CHAPTER – 12
HERON’S FORMULA
Exercise 12.1
Page number 12.8
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Question 1: Find the area of a triangle whose sides are respectively
150 cm, 120 cm and 200 cm.
Solution:
We know, Heron’s Formula
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Here, a = 150 cm
b = 120 cm
c = 200 cm
Step 1: Find s
s=
s=
(𝑎+𝑏+𝑐)
2
(150+200+120)
2
s = 235 cm
Step 2: Find Area of a triangle
Area = √235 × (235 − 150) × (235 − 120) × (235 − 200)
= √235 × (85) × (115) × (35)
= √80399375
= 8966.56
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Area of triangle is 8966.56 sq.cm.
Question 2: Find the area of a triangle whose sides are respectively 9
cm, 12 cm and 15 cm.
Solution:
We know, Heron’s Formula
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Here, a = 9 cm
b = 12 cm
c = 15 cm
Step 1: Find s
s=
s=
(𝑎+𝑏+𝑐)
2
(9+12+15)
2
s = 18 cm
Step 2: Find Area of a triangle
Area = √18 × (18 − 9) × (18 − 12) × (18 − 15)
= √18 × 9 × 6 × 3
= √2916
= 54
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Area of triangle is 54 sq.cm.
Question 3: Find the area of a triangle two sides of which are 18 cm
and 10 cm and the perimeter is 42 cm.
Solution:
Given:
a = 18 cm, b = 10 cm, and perimeter = 42 cm
Let c be the third side of the triangle.
Step 1: Find third side of the triangle, that is c
We know, perimeter = 2s,
2s = 42
s = 21
Again, s =
(𝑎+𝑏+𝑐)
2
Put the value of s, we get
21 =
(18+10+𝑐)
2
42 = 28 + c
c = 14 cm
Step 2: Find area of triangle
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
= √21 × (21 − 18) × (21 − 10) × (21 − 14)
= √21 × 3 × 11 × 7
= √4851
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= 21√11
Area = 21√11 square cm.
Question 4: In a triangle ABC, AB = 15cm, BC = 13cm and AC =
14cm. Find the area of triangle ABC and hence its altitude on AC.
Solution:
Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.
Here, a = 15 cm
b = 13 cm
c = 14 cm
From Heron’s Formula;
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
S=
(15+13+14)
2
= 21
Area = √21 × (21 − 13) × (21 − 14) × (21 − 15)
= √21 × 8 × 7 × 6
= √7056
= 84
Area = 84 cm2
Let, BE is a perpendicular on AC
1
Now, area of triangle = × Base × Height
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2
1
= × BE × AC = 84
2
BE = 12 cm
Hence, altitude is 12 cm.
Question 5: The perimeter of a triangular field is 540 m and its sides
are in the ratio 25:17:12. Find the area of the triangle.
Solution:
Let the sides of a given triangle be a = 25x, b = 17x, c = 12x respectively,
Given, Perimeter of triangle = 540 cm
2s = a + b + c
a + b + c = 540 cm
25x + 17x + 12x = 540 cm
54x = 540 cm
x = 10 cm
So, the sides of a triangle are
a = 250 cm
b = 170 cm
c = 120 cm
Semi perimeter, s =
=
(𝑎+𝑏+𝑐)
2
540
2
= 270
s = 270 cm
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From Heron’s Formula;
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
= √270(270 − 250)(270 − 170)(270 − 120)
= √270 × 20 × 100 × 150
= √81000000
= 9000
Hence, the area of the triangle is 9000 cm2.
Question 6: The perimeter of a triangle is 300 m. If its sides are in the
ratio 3: 5: 7. Find the area of the triangle.
