rd sharma solutions class 9 herons formula ex …...rd sharma solutions class 9 herons formula ex...

RD Sharma Solutions Class 9 Herons Formula Ex 12.1

RD Sharma Solutions Class 9 Chapter 12 Ex 12.1

Q 1. Find the area o f a triangle whose sides are respectively 150 cm, 120 cm and200 cm.

Solution:

Let the sides of the given triangle be a, b, c respectively.

So given,

a = 150 cm

b = 120 cm

c = 200 cm

By using Heron's Formula

The area of the triangle = y /s x (s — a) x (s — b) x (s — c)

Semi perimeter of a triangle = s

2s = a + b + c

(a+5+c)S 2

(150+200+120)S 2

s = 235 cm

Therefore, area of the triangle = y /s x (s — a) x (s — b) x (s — c)

= -^235 x (235 - 150) x (235 - 200) x (235 - 120)= 8966.56cm2

Q2. Find the area o f a diangle whose sides are respectively 9 cm, 12 cm and 15 cm.

Solution:

Let the sides of the given triangle be a, b, c respectively.

So given,

a = 9 cm

b = 12 cm

c = 15 cm


The area of the triangle = -\/s x (s — a) x (s — b) x (s — c)

Semi perimeter of a triangle = s

2s = a + b + c

(o+5+c)S = ^ —

(9+12+15)S = --------2--------

s = 18 cm

Therefore, area of the triangle = y / s x (s — a) x (s — b) x (s — c)

= y/18 x (18 - 9) x (18 - 12) x (18 - 15)= 54cm2

Q3. Find the area o f a diangle two sides o f which are 18 cm and 10 cm and the perim eter is 42cm .

Solution:

Whenever we are given the measurements of all sides of a triangle, we basically look for Heron's formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by:

A = i^/s x (s — a) x (s — b) x (s — c)

Where, s =(o+5+c)

2

We are given:

a = 18 cm

b = 10 cm, and perimeter = 42 cm

We know that perimeter = 2s,

So, 2s = 42

Therefore, s = 21 cm

(a+b+c)We know that, s = — —̂

_ (18+10+c)Z ” 2

42 = 28 + c

c = 14 cm

So the area of the triangle is:

a=V2T ir(2r^n^r(2nn^r(2r^i4)A = ^ /2 1 x (3) x (11) x (7)

A = -v/4851

A = 21 -v /ll cm0

Q4. In a triangle ABC, AB = 15cm, BC = 13cm and AC = 14cm. Find the area o f triangle ABC and hence Its altitude on AC.

Solution:

Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.

So given,

a = 15 cm

b = 13 cm

c = 14 cm


The area of the triangle = ^ s x (s — a) x (s — b) x (s — c)

Semi perimeter of a triangle^ 2s

2s = a + b + c

(a+b+c)s = Hr^(15+13+14)

S= 2

s = 21 cm

Therefore, area of the triangle = y /s x (s — a) x (s — b) x (s — c)

= -^21 x (21 - 15) x (21 - 13) x (21 - 14)

= 84cm2

BE is a perpendicular on AC

Now, area of triangle = 84cm2

\ x BE x AC = 84cm2

BE = 12cm

The length of BE is 12 cm

Q 5. The perimeter o f a triangular field Is 540 m and Its sides are In die ratio 25:17:12. Find the area o f Wangle.

Solution:

Let the sides of the given triangle be a = 25x, b = 17x, c = 12x respectively.

So,

a = 25x cm

b = 17x cm

c = 12x cm

Given Perimeter = 540 cm

2s = a + b + c

a + b + c = 540 cm

25x + 17x + 12x = 540 cm

54x = 540 cm

x = 10 cm

Therefore, the sides of a triangle are

a = 250 cm

b = 170 cm

c = 120 cm

, , „ . . (o+6+c)Now, Semi perimeter s = — -—

_ 540 ” 2

= 270 cm


The area of the triangle = y /s x (s — a) x (s — b) x (s — c)

= y/270 x (270 - 250) x (270 - 170) x (270 - 120)

= 9000cm2

Therefore, the area of the triangle is 9000cm2

Q6. The perimeter o f a triangle Is 300m. I f its sides are In the ratio o f 3 :5 :7 . Find die area o f the triangle.

Solution:

Given the perimeter of a triangle is 300 m and the sides are in a ratio of 3: 5:7

Let the sides a, b, c of a triangle be 3x, 5x, 7x respectively

So, the perimeter = 2s = a + b + c

200 = a + b + c

300 = 3x + 5x + 7x

300 = 15x

Therefore, x = 20 m

So, the respective sides are

a = 60 m

b = 100 m

c = 140 m

Now, semi perimeter s = °+ +̂c

_ 60+100+140 - 2

= 150 m


The area of a triangle = y /s x (s — a) x (s — b) x (s — c)

= y/150 x (150 - 60) x (150 - 100) x (150 - 140)

= 1500V3m2

Thus, the area of a triangle is 1500-\/3m2

Q7. The perimeter o f a triangular field is 240 dm. I f two o f its sides are 78 dm and 50 dm, find the length o f the perpendicular on the side o f length 50 dm from the opposite vertex.

Solution:

Given,

In a triangle ABC, a = 78 dm = AB, b = 50 dm = BC

Now, Perimeter = 240 dm

Then, AB + BC + AC = 240 dm

78 + 50 + AC = 240

AC = 240-(78+50)

AC = 112 dm = c

Now, 2s = a + b + c

2s = 78+ 50+ 112

s = 120 dm

Area of a triangle ABC = y/s x (s — a) x (s — b) x (s — c)

= ^120 x (120 - 78) x (120 - 50) x (120 - 112)= 1680dm2Let AD be a perpendicular on BC

Area of the triangle ABC = ^ x AD x BC

} x AD x B C = 1680dm2

AD = 67.2 dm

Q8. A triangle has sides 35 cm, 54 cm, 61 cm long. Find Its area. Also, find die smallest o f Its altitudes?

