CBSE Class 10 Mathematics Quadrilaterals MCQs Set 14

Answer: (c) 2BQ

Explanation:
1. PBQD is a parallelogram (PD ∥ BQ and PB ∥ DQ), so DQ = PB = AB/2 = DC/2 and Q is the mid-point of DC
2. In triangle RAB, PD ∥ BR and P is the mid-point of AB, so D is the mid-point of AR; then DC ∥ AB makes Q the mid-point of BR
3. Hence BR = 2BQ.

Teacher's Note:
1. Use the mid-point theorem and its converse.
2. Look for a smaller parallelogram first.

Answer: (a) QY

Explanation:
1. PX and QY are both perpendicular to the line SR, so PX ∥ QY
2. PQ ∥ XY, so PXYQ is a rectangle
3. Opposite sides of a rectangle are equal, so PX = QY.

Teacher's Note:
1. The distance between two parallel lines is the same everywhere.
2. Here PQ ∥ SR, so PX and QY are both that distance.

Answer: (c) parallelogram

Explanation:
1. AXCY and XBYD are parallelograms (AX = YC and AX ∥ YC; XB = DY and XB ∥ DY)
2. So DX ∥ YB and AY ∥ XC, which gives both pairs of opposite sides of PXQY parallel
3. Hence PXQY is a parallelogram.

Teacher's Note:
1. Opposite sides of PXQY lie along the parallel pairs DX, YB and AY, XC.
2. No extra condition makes it a rectangle or rhombus.

Answer: (d) 62°

Explanation:
1. Adjacent angles of a rhombus are supplementary, so ∠BCD = 180° − 56° = 124°
2. Diagonal AC bisects ∠BCD, so ∠ACD = 62°.

Teacher's Note:
1. The diagonals of a rhombus bisect its angles.
2. Adjacent angles add up to 180°.

Answer: (a) 45°

Explanation:
1. The diagonal of a square bisects the 90° angle at the vertex
2. ∠SRP = 45°.

Teacher's Note:
1. Diagonals of a square bisect its angles.
2. Each angle of a square is 90°.

Answer: (a) rhombus

Explanation:
1. PQ = AD/2, QR = BC/2, RS = AD/2, SP = BC/2 by the mid-point theorem
2. Opposite sides are equal and parallel, so PQRS is a parallelogram, and AD = BC makes all four sides equal
3. So PQRS is a rhombus (it is not a rectangle because its diagonals are not equal in general).

Teacher's Note:
1. Apply the mid-point theorem to triangles ABD, BDC, ADC and ABC.
2. AD = BC gives equal adjacent sides.

Answer: (c) 51 cm

Explanation:
1. RP ∥ AC and RP = AC/2, PQ ∥ AB and PQ = AB/2, so ARPQ is a parallelogram
2. AR = AB/2 = 15 cm and AQ = AC/2 = 10.5 cm
3. Perimeter = 2(15 + 10.5) = 51 cm.

Teacher's Note:
1. The mid-point theorem gives a parallelogram at vertex A.
2. BC is not needed for this perimeter.

Answer: (d) 70°, 110°, 70°, 110°

Explanation:
1. Let the smallest angle be x and the adjacent angle 2x − 30°
2. Adjacent angles are supplementary: x + 2x − 30 = 180, so x = 70°
3. The angles are 70°, 110°, 70°, 110°.

Teacher's Note:
1. Adjacent angles of a parallelogram add up to 180°.
2. Opposite angles are equal.

Answer: (a) parallelogram

Explanation:
1. AX = AB/2 = CD/2 = YC and AX ∥ YC
2. One pair of opposite sides is equal and parallel, so AXCY is a parallelogram.

Teacher's Note:
1. One pair of opposite sides equal and parallel is enough.
2. It is a rhombus only when AB = AD.

Answer: (b) 90°

Explanation:
1. ∠BEC = ∠ECD (alternate angles) = ∠BCE, so BE = BC
2. Then AE = AB/2 = BC = AD, so triangle ADE is isosceles and DE bisects ∠ADC
3. ∠CDE + ∠DCE = ½(∠D + ∠C) = 90°, so ∠DEC = 90°.

Teacher's Note:
1. An isosceles triangle appears from the alternate angles.
2. Co-interior angles of the parallelogram add up to 180°.

Answer: (c) AB ∥ DC

Explanation:
1. The angles are 36°, 72°, 108°, 144° (sum 360°)
2. ∠A + ∠D = 36° + 144° = 180° and ∠B + ∠C = 72° + 108° = 180°
3. These are co-interior angles for AB and DC with transversals AD and BC, so AB ∥ DC.

Teacher's Note:
1. Divide 360° in the given ratio.
2. Look for pairs of angles that add up to 180°.

Answer: (b) 162°

Explanation:
1. AE = BC and AE ∥ BC, so ABCE is a parallelogram and CE = AB; also ∠BCE = ∠A = 102°
2. CE = CD = ED, so triangle CED is equilateral and ∠ECD = 60°
3. ∠BCD = 102° + 60° = 162°.

Teacher's Note:
1. Opposite angles of a parallelogram are equal.
2. Three equal sides make an equilateral triangle.

Answer: (c) QM

Explanation:
1. PS ∥ QR and LM ∥ PQ, so PLMQ is a parallelogram
2. Opposite sides of a parallelogram are equal, so PL = QM.

Teacher's Note:
1. LM ∥ PQ and PL ∥ QM give a parallelogram.
2. The bisectors give extra equal segments but are not needed.

Answer: (c) 2∠CED

Explanation:
1. In triangle CED: ∠CED = 180° − (∠C + ∠D)/2
2. ∠A + ∠B + ∠C + ∠D = 360°, so (∠C + ∠D)/2 = 180° − (∠A + ∠B)/2
3. So ∠CED = (∠A + ∠B)/2 and ∠A + ∠B = 2∠CED.

Teacher's Note:
1. Use the angle sum of a quadrilateral and of a triangle.
2. Half angles are involved because of the bisectors.

Answer: (c) ½ (AB − CD)

Explanation:
1. The line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides
2. Its length is half the difference of the parallel sides (AB > CD): PQ = ½(AB − CD).

Teacher's Note:
1. This is a standard trapezium theorem.
2. Compare with the mid-line, whose length is half the sum.

Answer: (b) AP

Explanation:
1. The diagonals meet at O, the mid-point of BD; P lies on AO with AP = (2/3)AO, Q lies on CO with CQ = (2/3)CO
2. AC = 2AO, so PQ = 2AO − (4/3)AO = (2/3)AO = AP.

Teacher's Note:
1. The centroid divides a median in the ratio 2 : 1.
2. Both triangles share the median to BD.

Answer: (b) rectangle

Explanation:
1. The bisectors of two adjacent angles meet at 90°, since (∠A + ∠B)/2 = 90°
2. All four corners of the enclosed figure are right angles, and it is not a square in general.

Teacher's Note:
1. Adjacent angles of a parallelogram are supplementary.
2. Their half-angles add up to 90°.

Answer: (c) 24 m²

Explanation:
1. Half-diagonals are 4 m and 3 m, and 3² + 4² = 5², so the diagonals are perpendicular and the parallelogram is a rhombus
2. Area = ½ × 8 × 6 = 24 m².

Teacher's Note:
1. Check whether the half-diagonals and the side form a right triangle.
2. The area of a rhombus is half the product of the diagonals.

Answer: (b) 1 : 2

Explanation:
1. In triangle AOD, EF joins mid-points of AO and DO, so EF = AD/2; similarly FG = DC/2, GH = CB/2, HE = BA/2
2. So the perimeters are in the ratio 1 : 2.

Teacher's Note:
1. Apply the mid-point theorem in the four triangles around O.
2. Each small side is half of the matching side.

Answer: (c) 90°

Explanation:
1. The bisector of ∠A is the axis of symmetry of the regular pentagon
2. It meets the opposite side CD at its mid-point at right angles, so ∠AMC = 90°.

Teacher's Note:
1. A regular polygon is symmetric about each such line.
2. The axis is perpendicular to the opposite side.

Answer: (c) ¼ (rectangle ABCD)

Explanation:
1. PQ ∥ AB and PQ is midway between AB and DC, so the height of triangle ARB is half of AD
2. Area = ½ × AB × (AD/2) = ¼ × AB × AD.

Teacher's Note:
1. The distance from any point on PQ to AB is the same.
2. The area of a triangle is half base times height.