CBSE Class 9 Mathematics Quadrilaterals MCQs Set 07

Answer: (a) 70°

Explanation:
1. The angles of a quadrilateral add up to 360°.
2. Fourth angle = 360° − (100° + 98° + 92°) = 360° − 290° = 70°.

Teacher's Note:
1. Use the angle sum property of a quadrilateral.
2. Subtract the three given angles from 360°.

Answer: (a) 30°, 60°, 120°, 150°

Explanation:
1. Let the angles be x, 2x, 4x and 5x. Then 12x = 360°, so x = 30°.
2. The angles are 30°, 60°, 120° and 150°.

Teacher's Note:
1. Represent the ratio with a common multiplier.
2. Equate the sum to 360°.

Answer: (a) 61°

Explanation:
1. Opposite angles of a rhombus are equal, so ∠ADC = 58°.
2. In triangle ACD, AD = DC (sides of a rhombus), so ∠DAC = ∠ACD.
3. ∠ACD = (180° − 58°) ÷ 2 = 61°.

Teacher's Note:
1. All sides of a rhombus are equal, so triangle ACD is isosceles.
2. Use the angle sum property of the triangle.

Answer: (a) 51 cm

Explanation:
1. By the mid-point theorem, RP ∥ AC and RP = ½ AC = 10.5 cm; PQ ∥ AB and PQ = ½ AB = 15 cm.
2. ARPQ is a parallelogram with AR = ½ AB = 15 cm and AQ = ½ AC = 10.5 cm.
3. Perimeter = 2(15 + 10.5) = 51 cm.

Teacher's Note:
1. Use the mid-point theorem on the triangle.
2. The sides of ARPQ come in equal pairs.

Answer: (a) isosceles

Explanation:
1. Each side of the new triangle is half of a side of the original triangle.
2. Two sides of the original are equal, so two sides of the mid-point triangle are equal.
3. So it is isosceles.

Teacher's Note:
1. Apply the mid-point theorem to each side.
2. Equal sides in the original give equal sides in the new triangle.

Answer: (a) a right triangle

Explanation:
1. The mid-point triangle is similar to the original triangle (each side is half of the corresponding side, parallel to it).
2. Similar triangles have equal angles, so one angle is 90°.
3. It is a right triangle.

Teacher's Note:
1. The mid-point triangle has sides parallel to those of the original.
2. Parallel sides give equal angles.

Answer: (a) If all the sides of a quadrilateral are equal, it is a parallelogram.

Explanation:
1. A quadrilateral with all four sides equal has both pairs of opposite sides equal, so it is a parallelogram (a rhombus): statement (a) is true.
2. (b), (c) and (d) are false: an isosceles trapezium, a kite with three equal sides and angles 100°, 100°, 100°, 60° are counter-examples.

Teacher's Note:
1. Both pairs of opposite sides equal is a sufficient condition.
2. Test the other statements with simple counter-examples.

Answer: (a) 4 cm

Explanation:
1. Since AB ∥ DC, ∠DEA = ∠EAB = ∠DAE, so DE = AD = 6 cm and EC = 10 − 6 = 4 cm.
2. ∠CEF = ∠DEA (vertically opposite) and ∠CFE = ∠DAE (corresponding angles, AD ∥ BF).
3. So ∠CEF = ∠CFE, hence CF = CE = 4 cm.

Teacher's Note:
1. The bisector of an angle of a parallelogram creates an isosceles triangle with the opposite side.
2. Compare the angles in triangle CEF.

Answer: (a) Rectangle

Explanation:
1. The diagonals of a rectangle are equal (the triangles ABC and DCB are congruent).
2. In a rhombus, a general parallelogram and a kite the diagonals are in general unequal.

Teacher's Note:
1. Think of the congruent triangles formed by the diagonals.
2. Equal diagonals in a parallelogram characterise a rectangle.

Answer: (a) 10 cm

Explanation:
1. The diagonals bisect each other at right angles, so half-diagonals are 8 cm and 6 cm.
2. Side = \( \sqrt{8^2+6^2}=10 \) cm.

Teacher's Note:
1. Each side is the hypotenuse of a right triangle made by half-diagonals.
2. Use Pythagoras.

