ID : gb-9-Herons-Formula [1]
Grade 9
Herons Formula
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Answer t he quest ions
(1)
Find the area of a quadrilateral whose sides are 3 cm, 4 cm, 5 cm and 6 cm and the angle
between f irst two sides is a right angle.
(2)
T he sides of a triangular f ield are 7 m, 15 m and 20 m. Find the number of rose beds that can be
prepared in the f ield, if each rose bed, on an average needs 400 cm2 space.
(3)
A f ield in the shape of trapezium has its parallel sides as 28 m and 25 m while the non parallel
sides are 25 m and 26 m. Find the amount of money f armer has to pay if the cost of sowing the
seeds per m2 area is £50.
(4) A carpenter has cut a board in the shape of trapezium. if the parallel sides of the trapezium are
31 cm and 28 cm and non parallel sides are 25 cm and 26 cm, f ind the area of the board.
(5)
Find the area of the parallelogram ABCD and the length of the altitude DE in the f igure below:
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ID : gb-9-Herons-Formula [2]
(6) Find the area of the unshaded region in the f igure below:
(7) T he sides of a triangle are 11 cm, 13 cm and 20 cm. T he altitude to the longest side is :
Choose correct answer(s) f rom given choice
(8)
T he perimeter of a triangular f ield is 64 and the ratio of the sides is 15:13:4. T he area of the
f ield in sq m is :
a. 16.970562748477
b. 6240
c. 288
d. 96
(9) If in the f igure below BC = 13 cm, CA = 5 cm and BD = 12 cm, f ind the area of the triangle ABC.
a. 60 cm2
b. 36 cm2
c. 30 cm2
d. 15 cm2
(10) Which triangle has the maximum area f or a given perimeter :
a. Isoceles T riangle
b. Can not be determined
c. Obtuse Angle T riangle
d. Equilateral T riangle
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ID : gb-9-Herons-Formula [3]
(11) A parallelogram has a diagonal of 9 cm. T he perpendicular distance of this diagonal f rom an
opposite vertex is 4 cm. Find the area of the parallelogram.
a. 18cm2
b. 36cm2
c. 9cm2
d. 13cm2
(12) T he base of an isosceles triangle is N cm and its perimeter is P cm. Find the area of the triangle.
cm2
a.
cm2
c.
cm2
b.
cm2
d.
(13) T he base of an isosceles triangle is 8 cm and perimeter is 9 cm. Find the area of the triangle.
a. 8.4 cm2
b. 12 cm2
c. 15.6 cm2
d. 16.8 cm2
(14) T he sides of a quadrilateral, taken in order are 13 cm, 10 cm, 12 cm and 5 cm respectively. T he
angle contained by the last two sides is a right angle. Find the area of the quadrilateral.
a. 180 cm2
b. 117 cm2
c. 60 cm2
d. 90 cm2
(15) An umbrella is made by stitching 9 triangular pieces of cloth each piece measuring 24 cm, 13 cm
and 13 cm. How much cloth is required f or this umbrella.
a. 540cm2
b. 54cm2
c. 810cm2
d. 1080cm2
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ID : gb-9-Herons-Formula [4]
Answers
(1)
18 cm2
Step 1
Let's ABCD is the quadrilateral with AB = 3 cm, BC = 4 cm, CD = 5 cm, DA = 6 cm, and angle
∠ABC = 90°, as shown in the f ollowing f igure.
Step 2
Let's draw the diagonal AC in the quadrilateral ABCD,
T he area of the right triangle ABC = (1/2) × AB × BC = 1/2 (3) (4) = 6 cm2
Step 3
AC = √(AB2 + BC2) = √(32 + 4 2) = 5 cm
Step 4
T he area of the triangle ACD can be calculated using Heron's f ormula.
S = (CD + DA + AC)/2 = (5 + 6 + 5)/2 = 8 cm
T he area of the triangle ACD = √[ S (S-CD) (S-DA) (S-AC) ]
= √[ 8 (8-5) (8-6) (8-5) ]
= 12 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ABC) + Area(ACD) = 6 + 12 = 18 cm2
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ID : gb-9-Herons-Formula [5]
(2)
1050
Step 1
Following f igure shows the triangular f ield ABC,
T he area of the triangular f ield ABC can be calculated using Heron's f ormula, since all sides
of the triangular f ield are known.
S = (AB + BC + CA)/2
= (7 + 15 + 20)/2
= 21 m.
Area(Δ ABC) = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 21(21 - 7) (21 - 15) (21 - 20) ]
= 42 m2.
= 42 × 10000
= 420000 cm2
Step 2
According to the question, the number of rose bed that can be prepared in 400 cm2 space
= 1 rose bed.
T he number of rose beds that can be prepared in 420000 cm2 space =
420000
= 1050 =
400
1050 rose beds.
