CBSE Class 9 Mathematics Heron's Formula MCQs Set 15

Multiple Choice Questions (MCQs) for Class 9 Mathematics: Heron's Formula

Review structured MCQ sets for Class 9 Mathematics Heron's Formula. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Heron's Formula MCQs for Class 9 Mathematics

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Multiple Choice Questions

Question 1: The area of a triangle with two sides \( 8 \) cm and \( 11 \) cm and perimeter \( 32 \) cm is
(a) \( 8\sqrt{30}\text{ cm}^2 \)
(b) \( 16\sqrt{30}\text{ cm}^2 \)
(c) \( 4\sqrt{30}\text{ cm}^2 \)
(d) \( 88\text{ cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 8\sqrt{30}\text{ cm}^2 \)

Explanation:
1. Third side \( =32-8-11=13 \) cm and \( s=16 \).
2. Area \( =\sqrt{16\cdot8\cdot5\cdot3}=\sqrt{1920}=8\sqrt{30} \) cm\( ^2 \).

Teacher's Note:
1. Find the third side from the perimeter.
2. Then use Heron's formula.

Question 2: The sides of a triangular plot are in the ratio \( 3:5:7 \) and its perimeter is \( 300 \) m. Its area is
(a) \( 750\sqrt3\text{ m}^2 \)
(b) \( 1500\sqrt3\text{ m}^2 \)
(c) \( 1500\text{ m}^2 \)
(d) \( 3000\sqrt3\text{ m}^2 \)
Show Answer & Explanation

Answer: (b) \( 1500\sqrt3\text{ m}^2 \)

Explanation:
1. Sides: \( 60,\ 100,\ 140 \) m, and \( s=150 \).
2. Area \( =\sqrt{150\cdot90\cdot50\cdot10}=\sqrt{6750000}=1500\sqrt3 \) m\( ^2 \).

Teacher's Note:
1. Divide the perimeter in the ratio 3:5:7.

Question 3: An isosceles triangle has perimeter \( 30 \) cm and each equal side \( 12 \) cm. Its area is
(a) \( 18\sqrt{15}\text{ cm}^2 \)
(b) \( 36\text{ cm}^2 \)
(c) \( 9\sqrt{15}\text{ cm}^2\ (\approx34.86\text{ cm}^2) \)
(d) \( 9\sqrt{5}\text{ cm}^2 \)
Show Answer & Explanation

Answer: (c) \( 9\sqrt{15}\text{ cm}^2\ (\approx34.86\text{ cm}^2) \)

Explanation:
1. Base \( =30-24=6 \) cm and \( s=15 \).
2. Area \( =\sqrt{15\cdot3\cdot3\cdot9}=9\sqrt{15}\approx34.86 \) cm\( ^2 \).

Teacher's Note:
1. Base is perimeter minus the two equal sides.

Question 4: The perimeter of an equilateral triangle whose area is \( 36\sqrt3 \) cm\( ^2 \) is
(a) \( 12 cm \)
(b) \( 24 cm \)
(c) \( 72 cm \)
(d) \( 36 cm \)
Show Answer & Explanation

Answer: (d) \( 36 cm \)

Explanation:
1. \( \frac{\sqrt3}4a^2=36\sqrt3 \) gives \( a^2=144 \), so \( a=12 \) cm.
2. Perimeter \( =3\times12=36 \) cm.

Teacher's Note:
1. Use area = (root 3 / 4) a squared.

Question 5: The base of a right triangle is \( 48 \) cm and its hypotenuse is \( 50 \) cm. Its area is
(a) \( 336\text{ cm}^2 \)
(b) \( 600\text{ cm}^2 \)
(c) \( 168\text{ cm}^2 \)
(d) \( 1200\text{ cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 336\text{ cm}^2 \)

Explanation:
1. Other side \( =\sqrt{50^2-48^2}=\sqrt{196}=14 \) cm.
2. Area \( =\frac12\times48\times14=336 \) cm\( ^2 \).

Teacher's Note:
1. Use the Pythagoras theorem.

Question 6: The height of an equilateral triangle is \( 6 \) cm. Its area is
(a) \( 6\sqrt3\text{ cm}^2 \)
(b) \( 12\sqrt3\text{ cm}^2 \)
(c) \( 36\sqrt3\text{ cm}^2 \)
(d) \( 18\text{ cm}^2 \)
Show Answer & Explanation

Answer: (b) \( 12\sqrt3\text{ cm}^2 \)

Explanation:
1. Height \( =\frac{\sqrt3}2a=6 \) gives \( a=4\sqrt3 \) cm.
2. Area \( =\frac{\sqrt3}4(4\sqrt3)^2=\frac{\sqrt3}4\cdot48=12\sqrt3 \) cm\( ^2 \).

Teacher's Note:
1. Find the side from the height.

Question 7: The area of an equilateral triangle is \( 81\sqrt3 \) cm\( ^2 \). Its height is
(a) \( 18\text{ cm} \)
(b) \( 9\text{ cm} \)
(c) \( 9\sqrt3\text{ cm} \)
(d) \( 18\sqrt3\text{ cm} \)
Show Answer & Explanation

Answer: (c) \( 9\sqrt3\text{ cm} \)

Explanation:
1. \( \frac{\sqrt3}4a^2=81\sqrt3 \) gives \( a^2=324 \), \( a=18 \) cm.
2. Height \( =\frac{\sqrt3}2\times18=9\sqrt3 \) cm.

