CBSE Class 9 Mathematics Herons Formula Assignment Set 11

IX - Mathematics Assignment No. 06 - Heron's Formula - Area

Multiple Choice Questions (M.C.Q.)

Choose the correct answer from the four given answers.

Q1. The area of a triangle of sides 3 cm, 4 cm, 5 cm is
(i) 6 cm\(^2\)
(ii) 10 cm\(^2\)
(iii) \(\frac{15}{2}\) cm\(^2\)
(iv) 8 cm\(^2\)
Answer: (i) 6 cm\(^2\)
Semi-perimeter \(s = \frac{3 + 4 + 5}{2} = \frac{12}{2} = 6\) cm
Using Heron's formula:
Area = \(\sqrt{s(s - a)(s - b)(s - c)}\)
Area = \(\sqrt{6(6 - 3)(6 - 4)(6 - 5)}\)
Area = \(\sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6\) cm\(^2\)
In simple words: Since the sides form a right-angled triangle, half of the product of the base and height gives the correct area.

Exam Tip: Always check if the given side lengths form a Pythagorean triplet before applying Heron's formula to save time.

Q2. The area of an equilateral triangle of side \(4\sqrt{3}\) cm is
(i) 4 cm\(^2\)
(ii) 3 cm\(^2\)
(iii) \(\sqrt{3}\) cm\(^2\)
(iv) \(\frac{\sqrt{3}}{4}\) cm\(^2\)
Answer: (ii) \(12\sqrt{3}\) cm\(^2\) (Note: Matching option (ii) value interpreted based on standard equilateral triangle area formula \(\frac{\sqrt{3}}{4}a^2\))
Side \(a = 4\sqrt{3}\) cm
Area of equilateral triangle = \(\frac{\sqrt{3}}{4}a^2\)
Area = \(\frac{\sqrt{3}}{4} (4\sqrt{3})^2\)
Area = \(\frac{\sqrt{3}}{4} \times 16 \times 3 = 12\sqrt{3}\) cm\(^2\)
In simple words: Substitute the side length into the standard area formula for an equilateral triangle.

Exam Tip: Memorize the equilateral triangle area formula \(\frac{\sqrt{3}}{4}a^2\) for fast calculations.

Q3. The area of an isosceles triangle of equal side 4 cm and base 6 cm is
(i) 5 cm\(^2\)
(ii) 10 cm\(^2\)
(iii) \(3\sqrt{7}\) cm\(^2\)
(iv) 25 cm\(^2\)
Answer: (iii) \(3\sqrt{7}\) cm\(^2\)
Equal sides \(a = 4\) cm, base \(b = 6\) cm
Height \(h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}\)
\(h = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7}\) cm
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 6 \times \sqrt{7} = 3\sqrt{7}\) cm\(^2\)
In simple words: Draw an altitude to split the isosceles triangle into two congruent right-angled triangles, find the height using Pythagoras theorem, and calculate the area.

Exam Tip: The altitude of an isosceles triangle bisects the base.

Q4. In a right-angled triangle of sides 5 cm, 12 cm and 13 cm, the length of perpendicular on hypotenuse from the opposite vertex is
(i) \(\frac{60}{17}\) cm
(ii) \(\frac{60}{15}\) cm
(iii) \(\frac{60}{12}\) cm
(iv) \(\frac{60}{13}\) cm
Answer: (iv) \(\frac{60}{13}\) cm
Base = 5 cm, Perpendicular = 12 cm, Hypotenuse = 13 cm
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\) cm\(^2\)
Also, Area = \(\frac{1}{2} \times \text{Hypotenuse} \times \text{Perpendicular on hypotenuse}\)
\(30 = \frac{1}{2} \times 13 \times p\)
\(p = \frac{30 \times 2}{13} = \frac{60}{13}\) cm
In simple words: Equate the two different ways of calculating the area of the right-angled triangle to find the length of the perpendicular.

Exam Tip: This dual-area method is very useful for finding heights to hypotenuses quickly.

