CBSE Class 9 Mathematics Quadrilaterals MCQs Set 04

Answer: (a) DE = 4 cm, EF = 6 cm, DF = 6.5 cm

Explanation:
1. In △ABC, D and E are mid-points of AB and BC, so DE = ½AC = 4 cm.
2. E and F are mid-points of BC and CA, so EF = ½AB = 6 cm.
3. F and D are mid-points of CA and AB, so DF = ½BC = 6.5 cm.

Teacher's Note:
1. Each side of the medial triangle is half of the side it is parallel to.
2. The side joining two mid-points is opposite the vertex that is common to the two sides being bisected.

Answer: (d) AB = 6 cm, BC = 4.5 cm

Explanation:
1. Since EF ∥ AB, the line EF cuts DA and DB in the same ratio, so EF : AB = DF : DB = 2 : 3.
2. Hence AB = 4 × 3/2 = 6 cm.
3. Similarly FG ∥ BC gives FG : BC = DF : DB = 2 : 3, so BC = 3 × 3/2 = 4.5 cm.

Teacher's Note:
1. A line parallel to one side of a triangle cuts the other two sides in the same ratio.
2. DF : DB = 2 : 3 because DF : FB = 2 : 1.

Answer: (b) 1 : 3

Explanation:
1. The mid-point triangle divides the original triangle into four congruent triangles.
2. So the new triangle is 1/4 of the whole and the remaining part is 3/4.
3. The required ratio is 1 : 3.

Teacher's Note:
1. The four small triangles are congruent by SSS because every side of each is half of a side of the big triangle.
2. Read the question carefully: the ratio asked is to the remaining part, not to the whole triangle.

Answer: (b) 90°

Explanation:
1. The diagonals of a square bisect each other at right angles.
2. So ∠POQ = 90°.

Teacher's Note:
1. A square has all the properties of both a rectangle and a rhombus.
2. Perpendicular diagonals come from the rhombus property.

Answer: (a) ¼ AC

Explanation:
1. D is the mid-point of BC and DF ∥ BE, so F is the mid-point of CE (converse of the mid-point theorem in △CBE).
2. Hence CF = ½CE.
3. E is the mid-point of AC, so CE = ½AC and CF = ¼AC.

Teacher's Note:
1. Apply the converse of the mid-point theorem in the small triangle CBE.
2. Do not stop at CF = ½CE, which is not one of the options in terms of AC.

Answer: (c) 96 cm

Explanation:
1. DF ⊥ BC, so in right △DFC, DF² = 20² − 16² = 144 and DF = 12 cm.
2. BFDE is a square, so BF = DF = 12 cm and BC = BF + FC = 12 + 16 = 28 cm.
3. Perimeter = 2(20 + 28) = 96 cm.

Teacher's Note:
1. Spot the right triangle formed by the side of the parallelogram, the side of the square and CF.
2. 12, 16, 20 is a multiple of the 3, 4, 5 triplet.

Answer: (a) 180°

Explanation:
1. AB ∥ DC, so ∠PAB = ∠APD (alternate angles).
2. AP = AD gives ∠APD = ∠ADP = ∠D, so ∠PAB = ∠D.
3. ∠D + ∠C = 180° (adjacent angles of a parallelogram), so ∠PAB + ∠BCD = 180°.

Teacher's Note:
1. Combine alternate angles with the isosceles triangle property.
2. Adjacent angles of a parallelogram are supplementary.

Answer: (d) A is false, but R is true.

Explanation:
1. Diagonals that bisect each other give a parallelogram, not necessarily a square, so the Assertion is false.
2. In △AOD and △COB: AO = CO, OD = OB and the vertical angles are equal, so they are congruent by SAS. The Reason is true.

Teacher's Note:
1. Bisecting diagonals prove a parallelogram only.
2. A square also needs equal and perpendicular diagonals.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Joining the mid-points of the sides of any quadrilateral gives a parallelogram, so the Assertion is true.
2. The four corner triangles cut off together make half of the rectangle, so the inner figure has half the area. The Reason is true.
3. But the area does not explain why the figure is a parallelogram (that comes from the mid-point theorem), so R is not the explanation.

Teacher's Note:
1. Both statements are true here, but they are independent facts.
2. The explanation of the parallelogram is that each side is parallel to a diagonal.

Answer: (c) A is true, but R is false.

Explanation:
1. A parallelogram with equal diagonals is a rectangle, so the Assertion is true.
2. In a general parallelogram the angles are not right angles, so the Reason is false.

Teacher's Note:
1. Equal diagonals in a parallelogram force a right angle, by congruent triangles ABD and DCA.
2. Do not confuse the property of a rectangle with that of every parallelogram.

Answer: (a) p + q + r + s = w + x + y + z

Explanation:
1. The angle sum of every quadrilateral is 360°.
2. So p + q + r + s = 360° = w + x + y + z.

Teacher's Note:
1. Split a quadrilateral by a diagonal into two triangles to see 2 × 180° = 360°.
2. This is independent of the shape or size.

