RD SHARMA Solutions for Class 9 Maths Chapter 17 - Heron's Formula

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Chapter 17 - Heron's Formula Exercise 17.24

Question 1

The side of triangle are 16 cm, 30 cm, 34 cm, its area is

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 1

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 2

The base of an isosceles right triangle is 30 cm. Its area is

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 3

The sides of a triangle are 7 cm, 9 cm, and 14 cm. Its area is

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 4

The sides of a triangle are 325 m, 300 m and 125 m. Its area is

(a) 18750 m2

(b) 37500 m2

(c) 97500 m2

(d) 48750 m2

Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 5

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is 

(a) 20 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

Solution 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Chapter 17 - Heron's Formula Exercise 17.25

Question 1

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

(a) 11 m

(b) 66 m

(c) 50 m

(d) 60 m

Solution 1

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 3

The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. it is area is:

(a) 25 cm2

(b) 28 cm2

(c) 30 cm2

(d) 40 cm2

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

(a) 4

(b) 6

(c) 8

(d) 10

Solution 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Chapter 17 - Heron's Formula Exercise Ex. 17.1

Question 1

Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Solution 1

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 5
The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.
Solution 5
The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
  Perimeter of this triangle = 540 m
             25x + 17x + 12x = 540 m
                         54x = 540 m
                            x = 10 m
  Sides of triangle will be 250 m, 170 m, and 120 m.
 
Semi-perimeter (s) = Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
By Heron's formula:
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
    So, area of the triangle is 9000 m2.
Question 6

The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.

Solution 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 7

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 8

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

Solution 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 9

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.

Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 10

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 11

Find the area of the shaded region in fig.12.12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Chapter 17 - Heron's Formula Exercise Ex. 17.2

Question 1
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
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC is a right angle triangle, right angled at point B.
Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABCRd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
            s = 7 cm
By Heron's formula
Area of triangle Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of ABCD = Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC + Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaACD
    = (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

Question 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 2

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 3

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 4
A park, in the shape of a quadrilateral ABCD, has Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula = 90o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution 4
Let us join BD.
In Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD applying Pythagoras theorem
BD2 = BC2 + CD2
       = (12)2 + (5)2
       = 144 + 25
BD2 = 169
  BD = 13 m
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
                  Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABD
                    Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of triangle  Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                         Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
      Area of park = Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABD + Area of Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
                         = 35.496 + 30 m2
                         = 65.496 m2
                         = 65. 5 m2 (approximately)

Question 5

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Solution 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 6

A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.

Solution 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 8

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Solution 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 9

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 10

Find the area of the blades of the magnetic compass shown in fig.

(Take √11 = 3.32)

 

 

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 11

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Solution 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 13

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 13

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Question 14

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Solution 14

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Herons Formula