Solution:
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle. If
we denote area of the triangle by A, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
We are given, 𝑎: 𝑏: 𝑐 = 3: 5: 7 and perimeter = 300 m
Here,
perimeter
𝑥=
2
300
2
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=
= 150 m
Using these data, we will find the sides of the triangle. Suppose the sides
of the triangle are as follows,
𝑎 = 3𝑥
𝑏 = 5𝑥
𝑐 = 7𝑥
Since 2𝑠 = 300, so
2𝑠 = 𝑎 + 𝑏 + 𝑐
300 = 3𝑥 + 5𝑥 + 7𝑥
300 = 15𝑥
𝑥 = 20
Now we know each side that is,
𝑎 = 3𝑥
= 3 × 20
= 60 m
𝑏 = 5𝑥
= 5 × 20
= 100 m
𝑐 = 7𝑥
= 7 × 20
= 140 m
Now we know all the sides. So we can use Heron’s formula.
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The area of the triangle is;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
= √150(150 − 60)(150 − 100)(150 − 140)
= √150(90)(50)(10)
= √10015 × 9 × 5
= √1005 × 3 × 3 × 3 × 5
= √100 × 3 × 53
= √15003 m2
Question 7: The perimeter of a triangular field is 240 dm. If two of its
sides are 78 cm and 50 dm, find the length of the perpendicular on
the side of length 50 dm from the opposite vertex.
Solution:
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle. If
we denote area of the triangle by A, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where, s =
(𝑎+𝑏+𝑐)
2
We are given two sides of the triangle and
.
That is a = 78 dm, b = 50 dm
We will find third side c and then the area of the triangle using Heron’s
formula.
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2𝑠 = perimeter
2𝑠 = 240
𝑠 = 120
Now,
s=
120 =
𝑎+𝑏+𝑐
2
78+50+𝑐
2
120 × 2 = 128 + c
c = 240 – 128
c = 112 dm
Use Heron’s formula to find out the area of the triangle. That is
𝐴 = √120(120 − 78)(120 − 50)(120 − 112)
= √120(42)(70)(8)
= √2822400
= 1680 dm
Consider the triangle ΔPQR in which
PQ=50 dm, PR=78 dm, QR=120 dm
Where RD is the desired perpendicular length
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Now from the figure we have
1
Area of ∆PQR = × base × height
2
1
= × 50 × RD
2
1
1680 = × 50 × RD
2
RD =
1680×2
50
RD = 67.2 dm
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:
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle.
If we denote area of the triangle by A, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where, s =
(𝑎+𝑏+𝑐)
2
We are given: a = 35 cm; b = 54 cm; c = 61 cm
s=
s=
35+54+61
2
150
2
s = 75 cm
The area of the triangle is:
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𝐴 = √75(75 − 35)(75 − 54)(75 − 61)
= √75(40)(21)(14)
= √882000
= 9390.14 cm²
Suppose the triangle is ΔPQR and focus on the triangle given below,
In which PD1, QD2 and RD3 are three altitudes
Where PQ = 35 cm, QR = 54 cm, PR = 61 cm
We will calculate each altitude one by one to find the smallest one.
Case 1
In case of ΔPQR:
1
Area of ∆PQR = × QR × PD1
2
1
939.14 = × 54 × PD1
2
PD1 =
939.14×2
54
= 34.78 cm
Case 2
1
Area of ∆PQR = × PR × QD2
2
1
939.14 = × 61 × QD2
2
939.14×2
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QD2 =
61
= 30.78 cm
Case 3
1
Area of ∆PQR = × PQ × RD3
2
1
939.14 = × 54 × RD3
2
RD3 =
939.14×2
35
= 53.66 cm
The smallest altitude is QD2.
The smallest altitude is the one which is drawn on the side of length 61
cm from apposite vertex.
Smallest altitude = 30.79 cm
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:
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle.
If we denote area of the triangle by A, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where, s =
(𝑎+𝑏+𝑐)
2
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We are given, 𝑎: 𝑏: 𝑐 = 3: 4: 5 and perimeter = 144 cm
Here,
s=
s=
perimeter
2
144
2
= 72 cm
Using these data, we will find the sides of the triangle. Suppose the sides
of the triangle are as follows,
𝑎 = 3𝑥
𝑏 = 4𝑥
𝑐 = 5𝑥
Since 2s = 144, so
2s = 𝑎 + 𝑏 + 𝑐
144 = 3𝑥 + 4𝑥 + 5𝑥
144 = 12𝑥
𝑥 = 12
Now we know each side that is,
𝑎 = 3𝑥
= 3 × 12
= 36 cm
𝑏 = 4𝑥
= 4 × 12
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= 48 cm
𝑐 = 5𝑥
= 5 × 12
= 60 cm
Now we know all the sides. So we can use Heron’s formula.