Answer: (a) 13 cm

Explanation:
1. Draw perpendiculars from A and B to DC. Each right triangle has hypotenuse 5 cm and height 4 cm, so its base is 3 cm.
2. DC = 3 + 7 + 3 = 13 cm.

Teacher's Note:
1. Two equal right triangles stand at the ends and a rectangle in the middle.
2. The condition that DC is the longer base removes the other possibility, DC = 1 cm.

Answer: (c) ∠D

Explanation:
1. In an isosceles trapezium the angles on the same base are equal.
2. DC is a base and ∠C, ∠D are the angles on it, so ∠C = ∠D.

Teacher's Note:
1. Angles on the same parallel side are equal, those on different parallel sides are supplementary.
2. ∠C and ∠B are supplementary here, not equal.

Answer: (b) 0°

Explanation:
1. Opposite angles of a parallelogram are equal.
2. So ∠A = ∠C and ∠A − ∠C = 0°.

Teacher's Note:
1. Opposite angles are equal, adjacent angles are supplementary.
2. This holds for rectangles, rhombuses and squares as well.

Answer: (c) 50°

Explanation:
1. The diagonal AC bisects ∠C, so ∠BCD = 80°.
2. Opposite angle ∠BAD = 80°, so ∠ADC = 180° − 80° = 100°.
3. The diagonal BD bisects ∠D, so ∠ADB = 50°.

Teacher's Note:
1. Diagonals of a rhombus bisect its angles.
2. Alternatively, the diagonals meet at 90°, so ∠ADB = 90° − ∠DAC = 90° − 40°.

Answer: (a) 40 cm, 50 cm

Explanation:
1. The sum of two adjacent sides is 180 ÷ 2 = 90 cm.
2. Let the sides be x and x + 10: 2x + 10 = 90, so x = 40.
3. The sides are 40 cm and 50 cm.

Teacher's Note:
1. Half the perimeter is the sum of adjacent sides.
2. Check: 2(40 + 50) = 180.

Answer: (c) rhombus

Explanation:
1. Each side of PQRS is half a diagonal of the rectangle (mid-point theorem), e.g. PQ = ½AC.
2. The diagonals of a rectangle are equal, so all four sides of PQRS are equal.
3. PQRS is therefore a rhombus (not a square, as its angles are not 90° in general).

Teacher's Note:
1. Mid-point quadrilateral: rectangle gives rhombus, rhombus gives rectangle, square gives square.
2. PQRS is a square only if ABCD is a square.

Answer: (b) 2 BD

Explanation:
1. Each side of PQRS is half the length of a diagonal, so each side = ½BD.
2. Perimeter = 4 × ½BD = 2BD.

Teacher's Note:
1. The diagonals AC and BD of a rectangle are equal.
2. Use the mid-point theorem in △ABD and △BCD etc.

Answer: (a) PR = AD

Explanation:
1. PR joins the mid-points of the opposite sides AB and CD, so PR ∥ AD and PR = AD (APRD is a rectangle).
2. Similarly SQ = AB, and PR ≠ SQ unless the rectangle is a square.

Teacher's Note:
1. The segment joining mid-points of opposite sides of a rectangle equals the other pair of sides.
2. Option (D) holds only for a square.

Answer: (a) parallelogram

Explanation:
1. A quadrilateral whose diagonals bisect each other is a parallelogram (converse of the diagonal property).
2. It need not be a square or a kite.

Teacher's Note:
1. This is a standard test for a parallelogram.
2. Add equal diagonals to get a rectangle.

Answer: (a) equal

Explanation:
1. In rectangle ABCD, △ABC ≅ △DCB (SAS), so AC = BD.
2. They also bisect each other, but are not perpendicular unless it is a square.

Teacher's Note:
1. A parallelogram is a rectangle iff its diagonals are equal.
2. Compare with a rhombus, whose diagonals are perpendicular.

Answer: (b) rhombus

Explanation:
1. Perpendicular diagonals that bisect each other make all four triangles congruent (SAS).
2. So all four sides are equal and the parallelogram is a rhombus.

Teacher's Note:
1. A parallelogram with perpendicular diagonals is a rhombus.
2. If the diagonals are also equal it is a square.

Answer: (a) square

Explanation:
1. Each side of the new figure is half a diagonal of the square, and the diagonals are equal, so it is a rhombus.
2. The sides are parallel to the diagonals, which are perpendicular, so each angle is 90°.
3. A rhombus with a right angle is a square.

Teacher's Note:
1. Square gives square, rotated by 45°.
2. Its area is half of the original square.

Answer: (b) rectangle

Explanation:
1. Opposite angles of a cyclic quadrilateral add to 180°.
2. Opposite angles of a parallelogram are equal, so each is 90°.
3. A parallelogram with a right angle is a rectangle (a square is a special case).

Teacher's Note:
1. Cyclic plus parallelogram forces right angles.
2. A rhombus is cyclic only when it is a square.