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ID : gb-9-Herons-Formula [6]
(3)
£31800
Step 1
Following picture shows the trapezium ABCD,
Step 2
Let's draw the line DE parallel to the line BC.
We know that the distance between the two parallel lines at every point must be equal.
T heref ore, EB = DC = 25 m
AE = AB - EB
= 28 - 25
=3m
Step 3
Now, we can see that, this trapezium consists of a triangle ΔADE and a parallelogram
BCDE.
T he area of the triangle ΔADE can be calculated using Heron's f ormula, since all sides of
the triangles are known.
S = (AE + DE + AD)/2
= (3 + 25 + 26)/2
= 27 m.
T he area of the ΔADE = √[ S (S - AE) (S - DE) (S - AD) ]
= √[ 27(27 - 3) (27 - 25) (27 - 26) ]
= 36 m2
Step 4
T he height of the triangle ΔADE and the parallelogram BCDE is equal.
Let's assume, the height of the triangle ΔADE be 'h', as shown in the f ollowing f igure.
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ID : gb-9-Herons-Formula [7]
We know that the area of a triangle = 1/2(Base × Height),
T heref ore, the height of the ΔADE =
2 × (T he area of the ΔADE)
Base
or h =
2 × 36
3
=
72
m
3
T he area of the parallelogram BCDE = EB × h
= 25 ×
72
3
= 600 m2
Step 5
T he area of the f ield = Area(ADE) + Area(BCDE) = 36 + 600 = 636 m2
Step 6
According to the question, the cost of sowing the seeds per m2 area = £50
T he cost of sowing the seeds in 636 m2 area = 636 × 50 = £31800
Step 7
T heref ore, the f armer has to pay £31800.
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ID : gb-9-Herons-Formula [8]
(4) 708 cm2
Step 1
Following picture shows the trapezium ABCD,
Step 2
Let's draw the line DE parallel to the line BC.
We know that the distance between the two parallel lines at every point must be equal.
T heref ore, EB = DC = 28 cm
AE = AB - EB
= 31 - 28
= 3 cm
Step 3
Now, we can see that, this trapezium consists of a triangle ΔADE and a parallelogram
BCDE.
T he area of the triangle ΔADE can be calculated using Heron's f ormula, since all sides of
the triangles are known.
S = (AE + DE + AD)/2
= (3 + 25 + 26)/2
= 27 cm.
T he area of the ΔADE = √[ S (S - AE) (S - DE) (S - AD) ]
= √[ 27(27 - 3) (27 - 25) (27 - 26) ]
= 36 cm2
Step 4
T he height of the triangle ΔADE and the parallelogram BCDE is equal.
Let's assume, the height of the triangle ΔADE be 'h', as shown in the f ollowing f igure.
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ID : gb-9-Herons-Formula [9]
We know that the area of a triangle = 1/2(Base × Height),
T heref ore, the height of the ΔADE =
2 × (T he area of the ΔADE)
Base
or h =
2 × 36
3
=
72
cm
3
T he area of the parallelogram BCDE = EB × h
= 28 ×
72
3
= 672 cm2
Step 5
T he area of the board = Area(ADE) + Area(BCDE) = 36 + 672 = 708 cm2
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ID : gb-9-Herons-Formula [10]
(5)
Area : 120 cm2
Altitude : 20 cm
Step 1
T he diagonal AC divides the parallelogram ABCD into two equal triangles, ΔABC and ΔACD.
T he area of the parallelogram ABCD = 2 × Area(ΔABC).
Step 2
T he area of the ΔABC can be calculated using Heron's f ormula, since all sides of the
triangle are known.
S = (AB + BC + CA)/2
= (6 + 25 + 29)/2
= 30 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 30(30 - 6) (30 - 25) (30 - 29) ]
= 60 cm2
Step 3
T he area of the parallelogram ABCD = 2 × Area(ΔABC)
= 2 × 60
= 120 cm2
Step 4
T he length of the altitude DE =
2 × Area(ΔABC)
AB
=
2 × 60
6
= 20 cm.
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ID : gb-9-Herons-Formula [11]
(6) 66 m2
Step 1
If we look at the f igure caref ully, we notice that, the area of the unshaded region = T he
area of the triangle ΔABC - T he area of the triangle ΔACD.
Step 2
T he area of the triangle ΔABC can be calculated using Heron's f ormula, since all sides of
the triangle are known.
S = (AB + BC + CA)/2
= (39 + 17 + 44)/2
= 50 m.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 50(50 - 39) (50 - 17) (50 - 44) ]
= 330 m2
Step 3
Similarly, the area of the triangle ΔACD can be calculated using Heron's f ormula.
S = (AC + CD + DA)/2
= (44 + 15 + 37)/2
= 48 m.
