Chapter 12: Heron's Formula (Exercise 12.1
Chapter 12: Heron's Formula (Exercise 12.1)
Question 1:
A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side a. Find the area of the signal board using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
1. Using Heron's Formula for side 'a':
* Semi-perimeter (s) = (a + a + a) / 2 = 3a / 2
* Area = √[s(s - a)(s - b)(s - c)]
Area = √[(3a/2) × (3a/2 - a) × (3a/2 - a) × (3a/2 - a)]
Area = √[(3a/2) × (a/2) × (a/2) × (a/2)]
Area = √[(3a⁴) / 16] = (√3 / 4) a²
2. Calculating Area when Perimeter = 180 cm:
* 3a = 180 cm ⇒ a = 60 cm
* Semi-perimeter (s) = 180 / 2 = 90 cm
* Area = (√3 / 4) × (60)² = (√3 / 4) × 3600 = 900√3 cm²
Question 2:
There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN". If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in colour.
Solution:
* Given sides of the triangle: a = 15 m, b = 11 m, c = 6 m.
1. Calculating Semi-perimeter (s):
s = (a + b + c) / 2
s = (15 + 11 + 6) / 2 = 32 / 2 = 16 m
2. Calculating Area using Heron's Formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[16 × (16 - 15) × (16 - 11) × (16 - 6)]
Area = √[16 × 1 × 5 × 10]
Area = √[800] = √(400 × 2) = 20√2 m²
Question 3:
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
* Given: a = 18 cm, b = 10 cm, Perimeter = 42 cm.
1. Finding the third side (c):
a + b + c = 42
18 + 10 + c = 42
28 + c = 42 ⇒ c = 14 cm
2. Calculating Semi-perimeter (s):
s = Perimeter / 2 = 42 / 2 = 21 cm
3. Calculating Area using Heron's Formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[21 × (21 - 18) × (21 - 10) × (21 - 14)]
Area = √[21 × 3 × 11 × 7]
Area = √[(7 × 3) × 3 × 11 × 7]
Area = √(7² × 3² × 11) = 7 × 3 × √11 = 21√11 cm²
Question 4:
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
* Let the sides of the triangle be 12x, 17x, and 25x.
1. Finding the value of x and side lengths:
12x + 17x + 25x = 540
54x = 540 ⇒ x = 10
* Side a = 12 × 10 = 120 cm
* Side b = 17 × 10 = 170 cm
* Side c = 25 × 10 = 250 cm
2. Calculating Semi-perimeter (s):
s = 540 / 2 = 270 cm
3. Calculating Area using Heron's Formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[270 × (270 - 120) × (270 - 170) × (270 - 250)]
Area = √[270 × 150 × 100 × 20]
Area = √[81,000,000] = 9000 cm²
Chapter 12: Heron's Formula (Exercise 12.1 - Remaining Questions)
Question 2:
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m. The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution:
* Given sides of the wall: a = 122 m, b = 22 m, c = 120 m.
1. Calculating Semi-perimeter (s):
s = (122 + 22 + 120) / 2 = 264 / 2 = 132 m
2. Calculating Area using Heron's Formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[132 × (132 - 122) × (132 - 22) × (132 - 120)]
Area = √[132 × 10 × 110 × 12] = √[1,742,400] = 1320 m²
3. Calculating Rent for 3 Months:
Rent = 1320 m² × ₹5000/m²/year × (3/12) years
Rent = 1320 × 5000 × 0.25 = ₹16,50,000
Question 6:
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
* Given: Equal sides a = 12 cm, b = 12 cm, Perimeter = 30 cm.
1. Finding the third side (c):
12 + 12 + c = 30 ⇒ 24 + c = 30 ⇒ c = 6 cm
2. Calculating Semi-perimeter (s):
s = 30 / 2 = 15 cm
3. Calculating Area using Heron's Formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[15 × (15 - 12) × (15 - 12) × (15 - 6)]
Area = √[15 × 3 × 3 × 9] = 3 × 3 × √15 = 9√15 cm²
Chapter 12: Heron's Formula (Exercise 12.2)
Question 1:
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m. How much area does it occupy?
