CBSE - Herons Formula

Solution

Length of the side of equilateral triangle = a
Perimeter of the signal board = 3a = 180 cm
∴ 3a = 180 cm ⇒ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
Using heron's formula,
Area of the signal board = √s (s-a) (s-b) (s-c)
                                       = √(3a/2) (3a/2 - a) (3a/2 - a) (3a/2 - a)
                                       = √3a/2 × a/2 × a/2 × a/2
                                       = √3a⁴/16
                                       = √3a²/4
                                       = √3/4 × 60 × 60 = 900√3 cm²

Solution

The sides of the triangle are 122 m, 22 m and 120 m.
Perimeter of the triangle is 122 + 22 + 120 = 264m
Semi perimeter of triangle (s) = 264/2 = 132 m
Using heron's formula,
Area of the advertisement = √s (s-a) (s-b) (s-c)
                                       = √132(132 - 122) (132 - 22) (132 - 120) m²
                                       = √132 × 10 × 110 × 12 m²
                                       = 1320 m²
Rent of advertising per year = ₹ 5000 per m²
Rent of one wall for 3 months = ₹ (1320 × 5000 × 3)/12 = ₹ 1650000

Solution
Sides of the triangular wall are 15 m, 11 m and 6 m.
Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m
Using heron's formula, 
Area of the message = √s (s-a) (s-b) (s-c)
                                       = √16(16 - 15) (16 - 11) (16 - 6) m²
                                       = √16 × 1 × 5 × 10 m² = √800 m²
                                       = 20√2 m²

Solution

Two sides of the triangle = 18cm and 10cm
Perimeter of the triangle = 42cm
Third side of triangle = 42 - (18+10) cm = 14cm
Semi perimeter of triangle = 42/2 = 21cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √21(21 - 18) (21 - 10) (21 - 14) cm²
                                       = √21 × 3 × 11 × 7 m²
                                       = 21√11 cm²

Solution

Ratio of the sides of the triangle = 12 : 17 : 25
Let the common ratio be x then sides are 12x, 17x and 25x
Perimeter of the triangle = 540cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
⇒ x = 10
Sides of triangle are,
12x = 12 × 10 = 120cm
17x = 17 × 10 = 170cm
25x = 25 × 10 = 250cm
Semi perimeter of triangle(s) = 540/2 = 270cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √270(270 - 120) (270 - 170) (270 - 250)cm²
                                       = √270 × 150 × 100 × 20 cm²
                                       = 9000 cm²

Solution

Length of the equal sides = 12cm
Perimeter of the triangle = 30cm
Length of the third side = 30 - (12+12) cm = 6cm
Semi perimeter of the triangle(s) = 30/2 cm = 15cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 12) (15 - 12) (15 - 6)cm²
                                       = √15 × 3 × 3 × 9 cm²
                                       = 9√15 cm²

Solution

∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
BD is joined. 

In ΔBCD,
By applying Pythagoras theorem,
BD² = BC² + CD²  
⇒ BD² = 12² + 5² 
⇒ BD² = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m²
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD  = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 13) (15 - 9) (15 - 8) m²
                                       = √15 × 2 × 6 × 7 m²
                                       = 6√35 m² = 35.5 m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m² + 35.5m² = 65.5m² 

Solution:

Length of the sides of the triangle section I = 5cm, 1cm and 5cm
Perimeter of the triangle = 5 + 5 + 1 = 11cm
Semi perimeter = 11/2 cm = 5.5cm
Using heron's formula,
Area of section I  = √s (s-a) (s-b) (s-c)
                                       = √5.5(5.5 - 5) (5.5 - 5) (5.5 - 1) cm²
                                       = √5.5 × 0.5 × 0.5 × 4.5 cm²
                                       = 0.75√11 cm² = 0.75 × 3.317cm² = 2.488cm² (approx)
Length of the sides of the rectangle of section I = 6.5cm and 1cm
Area of section II = 6.5 × 1 cm² =  6.5 cm²
Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.
Area of the trapezium = 3 × √3/4 × 1² cm² = 1.3 cm² (approx)
Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm
Area of region IV and V = 2 × 1/2 × 6 × 1.5cm² = 9cm²
Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm² = 19.3 cm²

Solution

Given,
Area of the parallelogram and triangle are equal. 

Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.
Perimeter of the triangle = 26 + 28 + 30 = 84 cm
Semi perimeter of the triangle = 84/2 cm = 42 cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √42(42 - 26) (46 - 28) (46 - 30) cm²
                                       = √46 × 16 × 14 × 16 cm²
                                       = 336 cm²Let height of parallelogram be h.
Area of parallelogram = Area of triangle 
28cm × h = 336 cm²
⇒ h = 336/28 cm
⇒ h = 12 cm
The height of the parallelogram is 12 cm.

Solution

Diagonal AC divides the rhombus ABCD into two congruent triangles of equal area. 

Semi perimeter of ΔABC = (30 + 30 + 48)/2 m = 54 m
Using heron's formula,
Area of the ΔABC = √s (s-a) (s-b) (s-c)
                                       = √54(54 - 30) (54 - 30) (54 - 48) m²
                                       = √54 × 24 × 24 × 6 cm²
                                       = 432 m²
Area of field = 2 × area of the ΔABC = (2 × 432)m² = 864 m²
Thus,
Area of grass field which each cow will be getting = 864/18 m² = 48 m²

Solution

Semi perimeter of each triangular piece = (50 + 50 + 20)/2 cm = 120/2 cm = 60cm
Using heron's formula,
Area of the triangular piece = √s (s-a) (s-b) (s-c)
                                       = √60(60 - 50) (60 - 50) (60 - 20) cm²
                                       = √60 × 10 × 10 × 40 cm²
                                       = 200√6 cm²
Area of triangular piece = 5 × 200√6 cm² = 1000√6 cm²

Solution

We know that,
As the diagonals of a square bisect each other at right angle.
Area of given kite = 1/2 (diagonal)²
                              = 1/2 × 32 × 32 = 512 cm²
Area of shade I = Area of shade II
⇒ 512/2 cm² = 256 cm²
So, area of paper required in each shade = 256 cm²
For the III section,
Length of the sides of triangle = 6cm, 6cm and 8cm 
Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm
Using heron's formula,
Area of the III triangular piece = √s (s-a) (s-b) (s-c)
                                       = √10(10 - 6) (10 - 6) (10 - 8) cm²
                                       = √10 × 4 × 4 × 2 cm²
                                       = 8√6 cm²

Solution

Semi perimeter of the each triangular shape = (28 + 9 + 35)/2 cm = 36 cm
Using heron's formula,
Area of the each triangular shape = √s (s-a) (s-b) (s-c)
                                       = √36(36 - 28) (36 - 9) (36 - 35) cm²
                                       = √36 × 8 × 27 × 1 cm²
                                       = 36√6 cm² = 88.2 cm²
Total area of 16 tiles = 16 × 88.2 cm² = 1411.2 cm²Cost of polishing tiles = 50p per cm²
Total cost of polishing the tiles = Rs. (1411.2 × 0.5) = Rs. 705.6

Solution

Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.
CM ⊥ AB and CE || AD.
In ΔBCE,
BC = 14m, CE = AD = 13 m and
BE = AB - AE = 25 - 10 = 15m 
Semi perimeter of the ΔBCE = (15 + 13 + 14)/2 m = 21 m
Using heron's formula,
Area of the ΔBCE = √s (s-a) (s-b) (s-c)
                                       = √21(21 - 14) (21 - 13) (21 - 15) m²
                                       = √21 × 7 × 8 × 6 m²
                                       = 84 m² 
also, area of the ΔBCE = 1/2 × BE × CM = 84 m² 
⇒ 1/2 × 15 × CM = 84 m² 
⇒ CM = 168/15 m² 
⇒ CM = 56/5 m² 
Area of the parallelogram AECD = Base × Altitude = AE × CM = 10 × 84/5 = 112 m² 
Area of the trapezium ABCD = Area of AECD + Area of ΔBCE
                                                = (112+ 84) m²  = 196 m²