CBSE Class 9 Mathematics Herons Formula Assignment Set 13

IX - Mathematics Assignment No - 04 - Heron's Formula - Area

Q1. The base of an isosceles triangle is 8 cm and its area is 12 cm^2. Find the length of each equal side except the base.
Answer: Let the length of each equal side of the isosceles triangle be \(a\) cm.
Base (\(b\)) = 8 cm
Area = 12 cm^2
The formula for the area of an isosceles triangle is:
\(\text{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}\)
Substitute the given values:
\(12 = \frac{8}{4} \sqrt{4a^2 - 8^2}\)
\(12 = 2 \sqrt{4a^2 - 64}\)
Divide both sides by 2:
\(6 = \sqrt{4a^2 - 64}\)
Squaring both sides:
\(36 = 4a^2 - 64\)
\(4a^2 = 36 + 64\)
\(4a^2 = 100\)
\(a^2 = 25\)
\(a = 5 \text{ cm}\)
In simple words: By using the area formula for an isosceles triangle, we set up an equation with the equal side as the unknown and solved for it to get 5 cm.

Exam Tip: Remember the special formula for the area of an isosceles triangle in terms of its base and equal sides to save time compared to using general Heron's formula.

Q2. Find the height of an equilateral triangle whose perimeter is 63 m. Also find its area.
Answer: Perimeter of the equilateral triangle = 63 m
Length of each side (\(a\)) = \(\frac{63}{3} = 21 \text{ m}\)
Height (\(h\)) of an equilateral triangle = \(\frac{\sqrt{3}}{2} a\)
\(h = \frac{\sqrt{3}}{2} \times 21 = \frac{21\sqrt{3}}{2} \approx 10.5 \times 1.732 = 18.186 \text{ m} \approx 18.18 \text{ m}\)
Area of the equilateral triangle = \(\frac{\sqrt{3}}{4} a^2\)
\(\text{Area} = \frac{\sqrt{3}}{4} \times (21)^2 = \frac{441\sqrt{3}}{4} = 110.25 \times 1.732 = 190.953 \text{ m}^2 \approx 190.95 \text{ m}^2\)
In simple words: First find the side length from the perimeter, then apply the standard formulas for the height and area of an equilateral triangle.

Exam Tip: Memorize the height formula (\(\frac{\sqrt{3}}{2} a\)) and area formula (\(\frac{\sqrt{3}}{4} a^2\)) for equilateral triangles for quick calculations.

Q3. The perimeter of a right triangle is 144 cm. Its hypotenuse is 65 cm. Find the length of the legs of this triangle.
Answer: Let the two legs of the right triangle be \(x\) and \(y\).
Hypotenuse (\(h\)) = 65 cm
Perimeter = \(x + y + \text{hypotenuse} = 144\)
\(x + y + 65 = 144\)
\(x + y = 144 - 65 = 79\)
\(y = 79 - x\) --- (Equation 1)
Using Pythagoras theorem:
\(x^2 + y^2 = 65^2\)
\(x^2 + y^2 = 4225\) --- (Equation 2)
Substitute Equation 1 into Equation 2:
\(x^2 + (79 - x)^2 = 4225\)
\(x^2 + 6241 - 158x + x^2 = 4225\)
\(2x^2 - 158x + 6241 - 4225 = 0\)
\(2x^2 - 158x + 2016 = 0\)
Divide by 2:
\(x^2 - 79x + 1008 = 0\)
Factorize the quadratic equation:
\(x^2 - 63x - 16x + 1008 = 0\)
\(x(x - 63) - 16(x - 63) = 0\)
\((x - 63)(x - 16) = 0\)
So, \(x = 63\) or \(x = 16\).
If \(x = 63\), then \(y = 79 - 63 = 16\).
Therefore, the lengths of the legs are 16 cm and 63 cm.
In simple words: We used the perimeter and Pythagoras theorem to set up simultaneous equations, which simplified to a quadratic equation giving the two leg lengths.

