ID : pk-9-Herons-Formula [1]
Grade 9
Herons Formula
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Answer t he quest ions
(1)
Find the area of a trapezium, if its height is 4 cm and the lengths of parallel sides are 15 cm and
20 cm.
(2)
T he base of an isosceles triangle is J cm and its perimeter is R cm. Find the area of the triangle.
(3)
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 area of the triangle.
(4) Kabir makes the kite using two pieces of paper. 1st piece of paper is cut in the shape of square
where one diagonal is of the length 26 cm. At one of the vertex of this square a second piece of
paper is attached which is of the shape of an equilateral triangle of length to give the shape of
a kite. T he length of the sides of triangle is a, such that a2 = 4√3. Find the area of this kite.
(5)
T he perimeter of a triangle is 32 cm. One side of a triangle is 9 cm longer than the smallest side
and the third side is 1 cm less than 4 times the smallest side. Find the area of the triangle.
Choose correct answer(s) f rom given choice
(6) Find the percentage increase in the area of a triangle if each side its increased to 3 times.
a. 1000%
b. 900%
c. 800%
d. 700%
(7) A parallelogram has a diagonal of 11 cm. T he perpendicular distance of this diagonal f rom an
opposite vertex is 4 cm. Find the area of the parallelogram.
(8)
a. 22cm2
b. 11cm2
c. 15cm2
d. 44cm2
A f ield in the shape of trapezium has its parallel sides as 27 m and 24 m while the non parallel
sides are 25 m and 26 m. Find the area of the f ield.
a. 6120m2
b. 612m2
c. 1224m2
d. 576m2
(9) 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. 135 cm2
b. 60 cm2
c. 180 cm2
d. 90 cm2
(10) Which triangle has the maximum area f or a given perimeter :
a. Can not be determined
b. Equilateral T riangle
c. Isoceles T riangle
d. Obtuse Angle T riangle
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ID : pk-9-Herons-Formula [2]
(11) If in the f igure below AB = 13cm, BC=22cm, CA = 11cm and BE = 26cm, f ind the area of the
trapezium BDCE.
a. 328.19cm2
b. 105.08cm2
c. 52.54cm2
d. 218.79cm2
(12) T he perimeter of an isosceles triangle is 72 cm. T he ratio of the equal side to its base is 5 : 8.
Find the area of the triangle.
a. 230.4 cm2
b. 192 cm2
c. 167 cm2
d. 384 cm2
(13) T he perimeter of a triangular f ield is 180 m and the ratio of the sides is 8:5:5. Which of the
f ollowing is the area of the f ield in sq m :
a. 840
b. 1200
c. 16000
d. 800
(14) T he perimeter of an isosceles triangle is 242 cm and its unequal side is 120 cm. Find the area of
the triangle.
a. 660cm2
b. 671cm2
c. 121cm2
d. 446520cm2
(15) A rhombus has all its internal angles equal. If one of the diagonals is 14 cm, f ind the length of
other diagonal and the area of the rhombus.
a. 14 cm, 98 cm2
b. 14 cm, 196 cm2
c. 98 cm, 14 cm2
d. 15 cm, 105 cm2
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ID : pk-9-Herons-Formula [3]
Answers
(1)
Area : 70 cm2
Step 1
T he f ollowing picture shows the trapezium ABCD,
Step 2
According to the question, the height of the trapezium ABCD = 4 cm
T he area of the trapezium ABCD = T he height of the trapezium ABCD ×
AB + CD
2
=4×
20 + 15
2
=4×
35
2
= 70 cm2
Step 3
T hus, the area of the trapezium is 70 cm2.
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ID : pk-9-Herons-Formula [4]
(2)
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 =
R
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 R = a + b + c = 2a + J
a=
R- J
2
Area =
=
= (S -
R- J
2
)
Substituting S =
R
in the equation we get
2
Area =
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(3)
Step 1
Following 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 =
------(1)
Step 3
T he area of the triangle AOB =
AB × OP
2
=
x×b
2
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ID : pk-9-Herons-Formula [6]
bx
=
2
Step 4
Similarly, the area of the triangle ΔBOC =
ax
2
and the area of the triangle ΔAOC =
cx
2
Step 5
T he the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
bx
=
ax
+
2
+
2
2
(a + b + c)x
=
cx
-----(2)
2
Step 6
By comparing equation (1) and (2), we get,
√3
(x)2 =
(a + b + c)x
4
2
⇒x=
2(a + b + c)
√3
Step 7
Now, Area(ΔABC) =
√3
(x)2
4
=
√3
(
4
=
√3
4
2(a + b + c)
)2
√3
(
4(a + b + c)2
)
3
=
Step 8
Hence, the area of the triangle is
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(4) 341 cm2
Step 1
Following f igure shows the kite, made by two pieces of paper,
Step 2
Now, we can see that, this kite consists of a square ABCD and a equilateral triangle BEF.
