
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
Exercise 12.1
Q.1 A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Sol.
To find the area of an equilateral triangle using Heron’s formula.
If a, b, c be the lengths of sides BC, CA and AB of Δ ABC and if s=12(a+b+c), then
area of Δ ABC = s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
Since Δ ABC is given as equilateral of side = a
Therefore here a = b = c (= a) and s=12(a+a+a)=3a2
Now , s−a=3a2−a=a2
s−b=3a2−a=a2
and s−c=3a2−a=a2
Therefore Area=3a2×a2×a2×a2−−−−−−−−−−−−−√
=a2×a23–√=3√a24 … (1)
To find the area when its perimeter = 180 cm
Here a + a + a = 180 cm
⇒ 3a = 180 cm
⇒ a = 60 cm
Therefore Required area =3√4×(60)2 cm^{2
}=3√4×3600 cm^{2}
[On putting a = 60 in (1)]
=9003–√cm2
Q.2 The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of Rs 5000 per m2peryear. A company hired one of its walls for 3 months. How much rent did it pay?
First we find the semi perimeter of triangular side measuring 122 m, 22m and 120 m.
Let a = 122 m, b = 22 m and c = 120 m
Therefore s=12(a+b+c)=12(122+22+120)m
=(12×264)m=132m
Now, s – a = (132 – 122) m = 10 m
s – b = (132 – 22) m = 110 m
and s – c = (132 – 120) m = 12 m
Therefore Area of triangular wall =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√ [By Heron’s formula]
=132×10×110×12−−−−−−−−−−−−−−−−√m2
=11×12×10×10×11×12−−−−−−−−−−−−−−−−−−−−−−√m2
=10×10×11×11×12×12−−−−−−−−−−−−−−−−−−−−−−√m2
=(10×11×12)m2=1320m2
Rent charges = Rs 5000 per m2 per year
Therefore Rent charged from a company for 3 month
=Rs(5000×1320×312)
=Rs16,50,000
Q.3 There is a slide in a park. One of its side walls has been painted in blue colour with a message “KEEP THE PARK GREEN AND CLEAN” (See figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
First find the semi perimeter of side wall measuring 15 m, 11 m and 6 m.
Let a = 15 m, b = 11 m, c = 6 m
and s=12(a+b+c)=12(15+11+6)m
=(12×32)m=16m
Now, s – a = (16 – 15)m = 1 m
s – b = (16 – 11) m = 5 m
and s – c = (16 – 6) m = 10 m
Area of side wall =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=16×1×5×10−−−−−−−−−−−−−√m2
=4×4×5×5×2−−−−−−−−−−−−−−√m2
=4×52–√m2
=202–√m2
Therefore Area painted in blue colour = Area of side wall
=202–√m2
Q.4 Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Sol.
Let a, b and c be the sides of a triangle such that
a = 18 cm, b = 10 cm and a + b + c = 42 cm
Therefore c = 42 – a – b
⇒ c = (42 – 18 – 10) cm = 14 cm
Now, s=12(a+b+c)=12×42cm=21cm
s−a=(21−18)cm=3cm
s−b=(21−10)cm=11cm
and s−c=(21−14)cm=7cm
Therefore Area of the triangle =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=21×3×11×7−−−−−−−−−−−−−√cm2
=3×7×3×11×7−−−−−−−−−−−−−−−√cm2
=3×3×7×7×11−−−−−−−−−−−−−−−√cm2
=3×711−−√cm2=2111−−√cm2
Q.5 Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Sol.
Let the sides be a, b and c of the triangle.
Therefore a : b : c = 12 : 17 : 25
⇒ a12=b17=c25=k(say)
⇒ a = 12 k, b = 17 k and c = 25 k
Also, perimeter = 540 cm
⇒ a + b + c = 540
⇒ 12k + 17k + 25 k = 540
⇒ 54 k = 540
k=54054 i.e., k = 10
Thus a = 12 × 10 = 120 cm, b = 17 × 10 cm = 170 and c = 25 × 10 = 250 cm
Now, s=12×540cm=270cm
s – a = (270 – 120) cm = 150 cm
s – b = (270 – 170) cm = 100 cm
and s – c = (270 – 250) cm = 20 cm
Area of the triangle
=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=270×150×100×20−−−−−−−−−−−−−−−−−√cm2
=10027×15×10×2−−−−−−−−−−−−−√cm2
=1003×3×3×3×5×5×2×2−−−−−−−−−−−−−−−−−−−−−−−√cm2
=100×3×3×5×2cm2=9000cm2
Q.6 An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Sol.
