ID : ae-9-Herons-Formula [1]
Grade 9
Herons Formula
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Answer t he quest ions
(1)
A rhombus has all its internal angles equal. If one of the diagonals is 15 cm, f ind the length of
other diagonal and the area of the rhombus.
(2)
Find the percentage increase in the area of a triangle if each side is increased by Z times.
(3)
T he adjacent sides of a parallelogram are 4 cm and 7 cm. T he ratio of their altitudes is :
(4) T he perimeter of a rhombus is 164 cm and one of its diagonals is 80 cm. What is the length of
other diagonal.
(5)
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.
(6) T he area of an equilateral triangle with a side of 12 cm is :
(7) Find the area of the unshaded region in the f igure below:
(8)
T he area of a trapezium is 140 cm2 and the height is 8 cm. Find the lengths of its two parallel
sides if one side is 5 cm greater than the other.
(9) T he two adjacent sides of a parallelogram are 13 cm and 13 cm. If one of its diagonals is 24 cm,
f ind the area of the triangle.
Choose correct answer(s) f rom given choice
(10) Which triangle has the maximum area f or a given perimeter :
a. Can not be determined
b. Equilateral T riangle
c. Obtuse Angle T riangle
d. Isoceles T riangle
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ID : ae-9-Herons-Formula [2]
(11) T he area of an equilateral triangle with a side of S cm is :
a.
cm2
b.
c.
cm2
d.
(12) In Heron's f ormula,
a.
axbxc
cm2
, S is equal to:
b.
2
c. Half of perimeter of the triangle
cm2
a+b+c
abc
d. a + b + c
(13) Fatima 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 22 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 4 cm to give
the shape of a kite. Find the area of this kite.
a. 242cm2
b. 367.5cm2
c. 245cm2
d. 490cm2
(14) A parallelogram has a diagonal of 15 cm. T he perpendicular distance of this diagonal f rom an
opposite vertex is 8 cm. Find the area of the parallelogram.
a. 60cm2
b. 23cm2
c. 30cm2
d. 120cm2
(15) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three
sides. T he lengths of the perpendiculars are 6 cm, 7 cm and 5 cm. Find the area of the triangle.
a. 108√3 cm2
b.
c. 324 cm2
d. 18 cm2
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cm2
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ID : ae-9-Herons-Formula [3]
Answers
(1)
15 cm, 112.5 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 15 cm in length
T he area of a rhombus/square is half the product of the diagonals
Area = (15 x 15) ÷ 2 = 225 ÷ 2 = 112.5 cm2
(2)
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 Z times, we get a new triangle XYZ
XYZ has sides Z a, Z b and Z c
By Heron's f ormula
Area of new triangle =
Where S1 =
Za + Zb + Zc
=Zx
2
a+b+c
= MS
2
Area of XYZ =
=
= Z 2x A1
T his means the area increases by
(3)
4:7
T he area of a parallelogram is base x height, where base ref ers to the side, and height is
indicated by the altitude
Area = b1 x h1 = b2 x h2
Here, let's take b1 = 4, and b2 = 7
4 x h1 = 7 x h2 T aking the ration we get h2:h1 = 4:7
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ID : ae-9-Herons-Formula [4]
(4) 18 cm
Step 1
One way to solve this is as f ollows
We know that the
a) T he sides of a rhombus are equal. T heref ore one side =
1
x 164 = 41
4
b) A diagonal of a rhombus divides the rhombus into 2 equal triangles
c) T he area of a rhombus is
1
(Diagonal1 x Diagonal2)
4
Step 2
T aking one of the two triangles f ormed by the diagonal with length 80 cm
Area (using Heron's f ormula) =
Where S =
2 x 41 + 80
=
2
162
= 81
2
Area =
= 720 (the details of this computation are
lef t the the student)
Step 3
From c) above, Area = 720 =
1
4
Diagonal2 = 2 x
720
(Diagonal1 x Diagonal2) =
1
(80 x Diagonal2)
4
= 18 cm
80
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ID : ae-9-Herons-Formula [5]
(5)
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 : ae-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
(6)
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, a=b=c=12 and S =
3
x a = 18
2
Step 3
T heref ore Area =
Area =
Area =
cm2
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ID : ae-9-Herons-Formula [7]
(7) 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 : ae-9-Herons-Formula [8]
(8)
Lengths : 15 cm and 20 cm.