The area of the triangle is;
𝐴 = √72(72 − 36)(72 − 48)(72 − 60)
= √72(36)(24)(12)
= √746496
= 864 cm²
We are asked to find out the height corresponding to the longest side of
the given triangle. The longest side is c and supposes the corresponding
height is H then,
1
Area of given triangle = × 𝑐 × 𝐻
2
1
864 = × 60 × 𝐻
2
H=
864
30
= 28.8 cm
Question 10: The perimeter of an isosceles triangle is 42 cm and its
𝟑
base is ( ) times each of the equal sides. Find the length of each side
𝟐
of the triangle, area of the triangle and the height of the triangle.
Solution:
3
We are given that perimeter = 42 cm and its base is ( ) times each of the
2
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equal sides. We are asked to find out the length of each side, area of the
triangle and height of the triangle. In this case ‘height’ is the perpendicular
distance drawn on the base from the opposite vertex.
In the following triangle ΔABC
BC = a, AC = b, AB = c and AB = AC
Let the length of each of the equal sides be x and a, b and c are the side of
the triangle. So,
3
𝑎= 𝑥
2
𝑏=𝑥
𝑐=𝑥
Since perimeter = 𝑎 + 𝑏 + 𝑐.This implies that,
3
42 = 𝑥 + 𝑥 + 𝑥
2
7
42 = 𝑥
2
𝑥=
42×2
7
𝑥 = 12 cm
Therefore, all the sides of the triangle are:
3
𝑎= 𝑥
2
3
= × 12
2
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= 18 cm
𝑏 = 12 cm
𝑐 = 12 cm
All the sides of the triangle are 18 cm, 12 cm, and 12 cm.
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle.
If we denote area of the triangle by Area, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where, s =
(𝑎+𝑏+𝑐)
2
To calculate area of the triangle we need to find s:
s=
=
𝑎+𝑏+𝑐
2
12+12+18
2
= 21 cm
The area of the triangle is:
𝐴 = √21(21 − 12)(21 − 12)(21 − 18)
= √21(9)(9)(3)
= √5103
= 71.43 cm²
Now we will find out the height, say H. See the figure, in which AD = H
So,
1
Area of the ∆ABC = × 𝑐 × 𝐻
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2
1
71.43 = × 18 × 𝐻
2
H=
71.43×2
18
= 7.94 cm
Height = 7.94 cm
Question 11: Find the area of the shaded region in the given figure.
Solution:
We are given the following figure with dimensions.
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Figure:
Let the point at which angle is 90° be D.
AC = 52 cm, BC = 48 cm, AD = 12 cm, BD = 16 cm
We are asked to find out the area of the shaded region.
Area of the shaded region=Area of triangle ΔABC−area of triangle
ΔABD
In right angled triangle ABD, we have
AB² = AD² + BD²
AB² = (12) ² + (16) ²
AB = √144 + 256
AB = 20 cm
Area of the triangle ΔABD is given by
1
Area of the ∆ABC = (base × height)
2
1
= × 𝐴𝐷 × 𝐵𝐷
2
1
= × 12 × 16
2
= 96 cm²
Whenever we are given the measurement of all sides of a triangle, we
basically look for Heron’s formula to find out the area of the triangle.
If we denote area of the triangle by Area, then the area of a triangle having
sides a, b, c and s as semi-perimeter is given by;
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where, s =
(𝑎+𝑏+𝑐)
2
s=
=
=
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Here a = 48 cm, b = 52 cm, c = 20 cm and
𝑎+𝑏+𝑐
2
48+52+20
2
120
2
= 60 cm
Therefore, the area of a triangle ΔABC is given by,
Area of ∆ABC = √60(60 − 20)(60 − 48)(60 − 52)
= √60(40)(12)(8)
= √230400
= 480 cm²
Now we have all the information to calculate area of shaded region, so
Area of shaded region = Area of ΔABC − Area of ΔABD
= 480 – 96
= 384 cm²
The area of the shaded region is 384 cm2.
