CBSE Class 9 Mathematics Heron's Formula MCQs Set 06

Answer: (b) 4 cm

Explanation:
1. BC² = AC² − AB² = 25 − 9 = 16.
2. BC = 4 cm.

Teacher's Note:
1. 3, 4, 5 is a Pythagorean triplet.
2. The hypotenuse is opposite the right angle, so AC = 5 cm.

Answer: (a) 12 cm

Explanation:
1. The altitude from A bisects BC, so the half-base is 5 cm.
2. Height = √(13² − 5²) = √144 = 12 cm.

Teacher's Note:
1. 5, 12, 13 is a Pythagorean triplet.
2. The altitude of an isosceles triangle on its base is also its median.

Answer: (a) 4√3 cm²

Explanation:
1. Side = 12 ÷ 3 = 4 cm.
2. Area = (√3/4) × 4² = 4√3 cm².

Teacher's Note:
1. Use the formula (√3/4)a².
2. Perimeter gives the side first.

Answer: (a) 28 m and 14 m

Explanation:
1. Let the altitude be h; then base = 2h and 2h × h = 392.
2. h² = 196, so h = 14 m and the base is 28 m.

Teacher's Note:
1. Area of a parallelogram = base × altitude.
2. Check: 28 × 14 = 392.

Answer: (b) 216 cm²

Explanation:
1. Area = ½ × d1 × d2 = ½ × 18 × 24 = 216 cm².

Teacher's Note:
1. The area of a rhombus is half the product of its diagonals.
2. The side would be 15 cm (9, 12, 15 triangle).

Answer: (a) 5

Explanation:
1. Convert to cm: sides 12.5 cm and 6.25 cm, altitude 2.5 cm on the 12.5 cm side.
2. Area = 12.5 × 2.5 = 31.25 cm².
3. Other altitude = 31.25 ÷ 6.25 = 5 cm.

Teacher's Note:
1. Convert all lengths to the same unit first.
2. The two (side × altitude) products are equal.

Answer: (c) 25% decrease

Explanation:
1. New length = 1.5 L and new breadth = 0.5 B, so new area = 0.75 LB.
2. Area decreases by 25%.

Teacher's Note:
1. Multiply the factors: 1.5 × 0.5 = 0.75.
2. A decrease of 25%, not 75%.

Answer: (d) 30.79 cm

Explanation:
1. s = 75 cm. Area = √(75 × 40 × 21 × 14) = √882000 ≈ 939.15 cm².
2. The smallest altitude falls on the longest side, 61 cm.
3. Altitude = 2 × 939.15 ÷ 61 ≈ 30.79 cm.

Teacher's Note:
1. Smallest altitude means longest base.
2. Do not forget the factor 2 in h = 2A / b.

Answer: (b) 1.6 g

Explanation:
1. 8² + 15² = 17², so it is a right triangle with area ½ × 8 × 15 = 60 cm².
2. Weight per cm² = 96 ÷ 60 = 1.6 g.

Teacher's Note:
1. Spot the 8, 15, 17 triplet to avoid Heron's formula.
2. Divide the weight by the area.

Answer: (d) 13 cm

Explanation:
1. The half-diagonals and the side form a right triangle.
2. Side = √(5² + 12²) = 13 cm.

Teacher's Note:
1. 5, 12, 13 is a Pythagorean triplet.
2. Diagonals of a rhombus meet at right angles.

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

Explanation:
1. Area = ½ (a + b) h, so for fixed parallel sides the area is proportional to h. The Assertion is true.
2. A diagonal gives equal areas only in a parallelogram, so the Reason is false for a general trapezium.

Teacher's Note:
1. Check the formula for the dependence on h.
2. The triangles on the diagonal have different bases (the parallel sides).

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

Explanation:
1. Area = ½ d1 d2, so d1 d2 = 2 × area, not half of it. The Assertion is false.
2. The Reason states the correct area formula, so it is true.

Teacher's Note:
1. Rearrange the area formula carefully.
2. Product of diagonals is double the area.

Answer: (d) 240 m²

Explanation:
1. BC = √(26² − 24²) = √100 = 10 m.
2. Area = 24 × 10 = 240 m².

Teacher's Note:
1. Use Pythagoras in the right triangle ABC.
2. 10, 24, 26 is the 5, 12, 13 triplet doubled.

Answer: (a) 306 cm²

Explanation:
1. 9² + 40² = 41², so △ABC is right-angled at B with area ½ × 9 × 40 = 180 cm².
2. For △ACD, s = (41 + 28 + 15) ÷ 2 = 42 and area = √(42 × 1 × 14 × 27) = 126 cm².
3. Total = 180 + 126 = 306 cm².

Teacher's Note:
1. Split the quadrilateral by its diagonal.
2. Use the right triangle for one part and Heron's formula for the other.

Answer: (b) 546 cm²

Explanation:
1. △ABD: s = 48, area = √(48 × 6 × 14 × 28) = 336 cm².
2. △BCD: 20² + 21² = 29², so it is right-angled at B with area ½ × 20 × 21 = 210 cm².
3. Total = 336 + 210 = 546 cm².

Teacher's Note:
1. Check right-angled triples such as 20, 21, 29 first.
2. Heron's formula handles the other triangle.

Answer: (c) 114 cm²

Explanation:
1. BC = √(17² − 15²) = 8 cm, so ar(△ABC) = ½ × 8 × 15 = 60 cm².
2. 9² + 12² = 15², so △ADC is right-angled at D with area ½ × 9 × 12 = 54 cm².
3. Total = 60 + 54 = 114 cm².

Teacher's Note:
1. Each triangle here is right-angled.
2. Use the 8, 15, 17 and 9, 12, 15 triplets.

Answer: (c) 3√2.5

Explanation:
1. The common hypotenuse squared is 6² + 3² = 45.
2. For the other triangle y² + y² = 45, so y² = 22.5.
3. y = √22.5 = 3√2.5.

Teacher's Note:
1. Both triangles use the same hypotenuse.
2. 9 × 2.5 = 22.5, so √22.5 = 3√2.5.

Answer: (c) 60 cm²

Explanation:
1. Area of rectangle = 10 × 8 = 80 cm².
2. Area of △ADE = ½ × 10 × 4 = 20 cm².
3. Shaded area = 80 − 20 = 60 cm².

Teacher's Note:
1. Subtract the unshaded part from the whole.
2. The triangle area is independent of where E is on its line.

Answer: (a) 1500√3 cm²

Explanation:
1. Sides = 60 cm, 100 cm and 140 cm; s = 150 cm.
2. Area = √(150 × 90 × 50 × 10) = √6750000 = 1500√3 cm².

Teacher's Note:
1. Share the perimeter in the given ratio.
2. 6750000 = 2250000 × 3.

Answer: (d) 144

Explanation:
1. That side is a diameter of length 24, so the angle opposite it is 90°.
2. The triangle is an isosceles right triangle with hypotenuse 24, so each leg is 12√2.
3. Area = ½ × 12√2 × 12√2 = 144.

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. Alternatively, area = ½ × base 24 × height 12.