CBSE Class 9 Mathematics Heron's Formula MCQs Set 12

Answer: (c) 21.6

Explanation:
1. AC = √(27² + 36²) = 45.
2. Area = ½ × 27 × 36 = ½ × 45 × BD, so BD = 27 × 36 ÷ 45 = 21.6.

Teacher's Note:
1. Equate the two expressions for the area.
2. Altitude to the hypotenuse = product of legs ÷ hypotenuse.

Answer: (c) 33

Explanation:
1. Ring between the squares: 7² − 5² = 24.
2. Innermost square: 3² = 9.
3. Shaded area = 24 + 9 = 33.

Teacher's Note:
1. Add the two shaded regions and leave the middle ring unshaded.
2. Differences of squares are quick: 49 − 25 = 24.

Answer: (c) 15 cm

Explanation:
1. Let the diagonals be d and 2d. Area = ½ × d × 2d = d² = 25, so d = 5.
2. Diagonals are 5 cm and 10 cm, sum = 15 cm.

Teacher's Note:
1. Area = ½ d1 d2.
2. Check: ½ × 5 × 10 = 25.

Answer: (c) 75 m

Explanation:
1. ½ × (sum of parallel sides) × 24 = 1440, so the sum is 120 m.
2. Longer side = 120 × 5/8 = 75 m.

Teacher's Note:
1. Find the sum of the parallel sides first.
2. Check: 75 + 45 = 120.

Answer: (d) 64√3 cm²

Explanation:
1. Diagonal = side × √2, so the side of the square is 12 cm and its perimeter is 48 cm.
2. The side of the triangle is 48 ÷ 3 = 16 cm.
3. Area = (√3/4) × 16² = 64√3 cm².

Teacher's Note:
1. Find the common perimeter first.
2. 256 ÷ 4 = 64.

Answer: (b) 336 m²

Explanation:
1. The diagonal divides it into two congruent triangles with sides 30, 14, 40; s = 42.
2. Area of one triangle = √(42 × 12 × 28 × 2) = 168 m².
3. Parallelogram = 2 × 168 = 336 m².

Teacher's Note:
1. Remember to double the triangle's area.
2. 42 × 12 × 28 × 2 = 28224 = 168².

Answer: (a) 4.5 m²

Explanation:
1. Let each leg be a; the hypotenuse is a√2, so 2a + a√2 = 6 + 3√2 = 3(2 + √2).
2. So a = 3 m and the area = ½ × 3 × 3 = 4.5 m².

Teacher's Note:
1. Factor the perimeter to compare with a(2 + √2).
2. Check: 6 + 3√2 = 3(2 + √2).

Answer: (d) 4 : 1

Explanation:
1. Area of a square = d²/2, so the area is proportional to the square of the diagonal.
2. Doubling the diagonal multiplies the area by 4, giving 4 : 1.

Teacher's Note:
1. Sides also double when the diagonals double.
2. Area scales as the square of the length.

Answer: (c) A > B

Explanation:
1. Let the common semi-perimeter be s. Square: A = (s/2)². Rectangle with sides x and s − x: B = x(s − x).
2. A − B = (s/2 − x)² > 0 when x ≠ s/2.
3. So A > B.

Teacher's Note:
1. Among all rectangles of a given perimeter the square has the maximum area.
2. If the rectangle were also a square the areas would be equal.

Answer: (c) 20000 m²

Explanation:
1. Diagonal = 4 km/h × (3/60) h = 0.2 km = 200 m.
2. Area = d²/2 = 40000 ÷ 2 = 20000 m².

Teacher's Note:
1. Convert km/h and minutes carefully.
2. Area of a square = d²/2.

Answer: (b) 25 m

Explanation:
1. In right △ADC, AC² = 7² + 24² = 625.
2. AC = 25 m.

Teacher's Note:
1. 7, 24, 25 is a Pythagorean triplet.
2. The same field is used in the next two questions.

Answer: (a) (12/25)(14 + 13√14) m

Explanation:
1. DF = 7 × 24 ÷ 25 = 168/25 m (altitude to the hypotenuse of △ADC).
2. In △ABC, s = 39, area = √(39 × 13 × 12 × 14) = 78√14 m², so BE = 2 × 78√14 ÷ 25 = 156√14/25 m.
3. BE + DF = (156√14 + 168)/25 = (12/25)(13√14 + 14) m.

Teacher's Note:
1. Use area = ½ × AC × altitude for each triangle.
2. Factor out 12/25 to match the option.

Answer: (b) 6(14 + 13√14) m²

Explanation:
1. ar(△ADC) = ½ × 7 × 24 = 84 m².
2. ar(△ABC) = √(39 × 13 × 12 × 14) = 78√14 m².
3. Total = 84 + 78√14 = 6(14 + 13√14) m².

Teacher's Note:
1. Add the two triangle areas.
2. 84 = 6 × 14 and 78 = 6 × 13.

Answer: (a) 36 cm²

Explanation:
1. Area = 9 × 4 = 36 cm².

Teacher's Note:
1. Area of a rectangle = length × breadth.
2. Do not confuse it with the perimeter 26 cm.

Answer: (c) 9√3 cm²

Explanation:
1. Side = 6 cm.
2. Area = (√3/4) × 36 = 9√3 cm².

Teacher's Note:
1. First find the side from the perimeter.
2. (√3/4) × 36 = 9√3.

Answer: (a) 36 cm²

Explanation:
1. Side = diagonal ÷ √2 = 6 cm.
2. Area = 36 cm² (also d²/2 = 72/2 = 36).

Teacher's Note:
1. Area of a square = d²/2.
2. The diagonal is side × √2.

Answer: (a) 4.5 cm

Explanation:
1. Altitude = (√3/2) × 3√3 = 9/2 = 4.5 cm.

Teacher's Note:
1. The altitude of an equilateral triangle of side a is (√3/2)a.
2. √3 × √3 = 3.

Answer: (a) 300 m

Explanation:
1. Area = 332.10 ÷ 24.60 = 13.5 hectares = 135000 m².
2. ½ × 3h × h = 135000, so h² = 90000 and h = 300 m (base 900 m).

Teacher's Note:
1. 1 hectare = 10000 m².
2. Check: ½ × 900 × 300 = 135000.

Answer: (a) 104 cm²

Explanation:
1. Area = base × height = 16 × 6.5 = 104 cm².

Teacher's Note:
1. The distance between opposite sides is the altitude.
2. Area = base × altitude.

Answer: (a) 24 cm

Explanation:
1. 10² + 24² = 26², so the triangle is right-angled with area ½ × 10 × 24 = 120 cm².
2. The longest altitude falls on the shortest side 10 cm: 2 × 120 ÷ 10 = 24 cm.

Teacher's Note:
1. Longest altitude is on the shortest side.
2. The altitude on the side 10 equals the other leg 24.