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.
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Solution:
Area of the quadrilateral ABCD = Area of △ABC + Area of △ADC ….
(1)
△ABC is a right-angled triangle which B.
1
Area of △ABC = × Base × Height
2
1
= × AB × BC
2
1
= ×3×4
2
=6
Area of △ABC = 6 cm2
Now, In △CAD,
Sides are given, apply Heron’s Formula.
… (2)
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Perimeter = 2s = AC + CD + DA
2s = 5 cm + 4 cm + 5 cm
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2s = 14 cm
s = 7 cm
Area of the ∆CAD = √7 × (7 − 5) × (7 − 4) × (7 − 5)
= √7 × 2 × 3 × 2
= 2√21
= 9.61
Area of the △CAD = 9.16 cm2
… (3)
Using equation (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, 24 m respectively. The angle contained by the last two
sides is a right angle. Find its area.
Solution:
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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 Pythagoras theorem;
AC2 = AD2 + CD2
AC2 = 142 + 72
AC = 25
Now, area of △ABC
All the sides are known, Apply Heron’s Formula.
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Perimeter of △ABC = 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
Area of a triangle = √39 × (39 − 25) × (39 − 26) × (39 − 27)
= √39 × 14 × 13 × 12
= √85176
= 291.84
Area of a triangle ABC = 291.84 m2
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Now, for area of △ADC, (Right angle triangle)
1
Area = × Base × Height
2
1
= × 7 × 24
2
= 84
Thus, the area of a △ADC is 84 m2
Therefore, 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,
15 meters respectively, and the angle contained by first two sides is a
right angle. Find its area.
Solution:
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Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join the diagonal AC.
1
Now, area of △ABC = × 𝐴𝐵 × 𝐵𝐶
2
1
= × 5 × 12 = 30
2
Area of △ABC is 30 m2
In △ABC, (right triangle).
From 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:
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Perimeter of △ADC = 2s = AD + DC + AC
2s = 15 m +14 m +13 m
s = 21 m
Area of the ∆ADC = √21 × (21 − 13) × (21 − 14) × (21 − 15)
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= √21 × 8 × 7 × 6
= 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 right △BCD,
From Pythagoras theorem;
BD2 = BC2 + CD2
BD2 = 122 + 52 = 144 + 25 = 169
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BD = 13 m
1
Area of △BCD = × 𝐵𝐶 × 𝐶𝐷
2
1
= × 12 × 5
2
= 30
Area of △BCD = 30 m2
Now, In △ABD,
All sides are known, Apply Heron’s Formula:
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
Perimeter of △ABD = 2s = 9 m + 8m + 13m
s = 15 m
Area of the ∆ABD = √15 × (15 − 9) × (15 − 8) × (15 − 13)
= √15 × 6 × 7 × 2
= 6√35
= 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.
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Question 5: Find the area of a rhombus whose perimeter is 80 m and
one of whose diagonal is 24 m.
Solution:
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)
1
Therefore, OA = × AC
2
OA = 12
In △AOB,
Using Pythagoras theorem:
OB2 = AB2 − OA2 = 202 −122 = 400 – 144 = 256
or OB = 16
Since diagonal of rhombus bisect each other at 90 degrees.
And OB = OD
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Therefore, BD = 2 OB = 2 × 16 = 32 m
1
1
2
2
Area of rhombus = × 𝐵𝐷 × 𝐴𝐶 = × 32 × 24 = 384
Area of rhombus = 384 m2.
