Class 9 Mathematics Chapter 10 Heron’s Formula

Key Concept: Semi-perimeter Calculation

c) 14 m

[Solution Description]

Calculate semi-perimeter using formula $s = \frac{a + b + c}{2}$. Substitute $a = 7$, $b = 9$, $c = 12$:

$s = \frac{7 + 9 + 12}{2} = \frac{28}{2} = 14$ m.

Key Concept: Heron’s Formula Expression

a) $\sqrt{\frac{(a+b+c)}{2} \cdot (\frac{a+b+c}{2}-a) \cdot (\frac{a+b+c}{2}-b) \cdot (\frac{a+b+c}{2}-c)}$

[Solution Description]

Heron’s formula states: Area = $\sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a + b + c}{2}$. This formula is used when all three side lengths are known.

Key Concept: Heron’s Formula

b) $9\sqrt{15}$ cm²

[Solution Description]

In an isosceles triangle with two equal sides of 12 cm each and a perimeter of 30 cm, the third side $c$ can be found as follows:

$12 + 12 + c = 30 \Rightarrow c = 6 \text{ cm}$

The semi-perimeter $s$ is:

$s = \frac{30}{2} = 15 \text{ cm}$

Using Heron’s formula, the area is calculated as:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{15(15 – 12)(15 – 12)(15 – 6)} = \sqrt{15 \times 3 \times 3 \times 9} = \sqrt{1215} = 9\sqrt{15} \text{ cm}^2$

Therefore, the area of the triangle is $9\sqrt{15}$ cm².

Key Concept: Heron’s Formula, Fencing Cost

c) \$300

[Solution Description]

First, calculate the perimeter of the triangle: $15 + 20 + 25 = 60$ m. Next, calculate the cost of fencing: $60 \times 5 = \$300$.

Key Concept: Heron’s Formula Application

a) 6 m$^2$

[Solution Description] First compute semi-perimeter: $s = \frac{3 + 4 + 5}{2} = 6$. Apply Heron\’s formula: $\sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \, \text{m}^2$.

Key Concept: Applicability of Heron’s Formula

c) When height is not available

[Solution Description]

Heron’s formula is used when the height of the triangle is unknown. It works for all triangle types (scalene, isosceles, etc.) by using only side lengths to calculate area. This removes dependency on base-height measurements.

Key Concept: Heron’s Formula

c) 54 m²

[Solution Description]

First, calculate the semi-perimeter $s$:

$s = \frac{9 + 12 + 15}{2} = 18 \text{ m}$

Now, apply Heron’s formula:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{18(18 – 9)(18 – 12)(18 – 15)} = \sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54 \text{ m}^2$

Key Concept: Heron’s Formula Components

d) $\frac{5 + 6 + 7}{2}$

[Solution Description]

Semi-perimeter $s = \frac{a + b + c}{2}$. Substitute values:

$s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9 \, \text{cm}$.

Key Concept: Heron’s Formula, Real-World Application

a) $6000\sqrt{6} + 900$ dollars

[Solution Description]

First, calculate the area of the park using Heron’s formula. The semi-perimeter $s$ is $\frac{50 + 60 + 70}{2} = 90$ m. The area is $\sqrt{90(90 – 50)(90 – 60)(90 – 70)} = \sqrt{90 \times 40 \times 30 \times 20} = \sqrt{2160000} = 600\sqrt{6}$ m². The cost of planting grass is $10 \times 600\sqrt{6} = 6000\sqrt{6}$ dollars. The perimeter of the park is $50 + 60 + 70 = 180$ m. The cost of fencing is $5 \times 180 = 900$ dollars. The total cost is $6000\sqrt{6} + 900$ dollars.

Key Concept: Heron’s Formula

a) $10\sqrt{6}$ cm$^2$

[Solution Description]

To find the area of the triangle using Heron’s formula, we first calculate the semi-perimeter $s$. The sides of the triangle are 7 cm, 7 cm, and 10 cm. So,

$s = \frac{7 + 7 + 10}{2} = \frac{24}{2} = 12 \text{ cm}$

Next, we apply Heron’s formula:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{12(12 – 7)(12 – 7)(12 – 10)}$

Simplifying inside the square root:

$\text{Area} = \sqrt{12 \times 5 \times 5 \times 2} = \sqrt{600} = 10\sqrt{6} \text{ cm}^2$

Therefore, the area of the triangle is $10\sqrt{6}$ cm$^2$.