Answer: (a) 6 m

Explanation:
1. Half of the given diagonal is 4 m.
2. Half of the other diagonal = \( \sqrt{5^2-4^2}=3 \) m, so the other diagonal is 6 m.

Teacher's Note:
1. The diagonals of a rhombus bisect each other at right angles.
2. Apply Pythagoras to one of the four right triangles.

Answer: (a) 90°

Explanation:
1. Adjacent angles A and B satisfy A + B = 180°.
2. In the triangle formed by the two bisectors and the side, the angles at the side are A/2 and B/2, so the angle between the bisectors = 180° − (A + B)/2 = 90°.

Teacher's Note:
1. Adjacent angles of a parallelogram are supplementary.
2. Use the angle sum of the triangle formed.

Answer: (a) rhombus

Explanation:
1. Each side of the new figure is half of a diagonal of the rectangle.
2. The diagonals of a rectangle are equal, so all four sides are equal: the figure is a rhombus (not a square, unless the rectangle is a square).

Teacher's Note:
1. The sides of the mid-point figure equal half of the diagonals.
2. Equal diagonals give equal sides.

Answer: (a) 65°

Explanation:
1. Fourth angle = 360° − (80° + 95° + 120°) = 360° − 295° = 65°.

Teacher's Note:
1. The angles of a quadrilateral sum to 360°.
2. Subtract the three known angles.

Answer: (a) 45°

Explanation:
1. Consecutive angles are supplementary: x + 3x = 180°, so x = 45°.
2. The smaller angle is 45° (the other is 135°).

Teacher's Note:
1. Adjacent angles of a parallelogram add up to 180°.
2. Use the ratio with a common multiplier.

Answer: (a) 180

Explanation:
1. ∠PRQ = 90° − 30° = 60°. In a rectangle OQ = OR, so ∠OQR = ∠ORQ = 60°: x = 60.
2. In triangle OQR, ∠ROQ = 180° − 120° = 60°, and ∠SOR = 180° − 60° = 120°: y = 120.
3. x + y = 180.

Teacher's Note:
1. The diagonals of a rectangle are equal and bisect each other, so triangle OQR is isosceles.
2. The angles ∠ROQ and ∠SOR form a linear pair.

Answer: (a) 55°

Explanation:
1. The diagonals of a rhombus intersect at right angles, so ∠AOB = 90°.
2. In triangle AOB: ∠OBA = 180° − 90° − 35° = 55°.

Teacher's Note:
1. The diagonals of a rhombus are perpendicular.
2. Use the angle sum of triangle AOB.

Answer: (a) 30

Explanation:
1. The diagonals of a rhombus are perpendicular, so ∠COD = 90°.
2. In triangle COD: x + 2x + 90° = 180°, so 3x = 90° and x = 30.

Teacher's Note:
1. Perpendicular diagonals give a right angle at O.
2. Apply the angle sum in triangle COD.

Answer: (a) 140°

Explanation:
1. Let the angles be 2x, 4x, 5x, 7x; 18x = 360°, so x = 20°.
2. The largest angle is 7 × 20° = 140° (the angles are 40°, 80°, 100°, 140°).

Teacher's Note:
1. Use a common multiplier for the ratio.
2. The largest angle corresponds to the largest ratio term.

Answer: (a) 45°

Explanation:
1. PS = SR and ∠PSR = 90° in the square, so triangle PSR is right-angled and isosceles.
2. ∠SRP = (180° − 90°) ÷ 2 = 45°.

Teacher's Note:
1. The diagonal of a square bisects its angles.
2. Use the isosceles right triangle PSR.

Answer: (a) 55°

Explanation:
1. ∠C = ∠A = 70° (opposite angles of a parallelogram).
2. In triangle CDB, CD = CB, so ∠CDB = ∠CBD = x with 2x + 70° = 180°, hence x = 55°.

Teacher's Note:
1. Sides of a rhombus are equal, giving an isosceles triangle.
2. Use the angle sum in triangle CDB.

Answer: (a) 62°

Explanation:
1. ∠ADC = ∠ABC = 56°, and the diagonal BD bisects it: ∠ODC = 28°.
2. The diagonals are perpendicular, so in triangle ODC: ∠ACD = 90° − 28° = 62°.

Teacher's Note:
1. The diagonals of a rhombus bisect the angles and are perpendicular.
2. Use the right triangle formed at O.