Question 6: 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:
Perimeter of a rhombus = 32 m
We know, Perimeter of a rhombus = 4 × side
⇒ 4 × side = 32
side = a = 8 m
Each side of rhombus is 8 m
AC = 10 m (Given)
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1
Then, OA = × 𝐴𝐶
2
1
OA = × 10
2
OA = 5 m
In right triangle AOB,
From Pythagoras theorem;
OB2 = AB2–OA2 = 82 – 52 = 64 – 25 = 39
OB = √39 m
And, BD = 2 × OB
BD = 2√39 m
1
1
2
2
Area of the sheet = × 𝐵𝐷 × 𝐴𝐶 = × (2√39 × 10) = 10√39
Area of the sheet is 10√39 m2
Therefore, cost of printing on both sides of the sheet, at the rate of Rs. 5
per m2
= Rs. 2 × (10√39 × 5)
= Rs. 625.
Question 7: Find the area of a quadrilateral ABCD in which AD = 24
cm, ∠BAD = 90° and BCD forms an equilateral triangle who’s each
side is equal to 26 cm.
Solution:
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We assume ABCD be the quadrilateral having sides AB, BC, CD, DA,
diagonal BD and angle BAD = 90° where BCD forms an equilateral
triangle having equal sides.
We need to find area of ABCD
In triangle BAD, we have
BD2 = BA2 + AD2. So
BA = √262 − 242
= 10 cm
Area of right angled triangle ABD, say A1 is given by
1
A1 = (Base × Height)
2
Where,
Base = BA = 10 cm; Height = AD = 24 cm
1
A1 = (10 × 24)
2
A1 = 120 cm²
Area of equilateral triangle BCD, say A2 having sides a, b, c is given by
A2 =
√3 2
𝑎 ,
2
where
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a = BC = CD = BD = 26 cm
A2 =
A2 =
√3
(26)2
4
1169.48
4
A2 = 292.37 cm²
Area of quadrilateral ABCD, say A
A = Area of triangle BAD + Area of triangle BCD
A = A1 + A2
A = 120 + 292.37
A = 412.37 cm²
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:
The quadrilateral ABCD having sides AB, BC, CD, DA and diagonal BD is
given, where BD divides ABCD into two triangles
ΔDBC and ΔDAB
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In triangle DBC, we can observe that
DC2 = DB2 + BC2
Therefore, it is a right angled triangle.
Area of right angled triangle DBC, say A1 is given by
1
A1 = (Base × Height)
2
Where,
Base = BC = 21 cm; Height = BD = 20 cm
1
A1 = (21 × 20)
2
A1 = 210 cm²
Area of triangle DAB, say A2 having sides a, b, c and s as semi-perimeter
is given by
A2 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
a = DB = 20 cm; b = AD = 34 cm; c = AB = 42 cm
s=
20+34+42
2
s=
96
2
s = 48
A2 = √48(48 − 20)(48 − 34)(48 − 42)
A2 = √48(28)(14)(6)
A2 = √112896
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A2 = 336 cm²
Area of quadrilateral ABCD, say A
A = Area of triangle DBC + Area of triangle DAB
A = A1 + A2
A = 210 + 336
A = 546 cm²
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:
We are given the measure of adjacent sides of a parallelogram AB and BC
that is the sides having same point of origin and the diagonal AC which
divides parallelogram ABCD into two congruent triangles ABC and
ADC.
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Area of triangle ABC is equal to Area of triangle ADC as they are
congruent triangles.
Area of parallelogram ABCD, say A is given by
A = 2 × Area of triangle ABC
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A2 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
s=
𝑎+𝑏+𝑐
2
Therefore, the area of a triangle, say A1 having sides 20 cm, 34 cm and 42
cm is given by
a = 20 cm; b = 34 cm; c = 42 cm
s=
s=
s=
𝑎+𝑏+𝑐
2
20+34+42
2
96
2
s = 48 cm
A1 = √48(48 − 20)(48 − 34)(48 − 42)
A1 = √48(28)(14)(6)
A1 = √112896
A1 = 336 cm²
Area of parallelogram ABCD, say A is given by
A = 2 + A1
A = 2 + 336
A = 672 cm²
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Question 10: Find the area of the blades of the magnetic compass shown
in the given figure. (Take √11 = 3.32)
Solution:
The blades of the magnetic compass are forming a rhombus having all
equal sides measuring 5 cm each. A diagonal measuring 1 cm is given
which is forming the triangular shape of the blades of the magnetic
compass and diving the rhombus into two congruent triangles, say triangle
ABC and triangle DBC having equal dimensions.