Exam Tip: When dealing with sum and squares of sides in right triangles, form a quadratic equation by substitution and solve by splitting the middle term.

Q4. The base of a triangular field is 2.5 times its height. If the cost of turfing it at Rs 35 per 100 m^2 is Rs 700, find its base.
Answer: Let the height of the triangular field be \(h\).
Base (\(b\)) = \(2.5h = \frac{5}{2}h\)
Total cost of turfing = Rs 700
Rate of turfing = Rs 35 per 100 m^2
Area of the field = \(\frac{\text{Total Cost}}{\text{Rate}} \times 100\)
\(\text{Area} = \frac{700}{35} \times 100 = 20 \times 100 = 2000 \text{ m}^2\)
Now, using the area formula of a triangle:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(2000 = \frac{1}{2} \times \left(\frac{5}{2}h\right) \times h\)
\(2000 = \frac{5}{4} h^2\)
\(h^2 = \frac{2000 \times 4}{5}\)
\(h^2 = 400 \times 4 = 1600\)
\(h = 40 \text{ m}\)
Base (\(b\)) = \(2.5 \times 40 = 100 \text{ m}\).
In simple words: Find the total area by dividing total cost by rate, then use the triangle area formula to find height and subsequently the base.

Exam Tip: Always calculate total area first from total cost and rate before applying geometric formulas.

Q5. The area of a right triangle is 50 m^2. If one of the legs is 20 m, find the other.
Answer: Area of the right triangle = 50 m^2
One leg (\(a\)) = 20 m
Let the other leg be \(b\).
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(50 = \frac{1}{2} \times 20 \times b\)
\(50 = 10 \times b\)
\(b = \frac{50}{10} = 5 \text{ m}\)
In simple words: In a right triangle, the two perpendicular sides act as base and height, so plug the given values into the area formula to find the missing side.

Exam Tip: For a right-angled triangle, the legs form the base and altitude, making area calculations straightforward.

Q6. The cost of cultivating a field in the form of a parallelogram at Rs 1.10 per square metre is Rs 1925. If the length of one pair of parallel sides is 100 m. Find the distance between that pair of parallel side.
Answer: Total cost of cultivation = Rs 1925
Rate = Rs 1.10 per m^2
Area of the parallelogram = \(\frac{1925}{1.10} = \frac{19250}{11} = 1750 \text{ m}^2\)
Length of the base (\(b\)) = 100 m
Area of parallelogram = \(\text{base} \times \text{height}\) (distance between parallel sides)
\(1750 = 100 \times h\)
\(h = \frac{1750}{100} = 17.5 \text{ m}\) (Note: Printed answer key mentions 35 m if base was taken differently, but standard calculation with base 100 m yields 17.5 m).
In simple words: Divide total cost by rate to find area, then divide area by the given base to find the height.

Exam Tip: Ensure decimal division is done carefully when converting total cost to area.

Q7. A parallelogram, the measure of whose sides are 60 m and 25 m has one diagonal as 65 m long. Find the area of itgm.
Answer: Let the sides of the parallelogram be \(a = 60\) m and \(b = 25\) m, and diagonal \(d = 65\) m.
A diagonal divides a parallelogram into two congruent triangles.
For the triangle with sides 60 m, 25 m, and 65 m:
Semi-perimeter (\(s\)) = \(\frac{60 + 25 + 65}{2} = \frac{150}{2} = 75 \text{ m}\)
Using Heron's Formula for the area of the triangle:
\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)
\(\text{Area} = \sqrt{75(75 - 60)(75 - 25)(75 - 65)}\)
\(\text{Area} = \sqrt{75 \times 15 \times 50 \times 10}\)
\(\text{Area} = \sqrt{(25 \times 3) \times (5 \times 3) \times (25 \times 2) \times 10}\)
\(\text{Area} = \sqrt{75 \times 15 \times 500} = \sqrt{562500} = 750 \text{ m}^2\)
Since the diagonal divides the parallelogram into two triangles of equal area:
Total Area of parallelogram = \(2 \times 750 = 1500 \text{ m}^2\) (Note: The answer key states 750 m^2 as the area of one triangle, which is a common shorthand in assignments).
In simple words: Find the area of one triangle formed by the sides and diagonal using Heron's formula, then multiply by 2 for the full parallelogram.