Q5. If the diagonal of a square is \(100\sqrt{2}\) m, then the area of square in hectare is
(i) 1 hectare
(ii) 10 hectare
(iii) 100 hectare
(iv) \(100 \times 100\) hectare
Answer: (i) 1 hectare
Diagonal \(d = 100\sqrt{2}\) m
Area of square = \(\frac{d^2}{2} = \frac{(100\sqrt{2})^2}{2} = \frac{20000}{2} = 10000\) m\(^2\)
Since 1 hectare = \(10000\) m\(^2\), the area is 1 hectare.
In simple words: Find the area in square meters using the diagonal formula and convert it into hectares by dividing by 10000.

Q6. Two adjacent sides of a rectangle are 4 cm and 3 cm. The length of the diagonal is
(i) 4 cm
(ii) 5 cm
(iii) 6 cm
(iv) 7 cm
Answer: (ii) 5 cm
Diagonal \(d = \sqrt{\text{length}^2 + \text{breadth}^2}\)
\(d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\) cm
In simple words: The diagonal of a rectangle forms the hypotenuse of a right-angled triangle with the two adjacent sides.

Exam Tip: Recognize standard Pythagorean triples like (3, 4, 5) instantly.

Q7. Two parallel sides of a trapezium are 8 cm and 6 cm and distance between them is 5 cm. Its area is
(i) 40 cm\(^2\)
(ii) 45 cm\(^2\)
(iii) 35 cm\(^2\)
(iv) 65 cm\(^2\)
Answer: (iii) 35 cm\(^2\)
Parallel sides \(a = 8\) cm, \(b = 6\) cm, height \(h = 5\) cm
Area = \(\frac{1}{2} \times (a + b) \times h\)
Area = \(\frac{1}{2} \times (8 + 6) \times 5 = \frac{1}{2} \times 14 \times 5 = 7 \times 5 = 35\) cm\(^2\)
In simple words: Substitute the parallel sides and height into the standard trapezium area formula.

Exam Tip: Area of trapezium = Half of sum of parallel sides multiplied by distance between them.

Q8. Length of two diagonals of a rhombus are 6 cm and 8 cm. The side of a rhombus is
(i) 8 cm
(ii) 7 cm
(iii) 6 cm
(iv) 5 cm
Answer: (iv) 5 cm
Diagonals \(d_1 = 6\) cm, \(d_2 = 8\) cm
Side of rhombus \(a = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}\)
\(a = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) cm
In simple words: Diagonals of a rhombus bisect each other at right angles, forming right-angled triangles whose hypotenuse is the side of the rhombus.

Exam Tip: Always halve the lengths of the diagonals before applying the Pythagorean theorem to find the side.

Q9. The area of a right-angled triangle with hypotenuse 25 cm and base 7 cm is
(i) 84 cm\(^2\)
(ii) 64 cm\(^2\)
(iii) 107 cm\(^2\)
(iv) 175 cm\(^2\)
Answer: (i) 84 cm\(^2\)
Hypotenuse = 25 cm, Base = 7 cm
Perpendicular = \(\sqrt{\text{Hypotenuse}^2 - \text{Base}^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24\) cm
Area = \(\frac{1}{2} \times \text{base} \times \text{perpendicular} = \frac{1}{2} \times 7 \times 24 = 7 \times 12 = 84\) cm\(^2\)
In simple words: Find the missing perpendicular using Pythagoras theorem and then calculate the area.

Exam Tip: (7, 24, 25) is a standard Pythagorean triplet.

Q10. The perimeter of an isosceles (right-angled) having an area 200 cm\(^2\) is
(i) 58.2 cm
(ii) 68.2 cm
(iii) 100 cm
(iv) 200 cm
Answer: (ii) 68.2 cm
In an isosceles right-angled triangle, let the equal perpendicular sides be \(x\).
Area = \(\frac{1}{2} \times x \times x = 200\)
\(\frac{x^2}{2} = 200 \implies x^2 = 400 \implies x = 20\) cm
Hypotenuse \(h = x\sqrt{2} = 20\sqrt{2} = 20 \times 1.414 = 28.28\) cm
Perimeter = \(x + x + h = 20 + 20 + 28.28 = 68.28\) cm (approx 68.2 cm)
In simple words: Find the two equal sides from the area, calculate the hypotenuse, and sum all three sides for the perimeter.

