CBSE Class 9 Mathematics Areas of Parallelograms and Triangles MCQs Set 08

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

Explanation:
1. Equal halves from both diagonals give ar(ABC) = ar(ABD) with a common base AB, so the altitudes are equal and DC ∥ AB; similarly AD ∥ BC. The Assertion is true.
2. The Reason is also a true theorem, but it does not prove the Assertion, so it is not the explanation.

Teacher's Note:
1. Triangles with the same base and equal areas lie between the same parallels.
2. Both statements are true but independent.

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

Explanation:
1. △ADQ and △ADC stand on the same base AD between the parallels AD and BQ, so ar(△ADQ) = ar(△ADC). Subtracting the common △ADP gives ar(△DPQ) = ar(△APC).
2. △APC and △BPC stand on the same base PC between the parallels AB and DC, so ar(△APC) = ar(△BPC). Hence ar(△BCP) = ar(△DPQ) and the Assertion is true.
3. A diagonal divides a parallelogram into two triangles of equal area, so the Reason is false.

Teacher's Note:
1. The Reason confuses the diagonal property (two equal triangles).
2. Use triangles on the same base between the same parallels to prove the Assertion.

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

Explanation:
1. The Assertion is the definition of similar figures, so it is true.
2. Congruent figures have the same shape, so they are similar, while similar figures can differ in size. The Reason is true but it does not define similarity.

Teacher's Note:
1. Congruent means same shape and same size.
2. Similar figures have equal angles and proportional sides.

Answer: (a) equal area

Explanation:
1. This is the theorem on parallelograms between the same parallels.
2. Each has area = base × common height, so the areas are equal.

Teacher's Note:
1. Perimeters differ because the slant sides differ.
2. The shared base and shared height give equal area.

Answer: (a) corresponding sides

Explanation:
1. For similar triangles, corresponding sides are in the same ratio k.
2. Their areas are in the ratio k², the square of the ratio of corresponding sides.

Teacher's Note:
1. The altitudes and perimeters are in ratio k as well.
2. Area scales with the square of length.

Answer: (a) altitude

Explanation:
1. Area = ½ × base × height, where the height is the altitude to that base.
2. The altitude is perpendicular to the base (or its extension).

Teacher's Note:
1. Any side can be chosen as the base.
2. The product base × height for a parallelogram is double that for a triangle.

Answer: (a) diagonals

Explanation:
1. The diagonals cut the rhombus into four congruent right triangles.
2. Each has area ½ × (d1/2) × (d2/2), so the total area is ½ × d1 × d2.

Teacher's Note:
1. Rhombus area = ½ d1 d2 = base × height.
2. A kite also has area ½ d1 d2.

Answer: (c) (√3/3) x²

Explanation:
1. In an equilateral triangle of side a the median is x = (√3/2)a, so a = 2x/√3.
2. Area = (√3/4)a² = (√3/4)(4x²/3) = (√3/3)x².

Teacher's Note:
1. The median, altitude and angle bisector coincide in an equilateral triangle.
2. Check with a = 2: x = √3 and area = √3 = (√3/3)(3).

Answer: (b) 5 sq units

Explanation:
1. The mid-point triangle divides △ABC into four congruent triangles.
2. ar(△PQR) = 20 ÷ 4 = 5 square units.

Teacher's Note:
1. The mid-point triangle has one-quarter of the area.
2. Its sides are half the original sides.

Answer: (b) 18 cm

Explanation:
1. Let the legs be a, b and hypotenuse c. Then ab = 80 and a + b = 40 − c.
2. (a + b)² = a² + b² + 2ab gives (40 − c)² = c² + 160.
3. 1600 − 80c = 160, so c = 18 cm.

Teacher's Note:
1. Square the perimeter relation to bring in a² + b².
2. Check: a + b = 22 and ab = 80 give real legs about 4.8 cm and 17.2 cm.

Answer: (b) AB² / PQ²

Explanation:
1. For similar triangles the ratio of areas is the square of the ratio of corresponding sides.
2. AB corresponds to PQ, so the ratio is AB² / PQ² (equally BC²/QR² or AC²/PR²).

Teacher's Note:
1. Match corresponding sides by the order of the letters.
2. AC² / AB² compares sides of the same triangle, so it is wrong.

Answer: (a) ar(△ANB) = ar(△PNQ)

Explanation:
1. △ABQ and △PBQ stand on the same base BQ between the parallels AP and BQ, so ar(△ABQ) = ar(△PBQ).
2. Subtracting the common △BNQ gives ar(△ANB) = ar(△PNQ).

Teacher's Note:
1. Subtract the common part from two equal triangles on the same base.
2. The other options compare triangles that have no common base or equal altitude.

