Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-12.2, (Part -2), Heron's Formula, Class 9, Maths

RD Sharma Solutions Ex-12.2, (Part -2), Heron's Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q8. Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90 and BCD forms an equilateral  triangle whose each side is equal to 26 cm. [Take √3 = 1.73]

Solution:

Given that, in a quadrilateral ABCD in which AD = 24 cm,

Angle BAD = 90∘

BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

In triangle BAD, by applying Pythagoras theorem,

BA2 = BD2−AD2

BA2 = 262+242

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the triangle RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the triangle RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the triangle BAD = 120 cm2

Area of the equilateral triangle RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the equilateral triangle QRS  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Area of the equilateral triangle BCD = 292.37 cm2

Therefore, the area of quadrilateral ABCD = area of triangle BAD + area of the triangle BCD

The area of quadrilateral ABCD = 120 + 292.37

= 412.37 cm2


Q9.  Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and the diagonal BD = 20 cm.

Solution:

Given

AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and the diagonal

BD = 20 cm.

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, for the area of triangle ABD

Perimeter of triangle ABD 2s = AB + BD + DA

2s = 34 cm + 42 cm + 20 cm

s = 48 cm/p>

By using Heron’s Formula,

Area of the triangle  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 336cm2

Now, for the area of triangle BCD

Perimeter of triangle BCD 2s = BC + CD + BD

2s = 29cm + 21cm + 20cm

s = 35 cm

By using Heron’s Formula,

Area of the triangle BCD RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 210cm2

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

Area of quadrilateral ABCD = 336 + 210

Area of quadrilateral ABCD = 546 cm2


Q10. Find the perimeter and the area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB = 90.

Solution:

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Given are the sides of the quadrilateral ABCD in which

AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and an angle ACB = 90

By using Pythagoras theorem

BC2 = AB2−AC2

BC2 = 172−152

BC = 8 cm

Now, area of triangle RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

area of triangle  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

area of triangle ABC = 60 cm2

Now, for the area of triangle ACD

Perimeter of triangle ACD 2s = AC + CD + AD

2s = 15 + 12 +9

s = 18 cm

By using Heron’s Formula,

Area of the triangle ACD RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 54cm2

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

Area of quadrilateral ABCD = 60 cm2 + 54cm2

Area of quadrilateral ABCD = 114 cm2


Q11. The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of parallelogram.

Solution:

Given,

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

Area of the parallelogram = Area of triangle ADC + Area of triangle ABC

Note: Diagonal of a parallelogram divides into two congruent triangles

Therefore,

Area of the parallelogram = 2 × (Area of triangle ABC)

Now, for area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 34 cm + 20 cm + 42 cm

s = 48 cm

By using Heron’s Formula,

Area of the triangle ABC =  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 336cm2

Therefore, area of parallelogram ABCD = 2 × (Area of triangle ABC)

Area of parallelogram = 2 × 336cm2

Area of parallelogram ABCD = 672 cm2


Q12. Find the area of the blades of the magnetic compass shown in figure given below.

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB

Now, for the area of triangle ADB

Perimeter = 2s = AD + DB + BA

2s = 5 cm + 1 cm + 5 cm

s = 5.5 cm

By using Heron’s Formula,

Area of the triangle  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 2.49cm2

Also, area of triangle ADB = Area of triangle CDB

Therefore area of the blades of the magnetic compass = 2  × area of triangle ADB

Area of the blades of the magnetic compass = 2 ×2.49

Area of the blades of the magnetic compass = 4.98 cm2


Q13. A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. 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 Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

Given that,

The sides of AOB

AO = 25 cm

OB = 25 cm

BA = 14 cm

Area of each strip = Area of triangle AOB

Now, for the area of triangle AOB

Perimeter = AO + OB + BA

2s = 25 cm +25 cm + 14 cm

s = 32 cm

By using Heron’s Formula,

Area of the triangle  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 168cm2

Also, area of each type of paper needed to make a fan = 5 ×  Area of triangle AOB

Area of each type of paper needed to make a fan = 5 × 168 cm2

Area of each type of paper needed to make a fan = 840cm2


Q14. 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 a parallelogram.

Solution:

The sides of the triangle DCE are

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the h be the height of parallelogram ABCD

Now, for the area of triangle DCE

Perimeter = DC + CE + ED

2s  = 15 cm + 13 cm + 14 cm

s = 21 cm

By using Heron’s Formula,

Area of the triangle  RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-12.2, (Part -2), Heron`s Formula, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 84cm2

Also, area of triangle DCE = Area of parallelogram ABCD ⇒ 84cm2

24 × h  = 84cm2

h = 6 cm.

The document RD Sharma Solutions Ex-12.2, (Part -2), 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 -2), 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 mathematical formula used to find the area of a triangle when the lengths of all three sides are known. It is named after Hero of Alexandria, who first described it. The formula is given by A = √(s(s-a)(s-b)(s-c)), where A represents the area of the triangle, a, b, and c are the lengths of the sides, and s is the semi-perimeter of the triangle (s = (a+b+c)/2).
2. How is Heron's formula derived?
Ans. Heron's formula can be derived using the concept of the semi-perimeter of a triangle and the area of a triangle using base and height. By applying the Law of Cosines to find the angles of the triangle, we can then use the sine function to find the height. Substituting the height and the base into the formula for the area of a triangle, we can simplify the equation to obtain Heron's formula.
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 scalene, isosceles, or equilateral. As long as the lengths of all three sides of the triangle are known, Heron's formula can be applied to find the area accurately.
4. Are there any limitations or conditions for using Heron's formula?
Ans. Yes, there are a few conditions to be met in order to use Heron's formula. The lengths of the sides of the triangle must be positive real numbers, and the triangle must be a valid triangle (i.e., the sum of the lengths of any two sides must be greater than the length of the third side). If these conditions are not met, Heron's formula cannot be used to find the area of the triangle.
5. How is Heron's formula different from other methods of finding the area of a triangle?
Ans. Heron's formula is different from other methods of finding the area of a triangle, such as using base and height, because it does not require the measurement of the height of the triangle. Instead, it directly uses the lengths of the sides of the triangle to calculate the area. This makes it a more versatile formula, as it can be applied to any type of triangle without the need for additional measurements.
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