Q8. The perimeter of a rhombus ABCD is 100 m. If the diagonal AC is 14 m. Find the diagonal BD and also area of rhombus.
Answer: Perimeter of the rhombus = 100 m
Side of the rhombus (\(a\)) = \(\frac{100}{4} = 25 \text{ m}\)
Diagonal \(AC = d_1 = 14 \text{ m}\)
Diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Let diagonals intersect at point O.
\(AO = \frac{d_1}{2} = \frac{14}{2} = 7 \text{ m}\)
In right triangle AOB:
\(AB^2 = AO^2 + BO^2\)
\(25^2 = 7^2 + BO^2\)
\(625 = 49 + BO^2\)
\(BO^2 = 625 - 49 = 576\)
\(BO = 24 \text{ m}\)
Length of diagonal \(BD = d_2 = 2 \times BO = 2 \times 24 = 48 \text{ m}\).
Area of the rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
\(\text{Area} = \frac{1}{2} \times 14 \times 48 = 7 \times 48 = 336 \text{ m}^2\)
In simple words: Using the property that diagonals of a rhombus bisect at right angles, we found half the second diagonal via Pythagoras theorem and then calculated the area.

Exam Tip: Always remember that the four half-diagonals and the side of a rhombus form right-angled triangles.

Q9. The area of rhombus is 150 m^2. If one side is 15 m. Find the height of rhombus.
Answer: Area of the rhombus = 150 m^2
Side of the rhombus (\(b\)) = 15 m
A rhombus is a special type of parallelogram where all sides are equal.
Area = \(\text{base} \times \text{height}\)
\(150 = 15 \times \text{height}\)
\(\text{Height} = \frac{150}{15} = 10 \text{ m}\)
In simple words: Since a rhombus is a parallelogram, its area is base multiplied by height.

Exam Tip: Do not confuse the height of a rhombus with its side length or diagonals.

Q10. The measures of the sides of a cyclic quad are 36cm, 77cm, 75cm and 40cm. Find its area.
Answer: Let the sides of the cyclic quadrilateral be \(a = 36\) cm, \(b = 77\) cm, \(c = 75\) cm, and \(d = 40\) cm.
Semi-perimeter (\(s\)) = \(\frac{a + b + c + d}{2} = \frac{36 + 77 + 75 + 40}{2} = \frac{228}{2} = 114 \text{ cm}\)
Using Brahmagupta's formula for the area of a cyclic quadrilateral:
\(\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}\)
\(\text{Area} = \sqrt{(114 - 36)(114 - 77)(114 - 75)(114 - 40)}\)
\(\text{Area} = \sqrt{78 \times 37 \times 39 \times 74}\)
Break down the numbers into prime factors:
\(78 = 2 \times 3 \times 13\)
\(37 = 37\)
\(39 = 3 \times 13\)
\(74 = 2 \times 37\)
Combine the factors under the square root:
\(\text{Area} = \sqrt{(2 \times 3 \times 13) \times 37 \times (3 \times 13) \times (2 \times 37)}\)
\(\text{Area} = \sqrt{2^2 \times 3^2 \times 13^2 \times 37^2}\)
\(\text{Area} = 2 \times 3 \times 13 \times 37\)
\(\text{Area} = 6 \times 481 = 2886 \text{ cm}^2\)
In simple words: Apply Brahmagupta's formula with the semi-perimeter and the four sides to directly find the area of the cyclic quadrilateral.

Exam Tip: Brahmagupta's formula is the cyclic quadrilateral equivalent of Heron's formula and is very useful for problems with four given sides inscribed in a circle.