T he area of the ΔACD = √[ S (S - AC) (S - CD) (S - DA) ]
= √[ 48(48 - 44) (48 - 15) (48 - 37) ]
= 264 m2
Step 4
T hus, the area of the unshaded region = Area(ABC) - Area(ACD)
= 330 - 264
= 66 m2
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ID : gb-9-Herons-Formula [12]
(7) 6.6 cm
Step 1
Let's assume the altitude to the longest side be 'h'.
Following picture shows the required triangle,
T he area of the triangle ΔABC can be calculated using Heron's f ormula, since all sides of
the triangles are known.
S = (AB + BC + CA)/2
= (20 + 13 + 11)/2
= 22 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 22(22 - 20) (22 - 13) (22 - 11) ]
= 66 cm2
Step 2
T he altitude to the longest side =
2 × (T he area of the ΔABC)
Base 'AB'
=
2 × 66
20
= 6.6 cm.
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(8)
d. 96
Step 1
Since we know the perimeter, we can use Heron's f ormula to help us compute the area
T he f ormula states that the area of a triangle with sides a, b and c, and perimeter 2S =
Step 2
Let us assume the 3 sides are of length a=15x, b=13x and c=4x (we know this because the
ratio of the sides is given as 15:13:4)
Step 3
We also know that a+b+c = 64.
= 64
(15 + 13 + 4)x = 64
32x = 64
x=
64
=2
32
Step 4
From this we see that a = 30 m, b = 26 m and c=8 m. Also S=32
Step 5
Putting these values into Heron's f ormula,
Area =
Area = 96 m2
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ID : gb-9-Herons-Formula [14]
(9) c. 30 cm2
Step 1
If we look at the f igure caref ully, we notice that, 'BD' is the altitude to the base of the
triangle ABC.
Step 2
According to the question, the altitude(BD) to the base of the triangle ABC = 12 cm,
T he base(CA) of the triangle ABC = 5 cm
T he area of the ΔABC =
BD × CA
2
=
12 × 5
2
= 30 cm2
(10) d. Equilateral T riangle
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(11) b. 36cm2
Step 1
Consider a parallelogram ABCD as shown in the f igure below
P is the point where the perpendicular f rom point D meets diagonal AC
Step 2
From the diagram, we see that ACD is a triangle. T he area of ACD is half the area of
parallelogram ABCD
T he area of ACD is
1
x base x height
2
Here base = length of diagonal = 9 cm
Height = length of DP = 4 cm
Area of ACD=
1
x 4 x 9 = 18
2
Step 3
Area of parallelogram ABCD = 2 x area of ACD = 2 x 18 = 36 cm2
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(12) c.
cm2
Step 1
As per Heron's f ormula, the area of a triangle with sides a, b and c, and perimeter 2S =
Here S =
P
2
Step 2
Here we have an isosceles triangle, so two sides are equal
Let's assume a=b, and c=C is the base
Also, perimeter P = a + b + c = 2a + N
a=
P- N
2
Area =
=
= (S -
P- N
2
)
Substituting S =
P
in the equation we get
2
Area =
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(13) b. 12 cm2
Step 1
As per Heron's f ormula, the area of a triangle with sides a, b and c, and perimeter 2S =
Step 2
Here we have an isosceles triangle, so two sides are equal
Let's assume a=b
Area =
=
Step 3
We are told that the base (side c) = 8 cm
Also, perimeter = 2S = 18
S=
18
=9
2
Also, Perimeter = a + b + c
18 = 2a + c = 2a + 8
a=
18 - 8
=5
2
Step 4
Substituting, we get
Area =
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=
= 12 cm2
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(14) d. 90 cm2
Step 1
Following picture shows the quadrilateral ABCD,
Step 2
Let's draw the line AC.
T he ΔACD is the right angled triangle.
T heref ore, AC2 = AD2 + DC2
⇒ AC = √[ AD2 + DC2 ]
= √[ 52 + 122 ]
= 13 cm
Step 3
T he area of the right angled triangle ΔACD =
AD × DC
2
=
5 × 12
2
= 30 cm2
Step 4
Now, we can see that, this quadrilateral consists of the triangles ΔACD and ΔABC.
T he area of the ΔABC can be calculated using Heron's f ormula, since all sides of the
triangle are known.
S = (AB + BC + CA)/2
= (13 + 10 + 13)/2
= 18 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
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= √[ 18(18 - 13) (18 - 10) (18 - 13) ]
= 60 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = 30 + 60 = 90 cm2
(15) a. 540cm2
Step 1
Following picture shows the triangular piece of cloth,
Since all sides of the triangle are known, the area of the triangle can be calculated using
Heron's f ormula.
S = (AB + BC + CA)/2
= (24 + 13 + 13)/2
= 25 cm
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - AC) ]
= √[ 25(25 - 24) (25 - 13) (25 - 13) ]
= 60 cm2
Step 2
According to the question, the umbrella is made by stitching 9 triangular pieces of cloth.
T he cloth required f or this umbrella = 9 × Area(ABC)
= 9 × 60
= 540 cm2
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