Here a = b = 12 cm and a + b + c = perimeter = 30 cm
⇒ c = (30 – a – b) cm
⇒ c = (30 –12 – 12) cm
=(30 – 24) cm
= 6 cm
Now , s=12×30cm=15cm
s−a=(15−12)cm=3cm
s−b=(15−12)cm=3cm
and s−c=(15−6)cm=9cm
Therefore, Area of the triangle =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=15×3×3×9−−−−−−−−−−−−√cm2
=5×3×3×3×3×3−−−−−−−−−−−−−−−−−√cm2
=3×35×3−−−−√cm2=915−−√cm2
Hence, the area of triangle is = 915−−√ cm^{2}.
Exercise 12.2
Q.1 A park, in the shape of a quadrilateral ABCD, has ∠C= 90º, AB = 9 m , BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Sol.
Area of ΔBCD=12×BC×CD
=(12×12×5)m2
=30m2
Using Pythagoras theorem, we have
BD2=BC2+CD2 ⇒ BD2=122+52
⇒ BD2=144+25 ⇒ BD2=169
⇒ BD=169−−−√=13m
For :ΔABD:
Let a = 13 m, b = 8 m and c = 9 m
Now, s=12(a+b+c)=12(13+8+9)m
=12×30m=15m
s−a=(15−13)m=2m
s−b=(15−8)m=7m
and s−c=(15−9)m=6m
Therefore AreaofΔABD=15×2×7×6−−−−−−−−−−−−√m2
=3×5×2×7×2×3−−−−−−−−−−−−−−−−−√m2
=2×2×3×3×5×7−−−−−−−−−−−−−−−−−√m2
=2×335−−√m2
=6×5.9m2(approx)
=35.4m2(approx)
Therefore Required area = ΔABD+ΔBCD
=35.4m2+30m2=65.4m2
Q.2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Sol.
Since AC2=AB2+BC2
(as52=32+42i.e.,25=9+16)
Therefore ∠ABC= 90º
Area of rt ∠dΔABC=12×AB×BC
=(12×3×4)cm2=6cm2
For ΔACD :
Let a = 5 cm, b = 4cm and c = 5 cm. Then ,
s=12(a+b+c)=12(5+4+5)cm
=12×14cm=7cm
Now, s−a=(7−5)cm=2cm
s−b=(7−4)cm=3cm
and s−c=(7−5)cm=2cm
Area of ΔACD=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=7×2×3×2−−−−−−−−−−−√cm2
=221−−√cm2
=2×4.6cm2(approx.)
=9.2cm2(approx.)
Area of quadrilateral ABCD
AreaofΔABC+AreaofΔACD
=(6+9.2)cm2=15.2cm2approx
Q.3 Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
Region I is Enclosed by a triangle of sides a = 5 cm, b = 5 cm and c = 1 cm
Let s be the perimeter of the triangle. Then,
Now, s=12(a+b+c)
=12(5+5+1)m
⇒s=112cm
AreaofregionI=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√cm2
⇒AreaofregionI=112×(112−5)×(112−5)×(112−1)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2
⇒AreaofregionI=112×12×12×92−−−−−−−−−−−−−√cm2=311√4cm2
⇒AreaofregionI=34×3.32=2.49cm2
Area of region II:
Region II is a rectangle of length 6.5 cm and breadth 1 cm.
AreaofregionII=6.5×1=6.5cm2
Area of region III:
Region III is an isosceles trapezium.
In ΔABE, We have
AB^{2} = AE^{2} + BE^{2}
⇒ 1 = 0.25 + BE^{2}
⇒ BE = 0.75−−−−√�=34−−√
AreaofregionIII=12(AD+BC)×BE=12(2+1)×34−−√cm2
⇒AreaofregionIII=33√4cm2=1.3cm2
Area of region IV:
Region IV forms a right triangle whose two sides are of lengths 6 cm and 1.5 cm.
AreaofregionIV=12×6×1.5cm2=4.5cm2
Area of region V:
Region IV and V are congruent.
Area of region V = 4.5 cm^{2}
Hence, total area of the paper used = (2.49 + 6.5 + 1.3 + 4.5 + 4.5) cm^{2}
= 19.29 cm^{2}
Q.4 A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Sol.