Key Concept: Heron’s Formula

b) 30 cm$^2$

[Solution Description]

First, calculate the semi-perimeter $s = \frac{5 + 12 + 13}{2} = 15$ cm. Then, apply Heron’s formula:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{15(15 – 5)(15 – 12)(15 – 13)}$

$= \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30 \text{ cm}^2$

Key Concept: Heron’s Formula Verification

b) 25$\sqrt{3}$ cm$^2$

[Solution Description]

First, calculate the semi-perimeter $s = \frac{10 + 10 + 10}{2} = 15$ cm. Now, apply Heron’s formula:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{15(15 – 10)(15 – 10)(15 – 10)} = \sqrt{15 \times 5 \times 5 \times 5} = \sqrt{1875} = 25\sqrt{3} \text{ cm}^2.$

Key Concept: Application Scenario

d) When all three sides are known but height is unknown

[Solution Description]

Heron’s formula is used when the height of the triangle is not directly known. The standard formula requires base and height, while Heron’s formula only needs all three side lengths. For example, in a scalene triangle with unequal sides and unknown height, Heron’s formula is necessary.

Key Concept: Cost Calculation

b) Rs.4320

[Solution Description] Check if right-angled: $18^2 + 24^2 = 324 + 576 = 900 = 30^2$. Area = $\frac{1}{2} \times 18 \times 24 = 216$ m². Total cost = 216 × 20 = Rs.4320.

Key Concept: Semi-perimeter

b) 9 cm

[Solution Description]

To find semi-perimeter (s): Add all sides and divide by 2.

Calculation:

s = (5 + 6 + 7)/2 = 18/2 = 9 cm

Key Concept: Heron’s Formula, Scalene Triangle

b) $420$ m$^2$

[Solution Description]

For a triangle with sides $a = 25$ m, $b = 39$ m, and $c = 56$ m, first find the semi-perimeter $s$:

$s = \frac{25 + 39 + 56}{2} = 60 \text{ m}$

Using Heron’s formula, the area is:

$\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{60(60-25)(60-39)(60-56)}$

$= \sqrt{60 \times 35 \times 21 \times 4} = \sqrt{176400} = 420 \text{ m}^2$

Key Concept: Mixed Units, Cost Calculation

b) Rs.475

[Solution Description] Convert sides to decimeters: 1300 mm = 13 dm, 2.2 m = 22 dm, 150 cm = 15 dm. Semi-perimeter $s = \frac{13 + 22 + 15}{2} = 25$ dm. Area $= \sqrt{25 \times (25 – 13) \times (25 – 22) \times (25 – 15)} = \sqrt{25 \times 12 \times 3 \times 10} = \sqrt{9000} \approx 94.87 \, \text{dm}^2$. Cost $= 94.87 \times 5 \approx Rs.475$.

Key Concept: Semi-Perimeter Calculation

a) 21 cm

[Solution Description]

To find the semi-perimeter ($s$) of an equilateral triangle, we add all three sides and divide by 2.

Step 1: Sum of sides = $14 \, \text{cm} + 14 \, \text{cm} + 14 \, \text{cm} = 42 \, \text{cm}$.

Step 2: Semi-perimeter $s = \frac{42}{2} = 21 \, \text{cm}$.

Key Concept: Historical Origin

d) Heron

[Solution Description]

The formula was derived by Heron (also called Hero), an ancient mathematician from Alexandria. As per the syllabus, he explained this formula in Book I of his geometrical works on mensuration.

Key Concept: Heron’s Formula

b) 84 m²

[Solution Description]

First, calculate the semi-perimeter $s$ of the triangle:

$s = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28 \text{m}$

Apply Heron’s formula to find the area:

$\text{Area} = \sqrt{28(28 – 7)(28 – 24)(28 – 25)} = \sqrt{28 \times 21 \times 4 \times 3}$

Simplify the expression inside the square root:

$\text{Area} = \sqrt{7056} = 84 \text{m}^2$

Therefore, the area of the garden is 84 \text{m}^2.

Key Concept: Area of a Triangle — by Heron’s Formula

c) 96 cm\textsuperscript{2}

[Solution Description]

First, we calculate the semi-perimeter $s$ of the triangle:

$s = \frac{12 + 16 + 20}{2} = 24 \, \text{cm}$

Next, we use Heron’s formula:

$\text{Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{24(24 – 12)(24 – 16)(24 – 20)}$

Simplifying inside the square root:

$\text{Area} = \sqrt{24 \times 12 \times 8 \times 4} = \sqrt{9216} = 96 \, \text{cm}^2$

Therefore, the area of the triangle is $96 \, \text{cm}^2$.

Key Concept: Fencing Cost

c) \$1200

[Solution Description]

First, calculate the perimeter of the triangle:

$\text{Perimeter} = 30 + 40 + 50 = 120 \, \text{m}$

Then, calculate the total cost of fencing:

$\text{Cost} = 120 \times 10 = \$1200$