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:

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 =

= 6 cm2

Perimeter = 2s = AC + CD + DA

2s = 5 cm+ 4 cm+ 5 cm

2s   = 14 cm

s = 7 cm

By using Heron’s Formula

= 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:

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.

By applying Pythagoras theorem

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

= 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 triangle

= 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.

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

= 30 m2

In triangle ABC, By applying Pythagoras theorem

AC = 13 m

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 =

= 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:

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 =

= 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 =

= 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:

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

By using Pythagoras theorem,

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

DE = 24 m

Area of trapezium

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

Let AC = 24 m

Therefore OA

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

Area of rhombus

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

By using Pythagoras theorem

OB2 = AB2–OA2

OB2 = 82–52

Area of the sheet

Area of the sheet

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

= 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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