ID : ca-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 5 cm, 12 cm, 10 cm and 13 cm and the angle
between f irst two sides is a right angle.
(2)
T he perimeter of a quadrilateral is 18 cm. If the f irst three sides of a quadrilateral, taken in order
are 6 cm, 5 cm and 4 cm respectively, and the angle between f ourth side and the third side is a
right angle, f ind the area of the quadrilateral.
(3)
Find the area of the f igure below. Also f ind the altitude to the base of the triangle.
(All measurements are in centimeters)
(4) T he area of a triangle with sides 15 cm, 13 cm and 4 cm is :
(5)
Find the area of the parallelogram ABCD and the length of the altitude DE in the f igure below:
(6) T he base of an isosceles triangle is 8 cm and perimeter is 18 cm. Find the area of the triangle.
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ID : ca-9-Herons-Formula [2]
(7) T he sides of a triangular f ield are 5 m, 5 m and 8 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.
(8)
If in the f igure below AB = 12cm, BC=20cm, CA = 10cm and BE = 25cm, f ind the area of the
trapezium BDCE.
Choose correct answer(s) f rom given choice
(9) T he perimeter of a triangular f ield is 36 and the ratio of the sides is 5:4:3. T he area of the f ield
in sq m is :
a. 54
b. 12.727922061358
c. 162
d. 1620
(10) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three
sides. If the lengths of the perpendiculars are a, b and c, f ind the altitude of the triangle.
a.
b.
c. a+b+c
d.
(11) T he perimeter of an isosceles triangle is 36 cm. T he ratio of the equal side to its base is 5 : 8.
Find the area of the triangle.
a. 96 cm2
b. 62.4 cm2
c. 48 cm2
d. 33 cm2
(12) An umbrella is made by stitching 10 triangular pieces of cloth each piece measuring 24 cm, 13 cm
and 13 cm. How much cloth is required f or this umbrella.
a. 300cm2
b. 900cm2
c. 600cm2
d. 60cm2
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(13) Find the area of a quadrilateral ABCD where AB = 9 cm, BC = 10 cm, CD = 8 cm, DB = 15 cm and
AC = 17 cm.
a. 60cm2
b. 960cm2
c. 36cm2
d. 96cm2
(14) T he sides of a quadrilateral, taken in order are 26 cm, 17 cm, 24 cm and 7 cm respectively. T he
angle contained by the last two sides is a right angle. Find the area of the quadrilateral.
a. 204 cm2
b. 374.4 cm2
c. 576 cm2
d. 288 cm2
(15) T he sides of a triangle are 7 cm, 15 cm and 20 cm. T he altitude to the longest side is :
a. 21 cm
b. 5.46 cm
c. 4.2 cm
d. 42 cm
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ID : ca-9-Herons-Formula [4]
Answers
(1)
90 cm2
Step 1
Let's ABCD is the quadrilateral with AB = 5 cm, BC = 12 cm, CD = 10 cm, DA = 13 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 (5) (12) = 30 cm2
Step 3
AC = √(AB2 + BC2) = √(52 + 122) = 13 cm
Step 4
T he area of the triangle ACD can be calculated using Heron's f ormula.
S = (CD + DA + AC)/2 = (10 + 13 + 13)/2 = 18 cm
T he area of the triangle ACD = √[ S (S-CD) (S-DA) (S-AC) ]
= √[ 18 (18-10) (18-13) (18-13) ]
= 60 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ABC) + Area(ACD) = 30 + 60 = 90 cm2
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ID : ca-9-Herons-Formula [5]
(2)
Area : 18 cm2
Step 1
Following picture shows the quadrilateral ABCD,
T he perimeter of the quadrilateral ABCD = 18 cm
T heref ore, AB + BC + CD + DA = 18
⇒ 6 + 5 + 4 + DA = 18
⇒ 15 + DA = 18
⇒ DA = 18 - 15
⇒ DA = 3 cm
Step 2
Let's draw the line AC.
T he ΔACD is the right angled triangle.
T heref ore, AC2 = DA2 + DC2
⇒ AC = √[ DA2 + DC2 ]
= √[ 32 + 4 2 ]
= 5 cm
Step 3
T he area of the right angled triangle ΔACD =
DA × DC
2
=
3×4
2
= 6 cm2
Step 4
Now, we can see that, this quadrilateral consists of the triangles ΔACD and ΔABC.
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ID : ca-9-Herons-Formula [6]
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 + 5 + 5)/2
= 8 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 8(8 - 6) (8 - 5) (8 - 5) ]
= 12 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = 6 + 12 = 18 cm2
(3)
Area : 42 cm2
Altitude : 12 cm
Step 1
T he area of the triangle can be calculated using Heron's f ormula, since all sides of the
triangle are known.
S = (7 + 15 + 20)/2 = 21 cm.
T he area of the triangle = √[ 21(21 - 7) (21 - 15) (21 - 20) ] = 42 cm2
Step 2
Let's assume, 'h' is the altitude to the base of the triangle,
T he altitude to the base of the triangle =
2 × (T he area of the triangle)
T he base of the triangle
=
2 × 42
7
= 12 cm
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ID : ca-9-Herons-Formula [7]
(4) 24 cm2
Step 1
T he area of a triangle with sides a,b, c is given by Heron's f ormula as
Area =
Where S is half of the perimeter, i.e S =
a+b+c
2
Step 2
Here S =
15 + 13 + 4
2
=
32
= 16
2
Step 3
Area =
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=
= 24 cm2
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ID : ca-9-Herons-Formula [8]
(5)
Area : 84 cm2
Altitude : 12 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
= (7 + 15 + 20)/2
= 21 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 21(21 - 7) (21 - 15) (21 - 20) ]
= 42 cm2
Step 3
T he area of the parallelogram ABCD = 2 × Area(ΔABC)
= 2 × 42
= 84 cm2
Step 4
T he length of the altitude DE =
2 × Area(ΔABC)
AB
=
2 × 42
7
= 12 cm.