T he area of the equilateral triangle ΔBEF can be calculated using Heron's f ormula f or
equilateral triangle,
Area =
√3 a2
4
=
√3 (4√3)
4
= 3 cm2
Step 3
T he diagonal of the square = 26 cm.
T he area of the square ABCD = (
1
) × (26)2 = 338 cm2.
2
Step 4
T hus, the area of the kite = Area(ABCD) + Area(BEF)
= 3 + 338
= 341 cm2.
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ID : pk-9-Herons-Formula [8]
(5)
Area : 24 cm2
Step 1
Let's assume the smallest side of the triangle be x cm.
Step 2
According to the question, one side of the triangle is 9 cm longer than the smallest side,
T he length of the side = x + 9
Step 3
T he third side is 1 cm less than 4 times the smallest side,
the length of the third side = 4x - 1
Step 4
T he perimeter of the triangle is 32 cm,
T heref ore, x + (x + 9) + (4x - 1) = 32
⇒ x + x + 9 + 4x - 1 = 32
⇒ 6x = 32 + 1 - 9
⇒x=
24
6
⇒x=4
Now, x + 9 = 4 + 9 = 13,
4x - 1 = (4 × 4) - 1 = 15
Step 5
T heref ore, all sides of the triangle are 4 cm, 13 cm and 15 cm.
Step 6
Following picture shows the triangle,
T he area of the ΔABC can be calculated using Heron's f ormula, since all sides of the
triangle are known.
S = 32/2 = 16 cm
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 16(16 - 4) (16 - 13) (16 - 15) ]
= 24 cm2
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(6) b. 900%
Step 1
Consider a triangle QRS with sides a, b and c. Let S =
a+b+c
2
Area of triangle QRS = A1
Step 2
Increasing the side of each side by 3 times, we get a new triangle XYZ
XYZ has sides 3a, 3b and 3c
By Heron's f ormula
Area of new triangle =
Where S1 =
3a + 3b + 3c
2
=3x
a+b+c
= MS
2
Area of XYZ =
=
= 32x A1
T his means the area increases by 32x100 = 900 %
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(7) d. 44cm2
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 = 11 cm
Height = length of DP = 4 cm
Area of ACD=
1
x 4 x 11 = 22
2
Step 3
Area of parallelogram ABCD = 2 x area of ACD = 2 x 22 = 44 cm2
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(8)
b. 612m2
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 = 24 m
AE = AB - EB
= 27 - 24
=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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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
= 24 ×
72
3
= 576 m2
Step 5
T he area of the f ield = Area(ADE) + Area(BCDE) = 36 + 576 = 612 m2
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(9) 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
(10) b. Equilateral T riangle
(11) d. 218.79cm2
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(12) b. 192 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.
5
T heref ore,
=
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 = 72 cm
T heref ore, x + x + b = 72
8x
⇒ 2x +
= 72 [From equation (1), b =
8x
5
⇒
]
5
10x + 8x
= 72
5
⇒ 10x + 8x = 360
⇒ 18x = 360
⇒ x = 20 cm
Step 3
Putting the value of x in equation (1), we get:
b=
160
= 32 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 = 72/2 = 36 cm
T he area of the isosceles triangle ABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 36(36 - 32) (36 - 20) (36 - 20) ]
= 192 cm2
Step 5
T hus, the area of the triangle is 192 cm2.
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ID : pk-9-Herons-Formula [16]
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(14) a. 660cm2
Step 1
Here we have an isosceles triangle with perimeter 242 cm and the unequal side of 120 cm.
T he combined size of the other two sides = 242 cm - 120 cm = 122 cm.
Since the triangle is isosceles, the other two sides are equal, and the length of each of
them will be:
122 ÷ 2
= 61 cm.
Step 2
Now we know that the three sides of the triangle are 61, 61 and 120 cm respectively. We
can now use the Heron's f ormula to calculate the area of the triangle:
Area of a triangle =
,
where a, b, and c are its sides and S =
a+b+c
.
2
Step 3
Let us f irst calculate the value of S:
S=
120 + 61 + 61
=
2
242
= 121
2
Step 4
= 660 cm2
Area of the triangle =
(15) a. 14 cm, 98 cm2
T he key thing to note is that all the internal angles of a rhombus add up to 360°
So if all the internal angles are equal, then one internal angle is 360° ÷ 4 = 90°
T his is a square with all sides equal. So the other diagonal is also 14 cm in length
T he area of a rhombus/square is half the product of the diagonals
Area = (14 x 14) ÷ 2 = 196 ÷ 2 = 98 cm2
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