Step 1
T he f ollowing picture shows the trapezium ABCD,
Let's assume, CD = x cm
and CD||AB.
According to the question, one side of the trapezium is 5 cm greater than the other.
T heref ore, AB = x + 5,
Area(ABCD) = 140 cm2,
Height of the trapezium ABCD = 8 cm.
Step 2
T he area of the trapezium ABCD = T he height of the trapezium ABCD ×
AB + CD
2
2 × Area(ABCD)
⇒ AB + CD =
T he height of the trapezium ABCD
⇒x+5+x=
2 × 140
8
⇒ 2x = 35 - 5
⇒ 2x = 30
⇒ x = 15 cm.
Now, CD = 15 cm,
AB = 15 + 5 = 20 cm.
Step 3
T hus, the lengths of its two parallel sides are 15 cm and 20 cm respectively.
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ID : ae-9-Herons-Formula [9]
(9) 120 cm2
T ake a look at the representative image below
We can see that ABC is a triangle
Also we know the length of all the sides. We can theref ore use Heron's f ormula
Area =
, where a, b, c are the sides, and S =
a+b+c
=
2
13+13+24
= 25
2
= 120 cm2
Area =
(10) b. Equilateral T riangle
(11) b.
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, a=b=c=S and S =
3
xa=
2
Step 3
T heref ore Area =
Area =
Area =
cm2
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ID : ae-9-Herons-Formula [10]
(12) c. Half of perimeter of the triangle
Step 1
In Heron's f ormula, S represents the semi-perimeter of the triangle.
Step 2
Semi-perimeter is def ined as the half of perimeter of the triangle
(13) c. 245cm2
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 = 22 cm.
T he area of the square ABCD = (
1
) × (22)2 = 242 cm2.
2
Step 4
T hus, the area of the kite = Area(ABCD) + Area(BEF)
= 3 + 242
= 245 cm2.
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ID : ae-9-Herons-Formula [11]
(14) d. 120cm2
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 = 15 cm
Height = length of DP = 8 cm
Area of ACD=
1
x 8 x 15 = 60
2
Step 3
Area of parallelogram ABCD = 2 x area of ACD = 2 x 60 = 120 cm2
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ID : ae-9-Herons-Formula [12]
(15) a. 108√3 cm2
Step 1
Following f igure shows the required triangle,
Let's assume the sides of the equilateral triangle ΔABC be 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 = √[S(S - AB)(S - BC)(S - CA) ]
= √[
3x
(3x/2 - x)(3x/2 - x)(3x/2 - x) ]
2
= √[
3x
(x/2)(x/2)(x/2) ]
2
= √[
3x
(x/2)3 ]
2
= √[ 3(x/2)4 ]
= √3[ (x/2)2 ]
=
√3
(x)2 ------(1)
4
Step 2
T he area of the triangle AOB =
AB × OP
2
=
'x' × 7
2
=
7x
2
Step 3
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ID : ae-9-Herons-Formula [13]
6x
Similarly, the area of the triangle ΔBOC =
2
and the area of the triangle ΔAOC =
5x
.
2
Step 4
T he the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
7x
=
6x
+
2
2
18x
=
+
5x
2
-----(2)
2
Step 5
By comparing equation (1) and (2), we get,
√3
(x)2 =
4
18x
2
⇒x=
36
√3
Step 6
Now, Area(ΔABC) =
√3
(x)2
4
=
√3
4
=
324
(
36
)2
√3
= 108 √3 cm2
√3
Step 7
Hence, the area of the triangle is 108√3 cm2.
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