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
s=
𝑎+𝑏+𝑐
2
Therefore, the area of a triangle ABC, say A1 having sides 5 cm, 5 cm and
1 cm is given by:
s=
s=
s=
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a = 5 cm; b = 5 cm; c = 1 cm
𝑎+𝑏+𝑐
2
5+5+1
2
11
2
s = 5.5 cm
A1 = √5.5(5.5 − 5)(5.5 − 5)(5.5 − 1)
A1 = √5.5(0.5)(0.5)(4.5)
A1 = √6.1875
A1 = 2.49 cm²
Area of blades of magnetic compass, say A is given by
A = 2 × Area of one of triangle ABC
A = 2 × A1
A = 2 × 2.49
A = 4.98 cm²
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:
It is given that the area of triangle and parallelogram are equal.
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We will calculate the area of triangle with the given values and it will also
give us the area of parallelogram as both are equal.
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
s=
𝑎+𝑏+𝑐
2
Therefore, the area of a triangle; say A, having sides 15 cm, 13 cm and 14
cm is given by
a = 15 cm; b = 13 cm; c = 14 cm
s=
s=
s=
𝑎+𝑏+𝑐
2
15+13+14
2
42
2
s = 21 cm
A1 = √21(21 − 15)(21 − 13)(21 − 14)
A1 = √21(6)(8)(7)
A1 = √7056
A1 = 84 cm²
We have to find the height of the parallelogram, say h
Area of parallelogram AECD say A1 is given by
A1 = Base × Height
Base = 60 cm; Height = h cm; A =A1 = 84 cm2
84 = 14 × ℎ
84
14
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ℎ=
ℎ = 6 cm
Question 12: 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 right △ADE,
From Pythagoras theorem,
DE2 = AD2 − AE2
DE2 = 252 − x2
… (2)
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In right △BCF
From Pythagoras theorem,
CF2 = BC2 − BF2
CF2 = 262 − (17−x) 2
[Using (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
1
Area of trapezium = × (60 + 77) × 24 = 1644
2
Area of trapezium is 1644 m2 (Answer)
Question 13: Find the perimeter and area of the
quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm,
∠ACB = 90° and AC = 15 cm.
Solution:
We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and
∠ACB = 90∘.
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We take a diagonal AC, where AC divides ABCD into two triangles
ΔACB and ΔADC
Since ∆ACB is right angled at C, we have
AC = 15 cm; AB = 17 cm
AB2 = AC2 + BC2
(17)2 = (15)2 + BC 2
BC 2 = 289 − 225
BC 2 = √64
BC = 8 cm
Area of right angled triangle ABC, say A1 is given by
1
A1 = (Base × Height), where,
2
Base = BC = 8 cm; Height = AC = 15 cm
1
A1 = (8 × 15)
2
A1 = 60 cm²
Area of triangle ADC, say A2 having sides a, b, c and s as semi-perimeter
is given by
A2 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
a = AD = 9 cm; b = DC = 12 cm; c = AC = 15 cm
s=
𝑎+𝑏+𝑐
2
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s=
9+12+15
2
s = 18 cm
A2 = √18(18 − 9)(18 − 12)(18 − 15)
A2 = √18(9)(6)(3)
A2 = √2916
A2 = 54 cm²
Area of quadrilateral ABCD, say A
A = Area of ∆ACB + Area of ∆ADC
A = A1 + A2
= 60 + 54
A = 114 cm²
Perimeter of quadrilateral ABCD, say P
P = 9 + 12 + 8 + 17
= 46 cm
P = 46 cm
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Question 14: A hand fan is made by stitching 10 equal size triangular
strips of two different types of paper as shown in Fig. 12.28. 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.
