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Quadrilaterals: CBSE Class 9 Mathematics practice questions

Practise 50 questions on Quadrilaterals, free, with an explanation for each answer.

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Sample questions

In parallelogram ABCD, ∠A=65∘\angle A = 65^\circ. Find ∠B\angle B.

A parallelogram ABCD with one angle marked
  1. 65∘65^\circ
  2. 295∘295^\circ
  3. 115∘115^\circ
  4. 25∘25^\circ
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C. 115∘115^\circ

AD∥BCAD \parallel BC with transversal AB, so ∠A+∠B=180∘\angle A + \angle B = 180^\circ (interior angles on the same side). Hence ∠B=180∘−65∘=115∘\angle B = 180^\circ - 65^\circ = 115^\circ. Traps: 65∘65^\circ is the opposite angle, not the adjacent one; 295∘295^\circ comes from using 360∘−65∘360^\circ - 65^\circ; 25∘25^\circ uses 90∘90^\circ instead of 180∘180^\circ.

In the quadrilateral shown, the two diagonals (solid lines) do not meet. They would meet only if one of them were extended. What can you say about this quadrilateral?

A four-sided figure with its two diagonals drawn
  1. It is convex, because the diagonals are drawn inside the figure
  2. It is a parallelogram, because the diagonals are not equal
  3. It is non-convex: one internal angle is more than 180∘180^\circ
  4. It is self-intersecting, because the diagonals do not cross
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C. It is non-convex: one internal angle is more than 180∘180^\circ

The diagonals of a convex quadrilateral always intersect; if they do not, the quadrilateral is non-convex and has a reflex angle (>180∘>180^\circ) at the dent. The figure has no crossing sides, so it is not self-intersecting, and a parallelogram is always convex (its diagonals bisect each other).

In trapezium ABCD, AD∥BCAD \parallel BC, AD=3AD = 3 cm and BC=5BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the area of AEFD to the area of EBCF.

A trapezium ABCD split by EF into a green part and an orange part
  1. 3:53 : 5
  2. 1:11 : 1
  3. 9:79 : 7
  4. 7:97 : 9
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D. 7:97 : 9

EF=3+52=4EF = \dfrac{3 + 5}{2} = 4 cm. EF cuts the height hh in half, so both parts have height h2\dfrac h2. Areas: AEFD =3+42⋅h2= \dfrac{3+4}{2}\cdot\dfrac h2 and EBCF =4+52⋅h2= \dfrac{4+5}{2}\cdot\dfrac h2, giving 7:97 : 9. Traps: 3:53 : 5 compares the bases only; 1:11 : 1 assumes EF splits the area equally; 9:79 : 7 is reversed.

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