Teacher's Note:
1. Side first, then height.

Question 8: In \( \triangle ABC \), \( BC=8 \) cm, \( AC=15 \) cm and \( AB=17 \) cm. The altitude drawn on \( AB \) is about
(a) \( 7.5 cm \)
(b) \( 7.04 cm \)
(c) \( 8 cm \)
(d) \( 7.06 cm \)
Show Answer & Explanation

Answer: (d) \( 7.06 cm \)

Explanation:
1. \( 8^2+15^2=64+225=289=17^2 \), so the triangle is right-angled at \( C \) and its area is \( \frac12\times8\times15=60 \) cm\( ^2 \).
2. Altitude on \( AB \): \( \frac12\times17\times h=60 \), so \( h=\frac{120}{17}\approx7.06 \) cm.

Teacher's Note:
1. Check for a right angle first.
2. Area = half base times height on AB.

Question 9: The semi-perimeter of \( \triangle ABC \) exceeds the sides by \( 8 \) cm, \( 7 \) cm and \( 5 \) cm respectively. Its area is
(a) \( 20\sqrt{14}\text{ cm}^2 \)
(b) \( 40\sqrt{14}\text{ cm}^2 \)
(c) \( 20\sqrt{7}\text{ cm}^2 \)
(d) \( 280\text{ cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 20\sqrt{14}\text{ cm}^2 \)

Explanation:
1. \( (s-a)+(s-b)+(s-c)=3s-2s=s \), so \( s=8+7+5=20 \).
2. Area \( =\sqrt{20\cdot8\cdot7\cdot5}=\sqrt{5600}=20\sqrt{14} \) cm\( ^2 \).

Teacher's Note:
1. The three differences add up to s.

Question 10: The ratio of the area of a square to the area of the square drawn on its diagonal is
(a) \( 1 : 4 \)
(b) \( 1 : 2 \)
(c) \( 2 : 1 \)
(d) \( 1 : 3 \)
Show Answer & Explanation

Answer: (b) \( 1 : 2 \)

Explanation:
1. If the side is \( a \), the diagonal is \( a\sqrt2 \).
2. Areas: \( a^2 \) and \( 2a^2 \), ratio \( 1:2 \).

Teacher's Note:
1. Diagonal = side times root 2.

Question 11: The adjacent sides of a parallelogram are \( 24 \) cm and \( 32 \) cm. If the distance between the longer sides is \( 17.4 \) cm, the distance between the shorter sides is
(a) \( 13.05 cm \)
(b) \( 17.4 cm \)
(c) \( 23.2 cm \)
(d) \( 46.4 cm \)
Show Answer & Explanation

Answer: (c) \( 23.2 cm \)

Explanation:
1. Area \( =32\times17.4=556.8 \) cm\( ^2 \).
2. Distance between the shorter sides \( =\frac{556.8}{24}=23.2 \) cm.

Teacher's Note:
1. Area is the same whichever side is the base.

Question 12: The sides of a triangle are in the ratio \( 4:3:5 \) and its perimeter is \( 144 \) cm. The height on the longest side is
(a) \( 24 cm \)
(b) \( 36 cm \)
(c) \( 14.4 cm \)
(d) \( 28.8 cm \)
Show Answer & Explanation

Answer: (d) \( 28.8 cm \)

Explanation:
1. Sides \( 48,\ 36,\ 60 \); since \( 36^2+48^2=60^2 \) the triangle is right-angled and its area is \( \frac12\cdot36\cdot48=864 \).
2. \( \frac12\times60\times h=864 \), so \( h=28.8 \) cm.

Teacher's Note:
1. Check for a right angle.

Question 13: The area of a triangle with base \( 12 \) cm and height \( 5 \) cm is
(a) \( 30\text{ cm}^2 \)
(b) \( 60\text{ cm}^2 \)
(c) \( 17\text{ cm}^2 \)
(d) \( 15\text{ cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 30\text{ cm}^2 \)

Explanation:
1. Area \( =\frac12\times12\times5=30 \) cm\( ^2 \).

Teacher's Note:
1. Half base times height.

Question 14: The area of a triangle whose sides are \( 9 \) cm, \( 12 \) cm and \( 15 \) cm is
(a) \( 108\text{ cm}^2 \)
(b) \( 54\text{ cm}^2 \)
(c) \( 67.5\text{ cm}^2 \)
(d) \( 45\text{ cm}^2 \)
Show Answer & Explanation

Answer: (b) \( 54\text{ cm}^2 \)

Explanation:
1. \( 9^2+12^2=15^2 \), so it is a right triangle with area \( \frac12\cdot9\cdot12=54 \).
2. (Heron's formula: \( s=18 \), \( \sqrt{18\cdot9\cdot6\cdot3}=54 \).)

Teacher's Note:
1. Spot the Pythagorean triple.