Solution:
We have to find the area of each type of triangular strips needed for the
fan.
There are 5 strips of each type having equal dimensions, so we will
calculate the area of a single strip and then multiply it by 5 to ascertain
the area of each type of strip needed.
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A2 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where
s=
𝑎+𝑏+𝑐
2
Therefore, the area of a triangular strip, say A1 having sides 25 cm, 25 cm
and 14 cm is given by:
a = 25 cm; b = 25 cm; c = 14 cm
s=
s=
𝑎+𝑏+𝑐
2
25+25+14
2
s=
64
2
s = 32 cm
A1 = √32(32 − 25)(32 − 25)(32 − 14)
A1 = √32(7)(7)(18)
A1 = √28224
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A1 = 168 cm²
Area of each type of strip needed, say A.
A = 5 × Area of each triangular strip
A = 5 × A2
= 5 × 168
A = 840 cm²
VERY SHORT ANSWER TYPES QUESTION (VSAQs)
Page number 12.23
Question 1: Find the area of a triangle whose base and altitude are 5
cm and 4 cm respectively.
Solution:
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Given: Base of a triangle = 5 cm and altitude = 4 cm
1
Area of triangle = × base × altitude
2
1
= ×5×4
2
= 10
Area of triangle is 10 cm2.
Question 2: Find the area of a triangle whose sides are 3 cm, 4 cm and
5 cm respectively.
Solution:
Given: Sides of a triangle are 3 cm, 4 cm and 5 cm respectively
Apply Heron’s Formula:
Area of triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Semi Perimeter, s =
(𝑎+𝑏+𝑐)
2
Where, a, b and c are sides of a triangle
S=
(3+4+5)
2
=6
Semi perimeter is 6 cm
Now,
Area = √6 × (6 − 3) × (6 − 4) × (6 − 5)
= √6 × 3 × 2 × 1
= √36
=6
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Area of given triangle is 6 cm2.
Question 3: Find the area of an isosceles triangle having the base x
cm and one side y cm.
Solution:
In right triangle APC,
Using Pythagoras theorem,
AC2 = AP2 + PC2
2
2
𝑥 2
y =h +( )
2
𝑥 2
or h2 = y2 – ( )
2
or, h = √𝑦 2 −
𝑥2
4
1
Now, area = × Base × Height
2
1
𝑥2
2
4
= × ( √𝑦 2 −
)
𝑥
= √4𝑦 2 − 𝑥 2
4
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Question 4: Find the area of an equilateral triangle having each side
4 cm.
Solution: Each side of an equilateral triangle = a = 4 cm
√3
4
Formula for Area of an equilateral triangle = ( ) × a²
√3
4
= ( ) × 4² = 4√3
Area of an equilateral triangle is 4√3 cm2.
Question 5: Find the area of an equilateral triangle having each side
x cm.
Solution:
Each side of an equilateral triangle = a = x cm
√3
4
Formula for Area of an equilateral triangle = ( ) × a²
√3
4
= ( ) × x²
= x2
Area of an equilateral triangle is
√3𝑥 2
4
cm2.
√3
4
Question 6: The perimeter of a triangular field is 144 m and the ratio
of the sides is 3: 4: 5. Find the area of the field.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where,
𝑎+𝑏+𝑐
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s=
2
It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter
=144 m
Therefore, a: b: c = 3: 4: 5
We will assume the sides of triangular field as
𝑎 = 3𝑥; 𝑏 = 4𝑥; 𝑐 = 5𝑥
2s = 144
s=
144
2
s = 72
72 =
3𝑥+4𝑥+5𝑥
2
72 × 2 = 12𝑥
𝑥=
144
12
𝑥 = 12
Substituting the value of x in, we get sides of the triangle as
𝑎 = 3𝑥 = 3 × 12
𝑎 = 36 cm
𝑏 = 4𝑥 = 4 × 12
𝑏 = 48 m
𝑐 = 5𝑥 = 5 × 12
𝑐 = 60 m
Area of a triangular field, say A having sides a, b, c and s as semiperimeter is given by
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a = 36 m: b = 48 m: c = 60 m
s = 72 m
A = √72(72 − 36)(72 − 48)(72 − 60)
A = √72(36)(24)(12)
A = √746496
A = 864 m²
Question 7: Find the area of an equilateral triangle having
altitude h cm.