Solution:
* Join diagonal BD to form right ΔBCD and ΔABD.
1. In right ΔBCD:
By Pythagoras Theorem: BD = √(BC² + CD²) = √(12² + 5²) = √(144 + 25) = 13 m
Area of ΔBCD = ½ × base × height = ½ × 12 × 5 = 30 m²
2. In ΔABD (sides 9 m, 8 m, 13 m):
s = (9 + 8 + 13) / 2 = 15 m
Area of ΔABD = √[15 × (15 - 9) × (15 - 8) × (15 - 13)] = √[15 × 6 × 7 × 2] = 6√35 m² ≈ 35.5 m²
3. Total Area of Quadrilateral ABCD:
Total Area = 30 m² + 35.5 m² = 65.5 m²
Question 2:
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.
Solution:
1. In ΔABC (sides 3 cm, 4 cm, 5 cm):
Since 3² + 4² = 5², ΔABC is a right-angled triangle at B.
Area of ΔABC = ½ × 3 × 4 = 6 cm²
2. In ΔACD (sides 5 cm, 5 cm, 4 cm):
s = (5 + 5 + 4) / 2 = 7 cm
Area of ΔACD = √[7 × (7 - 5) × (7 - 5) × (7 - 4)] = √[7 × 2 × 2 × 3] = 2√21 cm² ≈ 9.17 cm²
3. Total Area:
Total Area = 6 cm² + 9.17 cm² = 15.17 cm²
Question 3:
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm, and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
1. Calculating Area of Triangle:
Sides: a = 26 cm, b = 28 cm, c = 30 cm
s = (26 + 28 + 30) / 2 = 42 cm
Area = √[42 × (42 - 26) × (42 - 28) × (42 - 30)] = √[42 × 16 × 14 × 12] = 336 cm²
2. Calculating Height of Parallelogram:
Area of Parallelogram = Area of Triangle
Base × Height = 336
28 × Height = 336
Height = 336 / 28 = 12 cm
Question 4:
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
* Diagonal divides the rhombus into two congruent triangles with sides 30 m, 30 m, and 48 m.
1. Area of one triangle:
s = (30 + 30 + 48) / 2 = 54 m
Area = √[54 × (54 - 30) × (54 - 30) × (54 - 48)] = √[54 × 24 × 24 × 6] = 432 m²
2. Total Area of Rhombus:
Total Area = 2 × 432 m² = 864 m²
3. Area per Cow:
Area per cow = 864 / 18 = 48 m²
Question 5:
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?
Solution:
1. Area of 1 triangular piece:
Sides: 20 cm, 50 cm, 50 cm ⇒ s = (20 + 50 + 50) / 2 = 60 cm
Area = √[60 × (60 - 20) × (60 - 50) × (60 - 50)] = √[60 × 40 × 10 × 10] = 200√6 cm²
2. Cloth required for each colour (5 pieces each):
Area for each colour = 5 × 200√6 = 1000√6 cm²
Question 6:
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
* Draw a line parallel to one non-parallel side, dividing the trapezium into a parallelogram (sides 10 m, 13 m) and a triangle (sides 15 m, 13 m, 14 m).
1. Area of the triangle (sides 15 m, 13 m, 14 m):
s = (15 + 13 + 14) / 2 = 21 m
Area = √[21 × (21 - 15) × (21 - 13) × (21 - 14)] = √[21 × 6 × 8 × 7] = 84 m²
2. Finding height of triangle (and trapezium):
½ × base × height = 84
½ × 15 × height = 84 ⇒ height = 11.2 m
3. Area of Trapezium:
Area = ½ × (Sum of parallel sides) × height
Area = ½ × (25 + 10) × 11.2 = ½ × 35 × 11.2 = 196 m²