into two

5. If D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Then:

6. If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,

7. PQRS is a parallelogram. If X and Y are the mid-points of PQ and SR and diagonal SQ is joined, then ar(XQRY) : ar(ΔQSR) is

8. If E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively, then ar (EFGH) is equal to:

9. If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,

10. 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 the parallelogram will be:

11. The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is

12. If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:

13. ABCD is a trapezium in which AB || DC. If ar(ΔABD) = 24 cm² and AB = 8 cm, then height of ΔABC is

14. If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:

15. A, B, C and D are the mid-points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm², then ar(ABCD) is

16. D and E are the mid-points of BC and AD respectively. If ar(ΔABC) = 12 cm², then ar(ΔBDE) is

17. ABCD is a quadrilateral whose diagonal AC divides it in two parts of equal area, then ABCD is a

18. In quadrilateral PQRS, M is the mid-point of PR. If ar(SMQR) = 18 cm², then ar(PQMS) is

19. If ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 10 cm, AE = 6 cm and CF = 5 cm, then AD is equal to:

20. If Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Then ABCD is a: