CBSE Class 9 Mathematics Areas of Parallelograms and Triangles MCQs Set 05

Answer: (a) 37°, 143°, 37°, 143°

Explanation:
1. Opposite angles of a parallelogram are equal, so 3x − 2 = 50 − x.
2. 4x = 52, so x = 13.
3. Each of the two equal angles = 3(13) − 2 = 37°.
4. Adjacent angles are supplementary, so the other pair = 180° − 37° = 143°.

Teacher's Note:
1. Always equate opposite angles first, then use the supplementary rule for adjacent angles.
2. Check by adding all four angles: 37 + 143 + 37 + 143 = 360°.

Answer: (b) 108°, 72°, 108°, 72°

Explanation:
1. Let the adjacent angles be x and \( \frac{2}{3} \)x. They are supplementary.
2. x + \( \frac{2}{3} \)x = 180°, so \( \frac{5}{3} \)x = 180°, x = 108°.
3. The other angle = \( \frac{2}{3} \) × 108° = 72°.
4. Opposite angles are equal, so the angles in order are 108°, 72°, 108°, 72°.

Teacher's Note:
1. Adjacent angles of a parallelogram always add up to 180°.
2. The order matters: equal angles must be opposite, not adjacent, so option (c) cannot be a parallelogram.

Answer: (c) 68°, 112°, 68°, 112°

Explanation:
1. Let the smallest angle be x. The adjacent angle is 2x − 24.
2. Adjacent angles are supplementary: x + 2x − 24 = 180, so 3x = 204 and x = 68°.
3. The other angle is 180° − 68° = 112° (and 2x − 24 = 112°, consistent).
4. The angles are 68°, 112°, 68°, 112°.

Teacher's Note:
1. The smaller of two adjacent angles is the 'smallest angle' of the parallelogram.
2. Check: 2 × 68 − 24 = 112, which equals 180 − 68.

Answer: (d) 4.5 cm

Explanation:
1. Perimeter = 2 (longer side + shorter side) = 22 cm.
2. So longer side + shorter side = 11 cm.
3. Shorter side = 11 − 6.5 = 4.5 cm.

Teacher's Note:
1. Opposite sides of a parallelogram are equal, so the perimeter is twice the sum of two adjacent sides.
2. Divide the perimeter by 2 before subtracting.

Answer: (a) 45° and 135°

Explanation:
1. ∠A and ∠D are adjacent angles, so ∠A = 180° − 135° = 45°.
2. ∠B is opposite to ∠D, so ∠B = 135°.

Teacher's Note:
1. ∠D and ∠B are opposite angles, hence equal; ∠D and ∠A are adjacent, hence supplementary.
2. Name the vertices in order around the figure before deciding which angles are adjacent.

Answer: (b) 112°

Explanation:
1. Let the smallest angle be x. Then the adjacent angle is 2x − 24.
2. x + 2x − 24 = 180, so x = 68°.
3. The largest angle = 180° − 68° = 112°.

Teacher's Note:
1. In a parallelogram that is not a rectangle there are only two different angle values, one acute and one obtuse.
2. The largest angle is the obtuse one.

Answer: (c) 72°

Explanation:
1. Let the adjacent angles be x and \( \frac{2}{3} \)x with x + \( \frac{2}{3} \)x = 180°.
2. x = 108°, so the other angle is \( \frac{2}{3} \) × 108° = 72°.
3. The smallest angle is 72°.

Teacher's Note:
1. Set up one equation using the supplementary property.
2. Do not forget to compute both angles and then pick the smaller one.

Answer: (a) 45°

Explanation:
1. AD ∥ BC and BD is a transversal, so ∠ADB = ∠DBC = 60° (alternate angles).
2. In △ABD: ∠ABD = 180° − 75° − 60° = 45°.
3. AB ∥ DC and BD is a transversal, so ∠BDC = ∠ABD = 45° (alternate angles).

Teacher's Note:
1. Use alternate angles twice: once with AD ∥ BC and once with AB ∥ DC.
2. The angle sum of triangle ABD links the two.

Answer: (d) 1 : 1

Explanation:
1. Area of a parallelogram = base × height.
2. The bases are equal and the height is the distance between the same pair of parallels, so the heights are equal.
3. Hence the areas are equal and the ratio is 1 : 1.

Teacher's Note:
1. Equal base and equal height always give equal areas.
2. This idea is the core of the theorem on parallelograms between the same parallels.

Answer: (a) less than 1

Explanation:
1. Equal base and equal area mean equal heights, so both lie between the same parallels.
2. The slant side of the parallelogram is longer than the perpendicular distance between the parallels, which is the rectangle's side.
3. Both figures have the same base, so the rectangle has the shorter perimeter, and the ratio is less than 1.

Teacher's Note:
1. Of all parallelograms on a given base between two parallels, the rectangle has the least perimeter.
2. The perpendicular is the shortest distance between two parallel lines.

Answer: (b) \( \frac{1}{4} \) ar(parallelogram ABCD)

Explanation:
1. E and F are mid-points of opposite sides, so EF ∥ AB and ABEF is a parallelogram with half the area of ABCD.
2. △GAB and parallelogram ABEF stand on the same base AB and lie between the same parallels AB and EF.
3. So ar(△GAB) = \( \frac{1}{2} \) ar(ABEF) = \( \frac{1}{2} \) × \( \frac{1}{2} \) ar(ABCD) = \( \frac{1}{4} \) ar(ABCD).

Teacher's Note:
1. A triangle and a parallelogram on the same base between the same parallels: triangle = half the parallelogram.
2. The position of G on EF does not matter.

Answer: (c) 25 cm and 50 cm

Explanation:
1. Let the base be b cm, so the altitude is 2b cm.
2. Area = b × 2b = 2b² = 1250, so b² = 625 and b = 25.
3. Base = 25 cm, altitude = 50 cm.

Teacher's Note:
1. Write the area formula with one unknown.
2. Check: 25 × 50 = 1250.

Answer: (d) 15 cm²

Explanation:
1. △ABE and parallelogram ABCD stand on the same base AB and between the same parallels AB and DC, so ar(△ABE) = \( \frac{1}{2} \) × 30 = 15 cm².
2. The three triangles ADE, ABE and BCE make up the whole parallelogram.
3. ar(△ADE) + ar(△BCE) = 30 − 15 = 15 cm².

Teacher's Note:
1. Find the triangle on the long base first, then subtract from the whole.
2. The answer does not depend on where E lies on DC.

Answer: (a) 16 cm²

Explanation:
1. △AEB stands on base AB between the parallels AB and DC, so ar(△AEB) = \( \frac{1}{2} \) ar(ABCD), hence ar(ABCD) = 32 cm².
2. △BFC stands on base BC between the parallels AD and BC, so ar(△BFC) = \( \frac{1}{2} \) ar(ABCD) = 16 cm².

Teacher's Note:
1. Each triangle is half the parallelogram because each shares a base with it and lies between the same parallels.
2. Triangles AEB and BFC therefore have equal areas.

Answer: (b) 5.5 units

Explanation:
1. Let altitude = 2k and area = 11k.
2. Base = area ÷ altitude = 11k ÷ 2k = 5.5 units.

Teacher's Note:
1. Rearrange area = base × altitude to get base = area ÷ altitude.
2. The common factor k cancels.

Answer: (c) 16 cm

Explanation:
1. Area = longer side × distance between the longer sides = 24 × 12 = 288 cm².
2. The same area = shorter side × distance between the shorter sides = 18 × h.
3. h = 288 ÷ 18 = 16 cm.

Teacher's Note:
1. The area is the same whichever side is taken as the base.
2. The longer side has the shorter altitude.

