ID : ww-9-Herons-Formula [1]
Grade 9
Herons Formula
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Answer t he quest ions
(1)
Find the area of the unshaded region in the f igure below:
(2)
T he perimeter of a rhombus is 164 cm and one of its diagonals is 80 cm. What is the length of
other diagonal.
(3)
T he area of a triangle with sides 6 cm, 5 cm and 5 cm is :
(4) T he area of a trapezium is 75 cm2 and the height is 5 cm. Find the lengths of its two parallel
sides if one side is 6 cm greater than the other.
(5)
T he perimeter of a quadrilateral is 56 cm. If the f irst three sides of a quadrilateral, taken in order
are 17 cm, 16 cm and 15 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) Find the percentage increase in the area of a triangle if each side is increased by X times.
(7) T he perimeter of an isosceles triangle is 64 cm. T he ratio of the equal side to its base is 5 : 6.
Find the area of the triangle.
(8)
T he sides of a quadrilateral, taken in order are 17 cm, 16 cm, 15 cm and 8 cm respectively. T he
angle contained by the last two sides is a right angle. Find the area of the quadrilateral.
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ID : ww-9-Herons-Formula [2]
Choose correct answer(s) f rom given choice
(9) From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three
sides. T he lengths of the perpendiculars are 8 cm, 10 cm and 6 cm. Find the altitude of the
triangle.
a. 24 cm
b. 28 cm
c. 24√3 cm
d. 22 cm
(10) Which triangle has the maximum area f or a given perimeter :
a. Equilateral T riangle
b. Obtuse Angle T riangle
c. Can not be determined
d. Isoceles T riangle
(11) Find the area of a quadrilateral ABCD where AB = 10 cm, BC = 17 cm, CD = 13 cm, DB = 20 cm
and AC = 21 cm.
a. 126cm2
b. 210cm2
c. 84cm2
d. 2100cm2
(12) 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. 15 cm, 105 cm2
b. 14 cm, 98 cm2
c. 98 cm, 14 cm2
d. 14 cm, 196 cm2
(13) Sharon 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. 490cm2
b. 245cm2
c. 242cm2
d. 367.5cm2
(14) A parallelogram has a diagonal of 13 cm. T he perpendicular distance of this diagonal f rom an
opposite vertex is 7 cm. Find the area of the parallelogram.
a. 91cm2
b. 20cm2
c. 22.75cm2
d. 45.5cm2
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ID : ww-9-Herons-Formula [3]
(15) T he area of the parallelogram ABCD and the length of the altitude DE respectively in the f igure
below are
a. 86.4cm2, 31.2cm
b. 72cm2, 24cm
c. 54cm2, 27cm
d. 36cm2, 24cm
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Answers
(1)
90 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
= (28 + 17 + 39)/2
= 42 m.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 42(42 - 28) (42 - 17) (42 - 39) ]
= 210 m2
Step 3
Similarly, the area of the triangle ΔACD can be calculated using Heron's f ormula.
S = (AC + CD + DA)/2
= (39 + 16 + 25)/2
= 40 m.
T he area of the ΔACD = √[ S (S - AC) (S - CD) (S - DA) ]
= √[ 40(40 - 39) (40 - 16) (40 - 25) ]
= 120 m2
Step 4
T hus, the area of the unshaded region = Area(ABC) - Area(ACD)
= 210 - 120
= 90 m2
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(2)
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
(Diagonal1 x Diagonal2) =
4
Diagonal2 = 2 x
720
1
(80 x Diagonal2)
4
= 18 cm
80
(3)
12 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 =
6+5+5
=
2
16
=8
2
Step 3
Area =
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=
= 12 cm2
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ID : ww-9-Herons-Formula [6]
(4) Lengths : 12 cm and 18 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 6 cm greater than the other.
T heref ore, AB = x + 6,
Area(ABCD) = 75 cm2,
Height of the trapezium ABCD = 5 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+6+x=
2 × 75
5
⇒ 2x = 30 - 6
⇒ 2x = 24
⇒ x = 12 cm.
Now, CD = 12 cm,
AB = 12 + 6 = 18 cm.
Step 3
T hus, the lengths of its two parallel sides are 12 cm and 18 cm respectively.