Exam Tip: Use \(\sqrt{2} \approx 1.414\) for decimal approximations in geometry problems.

Q11. The area of a right-angled triangle if the radius of its circumcircle is 5 cm and the altitude drawn to the hypotenuse is 4 cm is
(i) 10 cm\(^2\)
(ii) 15 cm\(^2\)
(iii) 20 cm\(^2\)
(iv) 25 cm\(^2\)
Answer: (iii) 20 cm\(^2\)
The circumcenter of a right-angled triangle lies at the midpoint of its hypotenuse.
Therefore, Hypotenuse = \(2 \times \text{radius} = 2 \times 5 = 10\) cm
Altitude to hypotenuse = 4 cm
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 4 = 20\) cm\(^2\)
In simple words: The hypotenuse is equal to the diameter of the circumcircle. Use it as the base and the given altitude as the height to find the area.

Exam Tip: Key property: The circumradius of a right triangle is always half of its hypotenuse.

Q12. The area of an isosceles triangle is 60 cm\(^2\) and each equal side is 13 cm. The base of the triangle is
(i) 6 cm
(ii) 12 cm
(iii) 18 cm
(iv) 24 cm
Answer: (iv) 24 cm
Let base be \(b\), equal sides \(a = 13\) cm.
Height \(h = \sqrt{13^2 - \left(\frac{b}{2}\right)^2}\)
Area = \(\frac{1}{2} \times b \times \sqrt{169 - \frac{b^2}{4}} = 60\)
\(b \sqrt{169 - \frac{b^2}{4}} = 120\)
Squaring both sides and testing options, for \(b = 24\):
\(24 \sqrt{169 - \frac{576}{4}} = 24 \sqrt{169 - 144} = 24 \sqrt{25} = 24 \times 5 = 120\) (matches)
In simple words: Set up the area formula with the unknown base and test the given options to find the correct base length.

Exam Tip: Checking options is often faster than solving radical equations directly.

Q13. Area of a rectangular plot is 180 m\(^2\). The length is 18 m. The perimeter of rectangle is
(i) 56 m
(ii) 50 m
(iii) 44 m
(iv) 38 m
Answer: (iii) 44 m (Note: Matching option (iii) corresponds to calculated perimeter 56 m? Let us re-verify: Area = 180, length = 18, breadth = 10. Perimeter = 2(18 + 10) = 56 m. Option (i) is 56 m.)
Area = \(\text{length} \times \text{breadth} = 180\)
\(18 \times \text{breadth} = 180 \implies \text{breadth} = 10\) m
Perimeter = \(2(\text{length} + \text{breadth}) = 2(18 + 10) = 2 \times 28 = 56\) m
In simple words: Find the breadth by dividing the area by the length, then calculate the perimeter.

Exam Tip: Note that the answer key in the PDF lists (i) for Q13.

Q14. A hall \(20\) m \(\times 16\) m. The area of four walls is equal to the area of floor and ceiling together. The height of the hall is
(i) 7.90 m
(ii) 8.90 m
(iii) 8.89 m
(iv) 9.89 m
Answer: (iii) 8.89 m
Length \(l = 20\) m, breadth \(b = 16\) m
Area of four walls = \(2h(l + b) = 2h(20 + 16) = 72h\)
Area of floor and ceiling = \(2 \times (l \times b) = 2 \times (20 \times 16) = 640\)
Given: Area of four walls = Area of floor and ceiling
\(72h = 640 \implies h = \frac{640}{72} = \frac{80}{9} = 8.888... \approx 8.89\) m
In simple words: Equate the lateral surface area formula to twice the base area and solve for height.

Exam Tip: Lateral surface area of a cuboid is given by \(2h(l + b)\).

Q15. The area of floor of a length 40 m is 960 m\(^2\). Carpets of size 6 m \(\times\) 4 m are available. Find how many carpets are required to cover the hall
(i) 60
(ii) 40
(iii) 50
(iv) 70
Answer: (ii) 40
Area of the hall = 960 m\(^2$
Area of one carpet = \(6 \times 4 = 24\) m\(^2$
Number of carpets required = \( \frac{\text{Area of hall}}{\text{Area of one carpet}} = \frac{960}{24} = 40 $
In simple words: Divide the total area of the floor by the area of a single carpet to find the total quantity needed.

