CBSE Class 9 Mathematics Heron's Formula MCQs Set 22

Answer: (a) 5 cm

Explanation:
1. Let base = x cm, so altitude = (x + 7) cm and ½x(x + 7) = 30.
2. x² + 7x − 60 = 0, so (x − 5)(x + 12) = 0 and x = 5 (length cannot be negative).

Teacher's Note:
1. Reject the negative root of the quadratic.

Answer: (b) 12 cm

Explanation:
1. Base = 5 cm from ½x(x + 7) = 30.
2. Altitude = 5 + 7 = 12 cm. Check: ½ × 5 × 12 = 30.

Teacher's Note:
1. Always verify with the area formula.

Answer: (c) 40 m

Explanation:
1. Area = 900 ÷ 45 × 100 = 2000 m².
2. Base = 5h/2, so area = ½ × (5h/2) × h = 5h²/4 = 2000, h² = 1600 and h = 40 m.

Teacher's Note:
1. Turn the cost into area first.

Answer: (d) 29 cm

Explanation:
1. Joining the point to the vertices splits the triangle into three; the sum of areas gives ½ × side × (8 + 10 + 11) = ½ × side × height.
2. So the height equals 8 + 10 + 11 = 29 cm.

Teacher's Note:
1. In an equilateral triangle the sum of the perpendiculars from an interior point equals the height.

Answer: (a) \( 485.6 \text{ cm}^2 \)

Explanation:
1. Height h = 29 cm. For an equilateral triangle, h = (√3/2)a, so a = 58/√3.
2. Area = (√3/4)a² = h²/√3 = 841/1.732 ≈ 485.6 cm².

Teacher's Note:
1. Area = h²/√3 for an equilateral triangle of height h.

Answer: (b) 60 cm

Explanation:
1. Let the sides be x and x − 14: ½x(x − 14) = 120, so x² − 14x − 240 = 0 and x = 24.
2. Sides 24 and 10, hypotenuse = √(576 + 100) = 26. Perimeter = 24 + 10 + 26 = 60 cm.

Teacher's Note:
1. 24-10-26 is a multiple of the 5-12-13 triple.

Answer: (c) 300%

Explanation:
1. Semi-perimeter becomes 2s and each factor (s − a) becomes 2(s − a).
2. Area becomes √(16 s(s−a)(s−b)(s−c)) = 4 × area, an increase of 3 times the original, i.e. 300%.

Teacher's Note:
1. Area scales with the square of the scale factor.

Answer: (d) \( 1000\sqrt6 \text{ cm}^2 \)

Explanation:
1. s = 60 cm, so area of one piece = √(60 × 40 × 10 × 10) = 200√6 cm².
2. Each colour has 5 pieces: 5 × 200√6 = 1000√6 cm².

Teacher's Note:
1. Colours alternate, so each colour uses 5 of the 10 pieces.

Answer: (a) becomes four times

Explanation:
1. Area is proportional to the square of the linear scale.
2. Doubling the sides gives 2² = 4 times the area.

Teacher's Note:
1. Heron's formula confirms the factor 4.

Answer: (b) 3 : 1

Explanation:
1. The garden has 4 times the area of the bed.
2. Path = garden − bed = 4 − 1 = 3 parts, so path : bed = 3 : 1.

Teacher's Note:
1. Similar figures: area ratio is the square of the side ratio.

Answer: (c) 216 cm²

Explanation:
1. 18² + 24² = 324 + 576 = 900 = 30², so the triangle is right-angled.
2. Area = ½ × 18 × 24 = 216 cm² (Heron gives the same: s = 36, √(36 × 18 × 12 × 6) = 216).

Teacher's Note:
1. Check for a right angle before using Heron's formula.

Answer: (d) 24 cm

Explanation:
1. Area = 216 cm².
2. Altitude on 18 cm = 2 × 216 ÷ 18 = 24 cm.

Teacher's Note:
1. The smallest side has the largest altitude.

Answer: (a) 120 cm²

Explanation:
1. 10² + 24² = 26², so it is a right triangle.
2. Area = ½ × 10 × 24 = 120 cm².

