Q1: What is a quadrilateral? Mention 6 types of quadrilaterals.
Sol: A quadrilateral is a two-dimensional closed polygon having four sides and four vertices. The sum of its interior angles is 360°. Common types of quadrilaterals include:
- Rectangle
- Square
- Parallelogram
- Rhombus
- Trapezium
- Kite
Q2: Identify the type of quadrilaterals:
(i) The quadrilateral formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are perpendicular.
(ii) The quadrilateral formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are congruent.
Sol: (i) By Varignon’s theorem, the quadrilateral formed by joining the midpoints of the sides of any quadrilateral is a parallelogram whose adjacent sides are parallel to the diagonals of the original quadrilateral and have lengths equal to half those diagonals. If the original diagonals are perpendicular, the two adjacent sides of the midpoint parallelogram are perpendicular. A parallelogram with adjacent sides perpendicular is a rectangle. Hence the figure is a rectangle.
(ii) If the diagonals of the original quadrilateral are congruent (equal in length), then the corresponding adjacent sides of the midpoint parallelogram are equal in length (each equals half a diagonal). A parallelogram whose adjacent sides are equal is a rhombus. Hence the figure is a rhombus.
Q3: In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.
Sol: Let θ = 25° be the acute angle made by a diagonal with a side of the rectangle. The two diagonals of a rectangle are symmetric about the centre and make equal angles θ and -θ with the same side. Therefore the acute angle between the two diagonals is 2θ = 2 × 25° = 50°. Hence the acute angle between the diagonals is 50°.
Q4: Prove that the angle bisectors of a parallelogram form a rectangle.
Sol: Let LMNO be a parallelogram. Let the bisectors of ∠L, ∠M, ∠N and ∠O meet pairwise to form quadrilateral PQRS (so each vertex of PQRS is the intersection of bisectors of two consecutive angles of LMNO).
In a parallelogram, consecutive interior angles are supplementary, so ∠L + ∠M = 180°.
When ∠L and ∠M are bisected, their half-angles satisfy (∠L)/2 + (∠M)/2 = 90°.
At the intersection point Q of the bisectors of ∠L and ∠M, the angle ∠PQR equals (∠L)/2 + (∠M)/2 = 90°.
Similarly, each interior angle of PQRS is formed by the sum of half of two consecutive angles of the parallelogram, and each such sum equals 90°.
Hence all four angles of PQRS are right angles, so PQRS is a rectangle.
Q5: Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.
Sol: Let one angle be x and its adjacent angle be 2x. Opposite angles of a parallelogram are equal, so the four angles are x, 2x, x and 2x. Their sum is 360°:
x + 2x + x + 2x = 360°
6x = 360° ⇒ x = 60°.
Thus the angles are 60°, 120°, 60° and 120°.
Q6: The diagonals of which quadrilateral are equal and bisect each other at 90°?
Sol: A square. In a square the diagonals are equal in length, they bisect each other, and they intersect at right angles. A rhombus has diagonals that bisect at 90° but they are not equal unless it is a square; a rectangle has equal diagonals but they are not perpendicular unless it is a square. Therefore the required quadrilateral is a square.
Q7: Find all the angles of a parallelogram if one angle is 80°.
Sol: In a parallelogram opposite angles are equal. If one angle is 80°, the opposite angle is also 80°. Consecutive angles are supplementary, so each adjacent angle is 180° – 80° = 100°. Hence the four angles are 80°, 100°, 80° and 100°.
Q8: Is it possible to draw a quadrilateral whose all angles are obtuse angles?
Sol: No. An obtuse angle is greater than 90°. If all four angles of a quadrilateral were obtuse, their sum would be greater than 4 × 90° = 360°, but the sum of interior angles of any quadrilateral is exactly 360°. Therefore it is not possible to have all four angles obtuse.
Q9: In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°
Sol: In a trapezium with AB ∥ CD, each pair of interior angles on the same leg are supplementary. Thus ∠A + ∠D = 180° and ∠B + ∠C = 180°.
Given ∠A = 55°, so ∠D = 180° – 55° = 125°.
Given ∠B = 70°, so ∠C = 180° – 70° = 110°.
Therefore ∠C = 110° and ∠D = 125°.
Q10: Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.
Sol: Let the common factor be x. Then the angles are 2x, 5x, 4x and x. Their sum is 360°:
2x + 5x + 4x + x = 360° ⇒ 12x = 360° ⇒ x = 30°.
Therefore the angles are: 2x = 60°, 5x = 150°, 4x = 120°, and x = 30°.