Answer: (a) ar(△QMR)

Explanation:
1. △BCR and △QCR stand on the same base CR between parallels BQ and CR, so they have equal areas.
2. Subtract the common △MCR to get ar(△BMC) = ar(△QMR).

Teacher's Note:
1. Use the same trick as in the previous question.
2. The result holds for the lower pair of triangles just as for the upper pair.

Answer: (c) ar(△PBR)

Explanation:
1. Since B lies on AC, ar(△AQC) = ar(△ABQ) + ar(△BQC). Since Q lies on PR, ar(△PBR) = ar(△PBQ) + ar(△BQR).
2. ar(△ABQ) = ar(△PBQ) (same base BQ, between AP ∥ BQ) and ar(△BQC) = ar(△BQR) (same base BQ, between BQ ∥ CR).
3. Adding, ar(△AQC) = ar(△PBR).

Teacher's Note:
1. Parallel lines give equal heights; compare each triangle by base times height.
2. Check numerically with a trapezium of parallel sides 2 and 4.

Answer: (d) 4 : 1

Explanation:
1. BD = ½BC, so the side of △BDE is half the side of △ABC.
2. Similar triangles: area ratio = (2 : 1)² = 4 : 1.

Teacher's Note:
1. Doubling the side multiplies the area by 4.
2. Ratio is ABC to BDE, not BDE to ABC.

Answer: (d) 4 : 1

Explanation:
1. △ADE ~ △ABC with ratio of sides DE : BC = 1 : 2.
2. Ratio of areas = 2² : 1² = 4 : 1 (so ar(△ABC) = 100 sq cm).

Teacher's Note:
1. Parallel line to a side gives a similar triangle.
2. Square the side ratio to get the area ratio.

Answer: (b) 1 : 4

Explanation:
1. ar(△AFE) = ¼ar(△ABC), because △AFE is a corner triangle of the mid-point triangle construction.
2. △DFE has area ¼ar(△ABC), and △DHI is its corner triangle at D, so ar(△DHI) = ¼ × ¼ar(△ABC) = (1/16)ar(△ABC).
3. Ratio = (1/16) : (1/4) = 1 : 4.

Teacher's Note:
1. Apply the mid-point triangle one-quarter rule twice.
2. The corner triangle at D of △DEF has half the sides of △DEF.

Answer: (a) (√2 − 1) : √2

Explanation:
1. △APQ ~ △ABC and ar(△APQ) = ½ar(△ABC), so (AP/AB)² = ½ and AP/AB = 1/√2.
2. PB/AB = 1 − 1/√2 = (√2 − 1)/√2.

Teacher's Note:
1. Square-root of the area ratio gives the side ratio.
2. Remember PB = AB − AP.

Answer: (b) parallelogram

Explanation:
1. EF ∥ AC and EF = ½AC, and HG ∥ AC and HG = ½AC (mid-point theorem).
2. So EF and HG are equal and parallel, hence EFGH is a parallelogram.
3. Its sides are half of the diagonals, which are unequal in general, so it is not a rhombus or a rectangle.

Teacher's Note:
1. The mid-point quadrilateral of any quadrilateral is a parallelogram.
2. It is a rhombus only when the diagonals are equal and a rectangle only when they are perpendicular.

Answer: (c) equilateral

Explanation:
1. Every equilateral triangle has all angles equal to 60°.
2. Triangles with equal angles are similar, so all equilateral triangles are similar.

Teacher's Note:
1. Isosceles and right-angled triangles can have different angles.
2. Equal angles (AAA) are enough for similarity.

Answer: (b) need not

Explanation:
1. Congruent figures have the same shape and size, hence they are similar.
2. Similar figures may differ in size, so they need not be congruent.

Teacher's Note:
1. Congruence is similarity with ratio 1.
2. A small and a large square are similar but not congruent.

Answer: (b) △ACD

Explanation:
1. AB = AC, DB = DC and AD is common.
2. So △ABD ≅ △ACD by SSS.

Teacher's Note:
1. SSS needs all three pairs of sides.
2. The common side AD is easy to miss.

Answer: (a) ∠A (∠BAC)

Explanation:
1. △ABD ≅ △ACD gives ∠BAD = ∠CAD.
2. So AP bisects ∠BAC.

Teacher's Note:
1. Corresponding parts of congruent triangles are equal.
2. In an isosceles triangle the axis through the apex bisects the apex angle.

Answer: (a) BC

Explanation:
1. △ABP ≅ △ACP (SAS: AB = AC, ∠BAP = ∠CAP, AP common), so BP = CP and ∠APB = ∠APC = 90°.
2. So AP is the perpendicular bisector of BC.

Teacher's Note:
1. The apex-to-base line of an isosceles triangle is also its median and altitude.
2. Both A and D lie on the perpendicular bisector of BC.