Solution:
Altitude of a equilateral triangle, having side a is given by
Altitude =
√3
𝑎
2
Substituting the given value of altitude h cm, we get
h=
√3
𝑎
2
𝑎=
2
√3
ℎ cm
Area of 𝑎 equilateral triangle, say A having each side a cm is given by
√3 2
𝑎
4
A=
Area of the given equilateral triangle having each equal side equal to
2
√3
ℎ
cm is given by;
A=
√3 2
( ℎ
4 √3
A=
√3
4
cm)
4
× ℎ²
3
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A=
2
ℎ2
√3
cm²
Question 8: Let Δ be the area of a triangle. Find the area of a triangle,
who’s each side is twice the side of the given triangle.
Solution:
We are given assumed value
is the area of a given triangle ABC
We assume the sides of the given triangle ABC be a, b, c
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
∆ = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
…… (1)
Where,
s=
𝑎+𝑏+𝑐
2
2𝑠 = 𝑎 + 𝑏 + 𝑐
……. (2)
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of
previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semiperimeter is given by,
A1 = √𝑠1 (𝑠1 − 2𝑎)(𝑠1 − 2𝑏)(𝑠1 − 2𝑐), where
𝑠1 =
𝑠1 =
2𝑎+2𝑏+2𝑐
2
2(𝑎+𝑏+𝑐)
2
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𝑠1 = 𝑎 + 𝑏 + 𝑐
𝑠1 = 2𝑠
…….. (using (2))
Now,
A1 = √2𝑠(2𝑠 − 2𝑎)(2𝑠 − 2𝑏)(2𝑠 − 2𝑐)
A1 = √2𝑠 × 2(𝑠 − 𝑎) × 2(𝑠 − 𝑏) × 2(𝑠 − 𝑐)
A1 = 4√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
…….. (using (1))
A1 = 4∆
Question 9: If each side of a triangle is doubled, the find percentage
increase in its area.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given
by,
A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
Where,
s=
𝑎+𝑏+𝑐
2
2𝑠 = 𝑎 + 𝑏 + 𝑐
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of
previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and
perimeter is given by,
as semi-
A1 = √𝑠1 (𝑠1 − 2𝑎)(𝑠1 − 2𝑏)(𝑠1 − 2𝑐)
Where,
2𝑎+2𝑏+2𝑐
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𝑠1 =
𝑠1 =
2
2(𝑎+𝑏+𝑐)
2
𝑠1 = 𝑎 + 𝑏 + 𝑐
𝑠1 = 2𝑠
Now,
A1 = √2𝑠(2𝑠 − 2𝑎)(2𝑠 − 2𝑏)(2𝑠 − 2𝑐)
A1 = √2𝑠 × 2(𝑠 − 𝑎) × 2(𝑠 − 𝑏) × 2(𝑠 − 𝑐)
A1 = 4√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
A1 = 4𝐴
Therefore, increase in the area of the triangle
= A1 – A
= 4A – A
= 3A
Percentage increase in area
=
3𝐴
𝐴
× 100
= 300%
Question 10: If each side of an equilateral triangle is tripled then what
is the percentage increase in the area of the triangle?
Solution:
Area of an equilateral triangle having each side a cm is given by
√3𝑎2
4
A=
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Now, Area of an equilateral triangle, say A1 if each side is tripled is given
by
a = 3a
A1 =
√3 2
𝑎
4
A1 =
√3
(3𝑎)2
4
A1 =
9√3𝑎2
4
cm²
Therefore, increase in area of triangle
= A1 – A
=
=
9√3𝑎2
−
4
√3𝑎2
4
8√3𝑎2
4
Percentage increase in area
=
8√3𝑎2
4
√3𝑎2
4
× 100
= 800%