Question 15: The area of a triangle with sides \( 20 \) cm, \( 16 \) cm and \( 12 \) cm is
(a) \( 192\text{ cm}^2 \)
(b) \( 120\text{ cm}^2 \)
(c) \( 96\text{ cm}^2 \)
(d) \( 48\text{ cm}^2 \)
Show Answer & Explanation

Answer: (c) \( 96\text{ cm}^2 \)

Explanation:
1. \( 12^2+16^2=20^2 \), a right triangle.
2. Area \( =\frac12\times12\times16=96 \) cm\( ^2 \).

Teacher's Note:
1. Check whether the sides form a Pythagorean triple.

Question 16: The base of an isosceles triangle is \( 6 \) cm and each equal side is \( 5 \) cm. Its height is
(a) \( 3 cm \)
(b) \( 4.5 cm \)
(c) \( 5 cm \)
(d) \( 4 cm \)
Show Answer & Explanation

Answer: (d) \( 4 cm \)

Explanation:
1. The height bisects the base, so \( h=\sqrt{5^2-3^2}=4 \) cm.

Teacher's Note:
1. The altitude meets the base at its midpoint.

Question 17: Each equal side of an isosceles right triangle is \( 10 \) cm. Its area is
(a) \( 50\text{ cm}^2 \)
(b) \( 100\text{ cm}^2 \)
(c) \( 25\text{ cm}^2 \)
(d) \( 50\sqrt2\text{ cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 50\text{ cm}^2 \)

Explanation:
1. The equal sides are perpendicular, so area \( =\frac12\times10\times10=50 \) cm\( ^2 \).

Teacher's Note:
1. The legs are base and height.

Question 18: The perimeter of a right triangle is \( 450 \) m and its sides are in the ratio \( 13:12:5 \). Its area is
(a) \( 13500\text{ m}^2 \)
(b) \( 6750\text{ m}^2 \)
(c) \( 3375\text{ m}^2 \)
(d) \( 2700\text{ m}^2 \)
Show Answer & Explanation

Answer: (b) \( 6750\text{ m}^2 \)

Explanation:
1. Sides \( 195,\ 180,\ 75 \) (sum 450).
2. The legs are \( 180 \) and \( 75 \): area \( =\frac12\times180\times75=6750 \) m\( ^2 \).

Teacher's Note:
1. The side with ratio 13 is the hypotenuse.

Question 19: The area of an isosceles triangle with equal sides \( 16 \) cm and third side \( 10 \) cm is
(a) \( 10\sqrt{231}\text{ cm}^2 \)
(b) \( 5\sqrt{21}\text{ cm}^2 \)
(c) \( 5\sqrt{231}\text{ cm}^2 \)
(d) \( 80\text{ cm}^2 \)
Show Answer & Explanation

Answer: (c) \( 5\sqrt{231}\text{ cm}^2 \)

Explanation:
1. \( s=21 \), so area \( =\sqrt{21\cdot5\cdot5\cdot11}=5\sqrt{231} \) cm\( ^2 \).

Teacher's Note:
1. Heron's formula with s = 21.

Question 20: The perimeter of a right triangle is \( 144 \) cm and its hypotenuse is \( 65 \) cm. Its area is
(a) \( 1008\text{ cm}^2 \)
(b) \( 252\text{ cm}^2 \)
(c) \( 630\text{ cm}^2 \)
(d) \( 504\text{ cm}^2 \)
Show Answer & Explanation

Answer: (d) \( 504\text{ cm}^2 \)

Explanation:
1. Legs \( a,b \) satisfy \( a+b=79 \) and \( a^2+b^2=4225 \).
2. \( 2ab=79^2-4225=2016 \), so \( ab=1008 \) and the legs are \( 63 \) and \( 16 \).
3. Area \( =\frac12\times1008=504 \) cm\( ^2 \).

Teacher's Note:
1. Use (a + b) squared = a squared + b squared + 2ab.

Question 21: An isosceles right triangle has area \( 200 \) cm\( ^2 \). Its hypotenuse is
(a) \( 20\sqrt2\text{ cm} \)
(b) \( 20\text{ cm} \)
(c) \( 40\text{ cm} \)
(d) \( 10\sqrt2\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 20\sqrt2\text{ cm} \)

Explanation:
1. \( \frac12a^2=200 \) gives \( a=20 \) cm.
2. Hypotenuse \( =20\sqrt2 \) cm.

Teacher's Note:
1. Equal legs a: area is a squared over 2.

Question 22: The area of an equilateral triangle of side \( 12 \) cm is
(a) \( 72\sqrt3\text{ cm}^2 \)
(b) \( 36\sqrt3\text{ cm}^2 \)
(c) \( 36\text{ cm}^2 \)
(d) \( 144\sqrt3\text{ cm}^2 \)
Show Answer & Explanation

Answer: (b) \( 36\sqrt3\text{ cm}^2 \)

Explanation:
1. Area \( =\frac{\sqrt3}4a^2=\frac{\sqrt3}4\cdot144=36\sqrt3 \) cm\( ^2 \).

Teacher's Note:
1. Use the formula for an equilateral triangle.

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