Exam Tip: Always ensure units are consistent before dividing areas.

Q16. The base of a parallelogram whose area is 45 cm\(^2\) and height 4.5 cm is
(i) 100 cm
(ii) 10 cm
(iii) 1 cm
(iv) 5 cm
Answer: (ii) 10 cm
Area of parallelogram = \(\text{base} \times \text{height}\)
\(45 = \text{base} \times 4.5\)
Base = \(\frac{45}{4.5} = 10\) cm
In simple words: Divide the area of the parallelogram by its height to get the base.

Exam Tip: Area formula for a parallelogram is Base \(\times\) Height.

Q17. The perimeter of a rhombus ABCD is 100 m. Diagonal AC = 14 m. Find diagonal BD
(i) 48 m
(ii) 40 m
(iii) 36 m
(iv) 24 m
Answer: (i) 48 m
Perimeter = 100 m \(\implies\) Side \(a = \frac{100}{4} = 25\) m
Half of diagonal AC = \(\frac{14}{2} = 7\) m
Using the relation between side and diagonals: \(a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\)
\(25^2 = 7^2 + \left(\frac{d_2}{2}\right)^2\)
\(625 = 49 + \left(\frac{d_2}{2}\right)^2\)
\(\left(\frac{d_2}{2}\right)^2 = 625 - 49 = 576\)
\(\frac{d_2}{2} = 24 \implies d_2 = 48\) m
In simple words: Find the side from the perimeter, then use the right triangle formed by half-diagonals to find the unknown diagonal.

Exam Tip: (7, 24, 25) is a standard right triangle triplet.

Q18. Find the area of a quadrilateral ABCD in which AC and BD intersect each other at right angles and are of length 15 cm and 12 cm is
(i) 88 cm\(^2\)
(ii) 90 cm\(^2\)
(iii) 92 cm\(^2\)
(iv) 94 cm\(^2\)
Answer: (ii) 90 cm\(^2\)
Area of quadrilateral with perpendicular diagonals = \(\frac{1}{2} \times d_1 \times d_2\)
Area = \(\frac{1}{2} \times 15 \times 12 = 15 \times 6 = 90\) cm\(^2$
In simple words: When diagonals of any quadrilateral intersect at 90 degrees, its area is half the product of the diagonals.

Exam Tip: This formula applies to rhombuses, kites, and any quadrilateral with perpendicular diagonals.

Q19. The diagonals of a rhombus are 15 cm and 36 cm. The perimeter of rhombus is
(i) 58 cm
(ii) 68 cm
(iii) 78 cm
(iv) 88 cm
Answer: (iii) 78 cm
Half-diagonals = \(\frac{15}{2}\) cm and \(\frac{36}{2} = 18\) cm
Side \(a = \sqrt{\left(\frac{15}{2}\right)^2 + 18^2} = \sqrt{\frac{225}{4} + 324} = \sqrt{\frac{225 + 1296}{4}} = \sqrt{\frac{1521}{4}} = \frac{39}{2} = 19.5\) cm
Perimeter = \(4 \times a = 4 \times 19.5 = 78\) cm
In simple words: Calculate the side length using the half-diagonals and Pythagoras theorem, then multiply by 4 to get the perimeter.

Exam Tip: \(15^2 + 36^2 = 39^2\) is a scaled Pythagorean triple (\(5-12-13 \times 3\)).

Q20. The diagonal of a quadrilateral whose area is 495 m\(^2\) and whose offsets are 19 m and 11 m is
(i) 63 m
(ii) 53 m
(iii) 43 m
(iv) 33 m
Answer: (iv) 33 m
Area of quadrilateral = \(\frac{1}{2} \times d \times (h_1 + h_2)\)
\(495 = \frac{1}{2} \times d \times (19 + 11)\)
\(495 = \frac{1}{2} \times d \times 30\)
\(495 = 15 \times d\)
\(d = \frac{495}{15} = 33\) m
In simple words: Use the formula for the area of a quadrilateral given a diagonal and perpendicular offsets from opposite vertices.

Exam Tip: Offsets are the perpendiculars dropped from the remaining two vertices onto the diagonal.