Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-12.2, (Part -1), Heron's Formula, Class 9, Maths

RD Sharma Solutions Ex-12.2, (Part -1), Heron's Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1. Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

For triangle ABC

AC2 = BC2+AB2

25 = 9 + 16

So, triangle ABC is a right angle triangle right angled at point R

Area of triangle ABC =  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 6 cm2

From triangle CAD

Perimeter = 2s = AC + CD + DA

2s = 5 cm+ 4 cm+ 5 cm

2s   = 14 cm

s = 7 cm

By using Heron’s Formula

Area of the triangle CAD RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 9.16cm2

Area of ABCD = Area of ABC + Area of CAD

= (6+9.16) cm2

= 15.16cm2


Q2. The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Here the length of the sides of the quadrilateral is given as

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

Diagonal AC is joined.

Now, in triangle ADC

By applying Pythagoras theorem

AC2 = AD2+CD2

AC2 = 142+72

AC = 25 m

Now area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m

By using Heron’s Formula

The area of a triangle  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 291.84m2

Thus, the area of a triangle is 291.84m2

Now for area of triangle ADC

Perimeter = 2s = AD + CD + AC

= 25 m + 24 m + 7 m

s = 28 m

By using Heron’s Formula

The area of a triangleRD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 84m2

Thus, the area of a triangle is 84m2

Therefore, Area of rectangular field ABCD

= Area of triangle ABC + Area of triangle ADC

= 291.84 + 84

= 375.8 m2


Q3. The sides of a quadrilateral, taken in order as 5 m, 12 m, 14 m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area.

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

Given that the sides of the quadrilateral are

AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m

Join AC

Now area of triangle ABC  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 30 m2

In triangle ABC, By applying Pythagoras theorem

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

AC = 13 m

Now in triangle ADC,

Perimeter = 2s = AD + DC + AC

2s = 15 m +14 m +13 m

s = 21 m

By using Heron’s Formula,

Area of the triangle PSR =  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 84m2

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

= (30 + 84) m2

= 114m2


Q4. A park in the shape of a quadrilateral ABCD, has angle C = 900, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?

Solution:

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.

Let us join BD

In triangle BCD , apply Pythagoras theorem

BD2 = BC2+CD2

BD2 = 122+52

BD = 13 m

Area of triangle BCD =  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 30 m2

Now, in triangle ABD

Perimeter = 2s = 9 m + 8m + 13m

s = 15 m

By using Heron’s Formula,

Area of the triangle ABD = RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 35.49m2

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= (35.496 + 30) m2

= 65.5m2.


Q5. Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium?

Solution:

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Given,

Two parallel sides of trapezium are AB = 77 m and CD = 60 m

The other two parallel sides of trapezium are BC = 26 m, AD = 25m

Join AE and CF

DE is perpendicular to AB and also, CF is perpendicular to AB

Therefore, DC = EF = 60 m

Let AE = x

So, BF = 77 – 60 – x

BF = 17 – x

In triangle ADE,

By using Pythagoras theorem,

DE2 = AD2−AE2

DE2 = 252−x2

In triangle BCF,

By using Pythagoras theorem,

CF2 = BC2−BF2

CF2 = 262−(17−x)2

Here, DE = CF

So, DE2 = CF2

252−x2 = 262−(17−x)2

252−x2 = 262−(172−34x+x2)

252−x2 = 262−172+34x+x2

252 = 262−172+34x

x = 7

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

DE = 24 m

Area of trapezium RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of trapezium = 1644m2


Q6. Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Solution:

Given,

Perimeter of a rhombus = 80 m

As we know,

Perimeter of a rhombus = 4×side = 4×a

4×a = 80 m

a = 20 m

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Let AC = 24 m

Therefore OA  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

OA = 12 m

In triangle AOB

OB2 = AB2−OA2

OB2 = 202−122

OB = 16 m

Also, OB = OD because diagonal of rhombus bisect each other at 90

Therefore, BD = 2 OB = 2 x 16 = 32 m

Area of rhombus  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of rhombus   RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of rhombus = 384 m2


Q7. A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per meter square. Find the cost of painting.

Solution:

Given that,

Perimeter of a rhombus = 32 m

We know that,

Perimeter of a rhombus = 4×side

4×side = 32 m

4×a = 32 m

a = 8 m

Let AC = 10 m

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

By using Pythagoras theorem

OB2 = AB2–OA2

OB2 = 82–52

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the sheet RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the sheet  RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Therefore, cost of printing on both sides at the rate of Rs. 5 per m2

RD Sharma Solutions Ex-12.2, (Part -1), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= Rs. 625

The document RD Sharma Solutions Ex-12.2, (Part -1), Heron's Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-12.2, (Part -1), Heron's Formula, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is Heron's formula?
Ans. Heron's formula is a formula used to find the area of a triangle when the lengths of all three sides are known. It is named after the ancient Greek mathematician Hero of Alexandria.
2. How is Heron's formula derived?
Ans. Heron's formula is derived using the concept of semiperimeter. The semiperimeter of a triangle is half the sum of its three sides. By using the semiperimeter, we can calculate the area of a triangle using the formula: Area = √(s(s-a)(s-b)(s-c)), where s is the semiperimeter and a, b, and c are the lengths of the sides.
3. Can Heron's formula be used for all types of triangles?
Ans. Yes, Heron's formula can be used to find the area of any type of triangle, whether it is equilateral, isosceles, or scalene. As long as the lengths of all three sides are known, Heron's formula can be applied.
4. Are there any limitations to using Heron's formula?
Ans. Yes, there are some limitations to using Heron's formula. It is only applicable when the lengths of all three sides are known. If the lengths of the sides are not given, Heron's formula cannot be used to find the area of the triangle.
5. How is Heron's formula useful in real-life applications?
Ans. Heron's formula is useful in various real-life applications, such as architecture, engineering, and physics. It can be used to calculate the area of irregularly shaped land or to determine the amount of material needed for construction projects. Additionally, it is used in navigation and surveying to calculate distances and angles.
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