For the triangle :
Let its sides be a, b and c such that
a = 26 cm, b = 28 cm and c = 30. Then,
s=12(26+28+30)cm
=12×84cm=42cm
Now, s−a=(42−26)cm=16cm
s−b=(42−28)cm=14cm
and s−c=(42−30)cm=12cm
Area of the triangle
=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=42×16×14×12−−−−−−−−−−−−−−√cm2
=2×3×7×4×4×2×7×2×2×3−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2
=2×2×2×2×3×3×4×4×7×7−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cm2
=(2×2×3×4×7)cm2=336cm2
For the parallelogram :
Area = base × height
Therefore Height=AreaBase
[Since Area of parallelogram = Area of Δ (given)
Therefore Area of parallelogram = 336 cm^{2} and its base = 28 cm]
=(33628)cm
=12cm
Q.5 A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Sol.
We know that the diagonals of the rhombus bisect each other at right angles. Using Pythagoras theorem, we have OD=AD2−AO2−−−−−−−−−−√=302−242−−−−−−−−√m
=(30+24)(30−24)−−−−−−−−−−−−−−−√m
=54×6−−−−−√m
=9×6×6−−−−−−−−√m
=(3×6)m=18m
Area of one ΔAOD=(12×24×18)m2=216m2
Therefore Area of rhombus =4×ΔAOD
=(4×216)m2=864m2
Grass area for 18 cows =864m2
Grass area for 1 cow =(86418)m2=48m2
Q.6 An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm
Now, s=12(a+b+c)=12(20+50+50)cm
=12×120cm=60cm
s−a=(60−20)cm=40cm
s−b=(60−50)cm=10cm
and s−c=(60−50)cm=10cm
Therefore Area of one triangular piece
=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=60×40×10×10−−−−−−−−−−−−−−√cm2=2006–√cm2
Therefore Area of 5 red triangles =10006–√cm2
and , area of 5 green triangles =10006–√cm2
Q.7 A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it ?
Sol.
ABCD is a square such that AC = BC = 32 cm and AEF is an isosceles triangle in which AE = AF = 6 cm
For the area of shades I and II
Clearly from the figure
Area of shade I = Area of shade II = Area of Δ CDB
=12×DB×CM [Since ABCD is in square shapes]
=12×32×16 sq cm
=256sqcm
For the area of shade III (i.e., Δ AEF)
EL=LF=12EF
=12×8=4cm
and AE=6cm(given)
Therefore AL=AE2−EL2−−−−−−−−−−√
=36−16−−−−−−√=20−−√=25–√cm
Area of shade III =12×EF×AL
=12×8×25–√sqcm
=85–√sqcm
=8×2.24sqcm(approx)
=17.92sqcm(approx)
Q.8 A floral design on floor is made up of 16 tiles which are triangular , the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per cm2.
For one triangular tile :
Let a = 9 cm, b = 28 cm and c = 35 cm
Now , s=12(a+b+c)=12(9+28+35)cm
=12×72cm=36cm
s−a=(36−9)cm=27cm
s−b=(36−28)cm=8cm
and s−c=(36−35)cm=1cm
Area of one tile =s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=36×27×8×1−−−−−−−−−−−−−√cm2
=9×4×9×3×4×2−−−−−−−−−−−−−−−−−√cm2
=2×3×4×4×9×9−−−−−−−−−−−−−−−−−√cm2=4×96–√cm2
=366–√cm2=36×2.45cm2approx.
=88.2cm2approx.
Therefore Area of 16 such tiles =(16×88.2)cm2
=1411.2cm2approx.
Cost of polishing @ 50 P per cm2=Rs(1411.2×50100)
=Rs705.60approx.
Q.9 A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Sol. Various length are as marked in the figure.
Clearly DM = CN
⇒ DM2=CN2
[Distance between parallel sides are always equal]
⇒ (13)2−x2=142−(15−x)2
⇒ 169−x2=196−(225−30x+x2)
⇒ 50x=169−196+625=598
⇒ x=59850=11.96
Now DM=AD2−AM2−−−−−−−−−−√=132−(11.96)2−−−−−−−−−−−√
=(13+11.96)(13−11.96)−−−−−−−−−−−−−−−−−−−−√
=24.96×1.04−−−−−−−−−−√=25.9584−−−−−−√=5.09
Area of the trapezium =12×(AB+CD)×DM
=12×(25+10)×5.09sq.m
=12×35×5.09sq.m
=89.075sq.m