Answer: (a) (3a + b) : (a + 3b)

Explanation:
1. The segment joining the mid-points of the non-parallel sides is parallel to the bases and EF = \( \frac{a + b}{2} \).
2. Both trapeziums have the same height h/2, where h is the height of ABCD.
3. ar(ABFE) = \( \frac{1}{2} \) × (a + \( \frac{a + b}{2} \)) × \( \frac{h}{2} \) and ar(EFCD) = \( \frac{1}{2} \) × (\( \frac{a + b}{2} \) + b) × \( \frac{h}{2} \).
4. The ratio is (3a + b) : (a + 3b).

Teacher's Note:
1. The mid-point segment is the average of the two parallel sides.
2. The longer base a belongs to the lower trapezium, so its area is larger when a is larger than b.

Answer: (b) 18 cm

Explanation:
1. Area of rectangle = 18 × 12 = 216 cm².
2. Area of triangle = \( \frac{1}{2} \) × 24 × h = 12h.
3. 12h = 216, so h = 18 cm.

Teacher's Note:
1. Area of a triangle is half of base times height.
2. Do not forget the factor \( \frac{1}{2} \).

Answer: (a) They bisect each other

Explanation:
1. In a general parallelogram the diagonals bisect each other.
2. They are equal only in a rectangle, perpendicular only in a rhombus, and bisect the vertex angles only in a rhombus.

Teacher's Note:
1. Only the bisecting property holds for every parallelogram.
2. The other three properties belong to special parallelograms.

Answer: (c) rectangle

Explanation:
1. Adjacent angles of a parallelogram are supplementary, so ∠A + ∠B = 180°.
2. With ∠A = ∠B, each is 90°.
3. A parallelogram with one right angle is a rectangle.

Teacher's Note:
1. Equal and supplementary adjacent angles must each be 90°.
2. A parallelogram with a right angle has all four angles right angles.

Answer: (b) AE

Explanation:
1. Let the diagonals meet at O, the mid-point of AC and BD. The medians AO of △ABD and CO of △BCD lie along AC.
2. The centroid divides a median in the ratio 2 : 1 from the vertex, so AE = \( \frac{2}{3} \) AO and CF = \( \frac{2}{3} \) CO.
3. With AO = CO = \( \frac{1}{2} \) AC: AE = CF = \( \frac{1}{3} \) AC, so EF = AC − AE − CF = \( \frac{1}{3} \) AC = AE.

Teacher's Note:
1. Both centroids lie on the diagonal AC.
2. The centroid is two-thirds of the way along each median from the vertex.

Answer: (c) ar(△ADC) = ar(△CBA)

Explanation:
1. A diagonal divides a parallelogram into two congruent triangles.
2. △ADC ≅ △CBA, so their areas are equal.

Teacher's Note:
1. Congruent figures have equal areas.
2. Each triangle has half the area of the parallelogram.

Answer: (d) an isosceles triangle

Explanation:
1. Each side of the mid-point triangle is half of the opposite side of the original triangle.
2. The two equal sides of the original triangle give two equal sides in the mid-point triangle.
3. Hence the mid-point triangle is isosceles.

Teacher's Note:
1. The mid-point theorem says each side of the inner triangle is half of the corresponding side of the outer triangle.
2. Equal sides stay equal after halving.

Answer: (a) a right-angled triangle

Explanation:
1. Each side of the mid-point triangle is parallel to a side of the original triangle and half as long.
2. So the inner triangle has the same angles as the original triangle.
3. A right angle in the original gives a right angle in the inner triangle.

Teacher's Note:
1. Parallel sides give equal corresponding angles.
2. The mid-point triangle is similar to the original one.

Answer: (b) parallelogram

Explanation:
1. Join one diagonal AC. In △ABC and △ADC the mid-point segments are each parallel to AC and half of AC.
2. So one pair of opposite sides of the inner figure is equal and parallel.
3. Hence it is a parallelogram.

Teacher's Note:
1. It becomes a rectangle only if the diagonals of the quadrilateral are perpendicular, and a rhombus only if they are equal.
2. The mid-point theorem is the whole proof.