RD Sharma Solutions for Class 9 Maths Chapter 12 Exercise 12.2 Heron’s Formula are given here. In this exercise, students will deal with the various applications of Heron’s formula. For example, if a landowner needs to find out the area of his land, which is in the shape of a quadrilateral, he needs to divide the quadrilateral into triangular parts and use heron’s formula for the area of a triangular part. Learn to apply heron’s formula easily by understanding the practice problems that are solved in RD Sharma Class 9 Solutions for Chapter 12, given here.
RD Sharma Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2
Access Answers to RD Sharma Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 Page Number 12.19
Question 1: 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:
Area of the quadrilateral ABCD = Area of △ABC + Area of △ADC ….(1)
△ABC is a right-angled triangle, which is right-angled at B.
Area of △ABC = 1/2 x Base x Height
= 1/2×AB×BC
= 1/2×3×4
= 6
Area of △ABC = 6 cm2 ……(2)
Now, In △CAD,
Sides are given, apply Heron’s Formula.
Perimeter = 2s = AC + CD + DA
2s = 5 cm + 4 cm + 5 cm
2s = 14 cm
s = 7 cm
Area of the △CAD = 9.16 cm2 …(3)
Using equations (2) and (3) in (1), we get
Area of quadrilateral ABCD = (6 + 9.16) cm2
= 15.16 cm2.
Question 2: The sides of a quadrilateral field, taken in order, are 26 m, 27 m, 7 m, and 24 m, respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution:
Here,
AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m
AC is the diagonal joined at A to C point.
Now, in △ADC,
From the Pythagoras theorem,
AC2 = AD2 + CD2
AC2 = 142 + 72
AC = 25
Now, the area of △ABC
All the sides are known, Apply Heron’s Formula.
Perimeter of △ABC= 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
= 291.84
Area of a triangle ABC = 291.84 m2
Now, for the area of △ADC, (Right angle triangle)
Area = 1/2 x Base X Height
= 1/2 x 7 x 24
= 84
Thus, the area of a △ADC is 84 m2
Therefore, the area of rectangular field ABCD = Area of △ABC + Area of △ADC
= 291.84 m2 + 84 m2
= 375.8 m2
Question 3: The sides of a quadrilateral, taken in order as 5, 12, 14, and 15 meters, respectively, and the angle contained by the first two sides is a right angle. Find its area.
Solution:
Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join the diagonal AC.
Now, the area of △ABC = 1/2 ×AB×BC
= 1/2×5×12 = 30
The area of △ABC is 30 m2
In △ABC (right triangle),
From the Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = 52 + 122
AC2 = 25 + 144 = 169
or AC = 13
Now in △ADC,
All sides are known, apply Heron’s Formula:
Perimeter of △ADC = 2s = AD + DC + AC
2s = 15 m +14 m +13 m
s = 21 m
= 84
Area of △ADC = 84 m2
Area of quadrilateral ABCD = Area of △ABC + Area of △ADC
= (30 + 84) m2
= 114 m2
Question 4: A park in the shape of a quadrilateral ABCD has ∠ C = 900, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?
Solution:
Here, AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.
And BD is a diagonal of ABCD.
In the right △BCD,
From the Pythagoras theorem;
BD2 = BC2 + CD2
BD2 = 122 + 52 = 144 + 25 = 169
BD = 13 m
Area of △BCD = 1/2×BC×CD
= 1/2×12×5
= 30
Area of △BCD = 30 m2
Now, In △ABD,
All sides are known, Apply Heron’s Formula:
Perimeter of △ABD = 2s = 9 m + 8m + 13m
s = 15 m
= 35.49
Area of the △ABD = 35.49 m2
Area of quadrilateral ABCD = Area of △ABD + Area of △BCD
= (35.496 + 30) m2
= 65.5m2.
Question 5: 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: AB = 77 m , CD = 60 m, BC = 26 m and AD = 25m
AE and CF are diagonals.
DE and CF are two perpendiculars on AB.
Therefore, we get, DC = EF = 60 m
Let’s say, AE = x
Then BF = 77 – (60 + x)
BF = 17 – x …(1)
In the right △ADE,
From the Pythagoras theorem,
DE2 = AD2 − AE2
DE2 = 252 − x2 ….(2)
In right △BCF
From the Pythagoras theorem,
CF2 = BC2 − BF2
CF2 = 262 − (17−x) 2
[Uisng (1)]Here, DE = CF
So, DE2 = CF2
(2) ⇒ 252 − x2 = 262 − (17−x)2
625 − x2 = 676 – (289 −34x + x2)
625 − x2 = 676 – 289 +34x – x2
238 = 34x
x =7
(2) ⇒ DE2 = 252 – (7)2
DE2 = 625−49
DE = 24
Area of trapezium = 1/2×(60+77)×24 = 1644
Area of trapezium is 1644 m2 (Answer)
Question 6: Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.
Solution:
The perimeter of a rhombus = 80 m (given)
We know, Perimeter of a rhombus = 4×side
Let a be the side of a rhombus.
4×a = 80
or a = 20
One of the diagonal, AC = 24 m (given)
Therefore OA = 1/2×AC
OA = 12
In △AOB,
Using the Pythagoras theorem:
OB2 = AB2 − OA2 = 202 −122 = 400 – 144 = 256
or OB = 16
Since the diagonal of the rhombus bisect each other at 90 degrees.
And OB = OD
Therefore, BD = 2 OB = 2 x 16 = 32 m
Area of rhombus = 1/2×BD×AC = 1/2×32×24 = 384
Area of rhombus = 384 m2.
Question 7: 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 m2. Find the cost of painting.
Solution:
The perimeter of a rhombus = 32 m
We know, Perimeter of a rhombus = 4×side
⇒ 4×side = 32
side = a = 8 m
Each side of the rhombus is 8 m
AC = 10 m (Given)
Then, OA = 1/2×AC
OA = 1/2×10
OA = 5 m
In the right triangle AOB,
From Pythagoras theorem;
OB2 = AB2–OA2 = 82 – 52 = 64 – 25 = 39
OB = √39 m
And, BD = 2 x OB
BD = 2√39 m
Area of the sheet = 1/2×BD×AC = 1/2 x (2√39 × 10 ) = 10√39
The area of the sheet is 10√39 m2
Therefore, the cost of printing on both sides of the sheet, at the rate of Rs. 5 per m2
= Rs. 2 x (10√39 x 5) = Rs. 625.
RD Sharma Solutions for Class 9 Maths Chapter 12 Exercise 12.2
RD Sharma Solutions Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 is based on applications of Heron’s formula. All questions provided in this exercise are solved by subject experts at BYJU’S to help students understand the concepts better. Find detailed RD Sharma Solutions for Class 9 Maths here and clear all your doubts.
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