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(5)
Area : 180 cm2
Step 1
Following picture shows the quadrilateral ABCD,
T he perimeter of the quadrilateral ABCD = 56 cm
T heref ore, AB + BC + CD + DA = 56
⇒ 17 + 16 + 15 + DA = 56
⇒ 48 + DA = 56
⇒ DA = 56 - 48
⇒ DA = 8 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 ]
= √[ 82 + 152 ]
= 17 cm
Step 3
T he area of the right angled triangle ΔACD =
DA × DC
2
=
8 × 15
2
= 60 cm2
Step 4
Now, we can see that, this quadrilateral consists of the triangles ΔACD and ΔABC.
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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
= (17 + 16 + 17)/2
= 25 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 25(25 - 17) (25 - 16) (25 - 17) ]
= 120 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = 60 + 120 = 180 cm2
(6)
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 X times, we get a new triangle XYZ
XYZ has sides Xa, Xb and Xc
By Heron's f ormula
Area of new triangle =
Where S1 =
Xa + Xb + Xc
= Xx
2
Area of XYZ =
a+b+c
= MS
2
=
= X2x A1
T his means the area increases by
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(7) 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 : 6.
5
T heref ore,
=
6
x
b
By cross multiplying, we get:
b=
6x
------(1)
5
Step 2
According to the question, the perimeter of the isosceles triangle ABC = 64 cm
T heref ore, x + x + b = 64
6x
⇒ 2x +
= 64 [From equation (1), b =
6x
5
⇒
]
5
10x + 6x
= 64
5
⇒ 10x + 6x = 320
⇒ 16x = 320
⇒ x = 20 cm
Step 3
Putting the value of x in equation (1), we get:
b=
120
= 24 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 = 64/2 = 32 cm
T he area of the isosceles triangle ABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 32(32 - 24) (32 - 20) (32 - 20) ]
= 192 cm2
Step 5
T hus, the area of the triangle is 192 cm2.
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(8)
Area : 180 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 ]
= √[ 82 + 152 ]
= 17 cm
Step 3
T he area of the right angled triangle ΔACD =
AD × DC
2
=
8 × 15
2
= 60 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
= (17 + 16 + 17)/2
= 25 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
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= √[ 25(25 - 17) (25 - 16) (25 - 17) ]
= 120 cm2
Step 5
T he area of the quadrilateral ABCD = Area(ΔACD) + Area(ΔABC) = 60 + 120 = 180 cm2
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(9) a. 24 cm
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' × 10
2
=
10x
2
Step 3
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8x
Similarly, the area of the triangle ΔBOC =
2
and the area of the triangle ΔAOC =
6x
.
2
Step 4
T he the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
10x
=
+
2
24x
=
8x
6x
+
2
2
-----(2)
2
Step 5
By comparing equation (1) and (2), we get,
√3
24x
(x)2 =
4
2
⇒x=
48
------(3)
√3
Step 6
Let H be altitude of triangle ΔABC.
Area(ΔABC) =
Base × Height
2
⇒
√3
(x)2 =
4
⇒H=
xH
2
√3
x
2
⇒H=
√3
2
(
48
) = 24 cm
√3
Step 7
Hence, the altitude of the triangle is 24 cm.
(10) a. Equilateral T riangle
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(11) b. 210cm2
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
= (10 + 17 + 21)/2
= 24 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - AC) ]
= √[ 24(24 - 10) (24 - 17) (24 - 21) ]
= 84 cm2
Step 5
Similarly, the area of the ΔACD can be calculated using Heron's f ormula.
S = (AC + CD + AD)/2
= (21 + 13 + 20)/2
= 27 cm.
T he area of the ΔACD = √[ S (S - AC) (S - CD) (S - AD) ]
= √[ 27(27 - 10) (27 - 17) (27 - 21) ]
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= 126 cm2
Step 6
T he area of the quadrilateral ABCD = T he area of the ΔABC + T he area of the ΔACD
= 84 + 126
= 210 cm2
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(13) b. 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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(14) a. 91cm2
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 = 13 cm
Height = length of DP = 7 cm
Area of ACD=
1
x 7 x 13 = 45.5
2
Step 3
Area of parallelogram ABCD = 2 x area of ACD = 2 x 45.5 = 91 cm2
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(15) b. 72cm2, 24cm
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
= (3 + 25 + 26)/2
= 27 cm.
T he area of the ΔABC = √[ S (S - AB) (S - BC) (S - CA) ]
= √[ 27(27 - 3) (27 - 25) (27 - 26) ]
= 36 cm2
Step 3
T he area of the parallelogram ABCD = 2 × Area(ΔABC)
= 2 × 36
= 72 cm2
Step 4
T he length of the altitude DE =
2 × Area(ΔABC)
AB
=
2 × 36
3
= 24 cm.
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