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ID : ca-9-Herons-Formula [9]
(6) 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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ID : ca-9-Herons-Formula [10]
(7) 300
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
= (5 + 5 + 8)/2
= 9 m.
Area(Δ ABC) = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 9(9 - 5) (9 - 5) (9 - 8) ]
= 12 m2.
= 12 × 10000
= 120000 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 120000 cm2 space =
120000
= 300 =
400
300 rose beds.
(8)
Area : 195.17 cm2
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ID : ca-9-Herons-Formula [11]
(9) a. 54
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=5x, b=4x and c=3x (we know this because the
ratio of the sides is given as 5:4:3)
Step 3
We also know that a+b+c = 36.
= 36
(5 + 4 + 3)x = 36
12x = 36
x=
36
=3
12
Step 4
From this we see that a = 15 m, b = 12 m and c=9 m. Also S=18
Step 5
Putting these values into Heron's f ormula,
Area =
Area = 54 m2
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(10) c. a+b+c
Step 1
T he f ollowing f igure shows the required triangle:
Step 2
Let's assume the side of the equilateral triangle ΔABC is x.
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
= (x + x + x)/2
= 3x/2 cm.
T he area of the ΔABC =
4
x2 ------(1)
Step 3
T he area of the triangle AOB =
AB × OP
2
=
x×b
2
=
bx
2
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ID : ca-9-Herons-Formula [13]
Step 4
Similarly, the area of the triangle ΔBOC =
ax
2
cx
and the area of the triangle ΔAOC =
2
Step 5
T he area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
bx
=
+
2
ax
2
(a + b + c)x
=
cx
+
2
-----(2)
2
Step 6
By comparing the equations (1) and (2), we get:
√3
x2 =
(a + b + c)x
4
2
⇒x=
2(a + b + c)
------(3)
√3
Step 7
Now, Area(ΔABC) =
√3
(x)2
4
Step 8
Area(ΔABC) =
'AB' × 'Altitude of the triangle ΔABC'
2
⇒
√3
(x)2 × 2 = 'x' × 'Altitude of the triangle ΔABC'
4
⇒
√3
(x) = 'Altitude of the triangle ΔABC'
2
By putting the value of 'x' f rom the equation (3), we get,
Altitude of the triangle ΔABC =
√3
(
2(a + b + c)
2
)
√3
⇒ Altitude of the triangle ΔABC = a+b+c
Step 9
Hence, the altitude of the triangle is a+b+c.
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(11) c. 48 cm2
Step 1
Let's assume, the lengths of the base and the equal sides of the isosceles triangle are b
cm and x cm respectively.
Following f igure shows the isosceles triangle ABC,
T he ratio of the equal side to its base is 5 : 8.
T heref ore,
5
=
8
x
b
By cross multiplying, we get:
b=
8x
------(1)
5
Step 2
According to the question, the perimeter of the isosceles triangle ABC = 36 cm
T heref ore, x + x + b = 36
8x
⇒ 2x +
= 36 [From equation (1), b =
8x
5
⇒
10x + 8x
]
5
= 36
5
⇒ 10x + 8x = 180
⇒ 18x = 180
⇒ x = 10 cm
Step 3
Putting the value of x in equation (1), we get:
b=
80
= 16 cm
5
Step 4
T he area of the isosceles triangle ABC can be calculated using Heron's f ormula, since all
sides of the triangle are known.
S = 36/2 = 18 cm
T he area of the isosceles triangle ABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 18(18 - 16) (18 - 10) (18 - 10) ]
= 48 cm2
Step 5
T hus, the area of the triangle is 48 cm2.
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(13) d. 96cm2
Step 1
Following picture shows the quadrilateral ABCD,
Step 2
Let's draw the diagonal AC in the quadrilateral ABCD,
Step 3
Now, we can see that, this quadrilateral consists of two triangles, i.e. ΔABC and ΔACD, the
area of each triangle can be calculated using Heron's f ormula, since all sides of the
triangles are known.
Step 4
T he area of the ΔABC can be calculated using Heron's f ormula.
S = (AB + BC + AC)/2
= (9 + 10 + 17)/2
= 18 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - AC) ]
= √[ 18(18 - 9) (18 - 10) (18 - 17) ]
= 36 cm2
Step 5
Similarly, the area of the ΔACD can be calculated using Heron's f ormula.
S = (AC + CD + AD)/2
= (17 + 8 + 15)/2
= 20 cm.
T he area of the ΔACD = √[ S (S - AC) (S - CD) (S - AD) ]
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= √[ 20(20 - 9) (20 - 10) (20 - 17) ]
= 60 cm2
Step 6
T he area of the quadrilateral ABCD = T he area of the ΔABC + T he area of the ΔACD
= 36 + 60
= 96 cm2
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(14) d. 288 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 ]
= √[ 7 2 + 24 2 ]
= 25 cm
Step 3
T he area of the right angled triangle ΔACD =
AD × DC
2
=
7 × 24
2
= 84 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
= (26 + 17 + 25)/2
= 34 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
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= √[ 34(34 - 26) (34 - 17) (34 - 25) ]
= 204 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = 84 + 204 = 288 cm2
(15) c. 4.2 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 + 15 + 7)/2
= 21 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 21(21 - 20) (21 - 15) (21 - 7) ]
= 42 cm2
Step 2
T he altitude to the longest side =
2 × (T he area of the ΔABC)
Base 'AB'
=
2 × 42
20
= 4.2 cm.
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