Teacher's Note:
1. 10-24-26 is twice the triple 5-12-13.

Answer: (b) 24 cm

Explanation:
1. Area = 120 cm². The longest altitude is on the shortest side, 10 cm.
2. Altitude = 2 × 120 ÷ 10 = 24 cm.

Teacher's Note:
1. Altitude is inversely proportional to its side.

Answer: (c) \( 2772 \text{ m}^2 \)

Explanation:
1. Third side = 324 − 85 − 154 = 85 m and s = 162 m.
2. Area = √(162 × 77 × 8 × 77) = 77 × 36 = 2772 m².

Teacher's Note:
1. Here 162 × 8 = 1296 = 36².

Answer: (d) 36 m

Explanation:
1. Area = ½ × 154 × h = 2772.
2. h = 2 × 2772 ÷ 154 = 36 m.

Teacher's Note:
1. The triangle is isosceles, so the foot is the midpoint of the 154 m side.

Answer: (a) \( \text{Rs. } 1219.68 \)

Explanation:
1. s = 231 m. Area = √(231 × 66 × 88 × 77) = 10164 m².
2. Cost = 10164 × 0.12 = Rs. 1219.68.

Teacher's Note:
1. Convert paise to rupees: 12 paise = Rs. 0.12.

Answer: (b) 162 cm

Explanation:
1. Height = 2 × 360 ÷ 80 = 9 cm; half of base = 40 cm.
2. Equal side = √(40² + 9²) = 41 cm. Perimeter = 80 + 41 + 41 = 162 cm.

Teacher's Note:
1. Drop the altitude to the base to get a right triangle.

Answer: (c) 12 cm

Explanation:
1. Let each equal side = a, so the base = 3a/2 and 2a + 3a/2 = 42.
2. 7a/2 = 42, a = 12 cm (base = 18 cm).

Teacher's Note:
1. Write the perimeter in terms of one unknown.

Answer: (d) \( 71.42 \text{ cm}^2 \)

Explanation:
1. s = 21 cm. Area = √(21 × 9 × 9 × 3) = 9√63 = 27√7.
2. 27 × 2.6458 ≈ 71.43 cm², about 71.4 cm².

Teacher's Note:
1. Two of the factors (s − a) are equal, so the root simplifies.

Answer: (a) 8 cm and 15 cm

Explanation:
1. Sum of the two legs = 40 − 17 = 23 and x² + y² = 289.
2. (x + y)² − 2xy = 289 gives xy = 120, so the legs are the roots of t² − 23t + 120 = 0: 8 and 15.

Teacher's Note:
1. 8-15-17 is a Pythagorean triple.

Answer: (b) 60 cm²

Explanation:
1. The legs are 8 cm and 15 cm.
2. Area = ½ × 8 × 15 = 60 cm².

Teacher's Note:
1. xy = 120 was found from the perimeter condition.

Answer: (c) 50 cm²

Explanation:
1. Area = ½ × 10 × 10 = 50 cm².

Teacher's Note:
1. Area = a²/2 for legs a.

Answer: (d) 34.14 cm

Explanation:
1. Hypotenuse = 10√2 = 14.14 cm.
2. Perimeter = 10 + 10 + 14.14 = 34.14 cm.

Teacher's Note:
1. Perimeter = a(2 + √2).

Answer: (a) 6 minutes

Explanation:
1. Area = 80 000 m², so the diagonal d satisfies d²/2 = 80 000, d = 400 m.
2. Time = 0.4 km ÷ 4 km/h = 0.1 h = 6 minutes.

Teacher's Note:
1. Area of a square = d²/2.

Answer: (b) 48 m²

Explanation:
1. Half of the long diagonal = 24 m, so half of the other = √(900 − 576) = 18 m and the short diagonal = 36 m.
2. Area = ½ × 48 × 36 = 864 m². Per cow = 864 ÷ 18 = 48 m².

Teacher's Note:
1. Rhombus area = ½ × product of diagonals.