Solution:

Given,

The sides of the triangle are

a = 35 cm

b = 54 cm

c = 61 cm

Perimeter 2s = a + b + c

2s = 35 + 54 + 61 cm

Semi perimeter s = 75 cm

By using Heron's Formula,

Area of the triangle = y /s x (s — a) x (s — b) x (s — c)

= y/75 x (75 - 35) x (75 - 54) x (75 - 61)= 939.14cm2

The altitude will be smallest provided the side corresponding to this altitude is longest.

The longest side = 61 cm

Area of the triangle = | x A x 61

} x h x 61 = 939.14cm2

h = 30.79cm

Hence the length of the smallest altitude is 30.79cm

Q9. The lengths o f the sides o f a triangle are In a ratio o f 3 : 4 : 5 and Its perimeter Is 144 on. Find the area o f the triangle and the height corresponding to die longest side?

Solution:

Given the perimeter of a triangle is 160m and the sides are in a ratio of 3 : 4 : 5

Let the sides a, b, c of a triangle be 3x, 4x, 5x respectively


144 = a + b + c

144 = 3x+ 4x+ 5x

Therefore, x = 12cm


a = 36cm

b = 48cm

c = 60cm

Now, semi perimeter s = — -—

_ 36+48+60 " 2

= 72 cm


The area of a triangle = ^ /s x (s — a) x (s — b) x (s — c)

= x (72 - 36) x (72 - 48) x (72 - 60)

= 864cm2

Thus, the area of a triangle is 864cm2



Area of the triangle = x h x 60

j x h x 60 = 864cm2

h = 28.8 cm

Hence the length of the smallest altitude is 28.8 cm

Q10. The perimeter o f an isosceles triangle is 42 cm and its base is | times each o f the equal side. Find the length o f each o f the triangle, area

o f the triangle and the height o f the triangle.

Solution:

Let 'x' be the length of two equal sides.

Therefore the base = x x

Let the sides a, b, c of a triangle be j x i , x and x respectively


42 = a + b + c

42 = | x i t x t x

Therefore, x = 12 cm


a = 12 cm

b = 12 cm

c = 18 cm

Now, semi perimeter s = °+ +̂c

_ 12+12+18 " 2

= 21 cm


The area of a triangle = ■s/s x (s — a) x (s — b) x (s — c)

= s j 2 \ x (21 - 12) x (21 - 12) x (21 - 18)

= 71.42cm2

Thus, the area of a triangle is 71.42cm2



Area of the triangle = x h x 18

} x h x 18 = 71.42cm2

h = 7.93 cm

Hence the length of the smallest altitude is 7.93 cm

Q11. Find die area o f the shaded region in fig. below

Solution:

Area of the shaded region = Area o f A ABC — Area o f AADB Now in triangle ADB

AB2 = AD2 + BD2....... (i)

Given, AD = 12 cm, BD =16 cm

Substituting the value of AD and BD in eq (i), we get

AB2 = 122 + 162= 400cm2

AB = 20 cm

Now, area of a triangle = x AD x BD

= 96cm2Now in triangle ABC,

s = | x (AB + BC + CA)

= \ x (52 + 48 + 20)

= 60 cm


The area of a triangle = x (s — a) x (s — b) x (s — c)

= ^60 x (60 - 20) x (60 - 48) x (60 - 52)= 480cm2Thus, the area of a triangle is 480cm2Area of shaded region = Area of triangle ABC - Area of triangle ADB

= (480 - 96) cm2

= 384 cm2

Area of shaded region = 384 cm2

RD Sharma Solutions Class 9 Herons Formula Ex 12.2

RD Sharma Solutions Class 9 Chapter 12 Ex 12.2

Q1 .Find the area o f the quadrilateral ABCD in which AB = 3 cm, BC = 4cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

For triangle ABC

AC2 = BC2 + AB2 25 = 9 + 16

So, triangle ABC is a right angle triangle right angled at point R

Area of triangle ABC = x AB x BC= \ x 3 x 4

= 6 cm 2

From triangle CAD

Perimeter = 2s = AC + CD + DA

2s = 5 cm+ 4 cm+ 5 cm

2s = 14 cm

s = 7 cm

By using Fleron's Formula

Area of the triangle CAD = -\/s x (s — a) x (s — b) x (s — c)

= y /7 x (7 - 5) x (7 - 4) x (7 - 5)

= 9.16cm2

Area of ABCD = Area of ABC + Area of CAD

= (6+9.16) cm 2

= 15.16 cm 2

Q2. The sides o f a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

D 7 m C

Here the length of the sides of the quadrilateral is given as

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

Diagonal AC is joined.

Now, in triangle ADC

By applying Pythagoras theorem

AC2 = AD2 + CD2 AC2 = 142 + 72

AC = 25 m

Now area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m


The area of a triangle = \ / s x (a — a) x (a — b) x (a — c)

= ^/39 x (39 - 26) x (39 - 27) x (39 - 25)

= 291.84m2

Thus, the area of a triangle is 291.84m2

Now for area of triangle ADC

Perimeter = 2s = AD + CD + AC

= 25m + 24m + 7m

s = 28 m


The area of a triangle = ^ a x (a — a) x (a — b) x (a — c)

= x/28x“(28^r 24y^ 2 8 ^ 7 7 x 1 2 8 ^ 2 5 )

= 84m2

Thus, the area of a triangle is 84m2

Therefore, Area of rectangular field ABCD

= Area of triangle ABC + Area of triangle ADC

= 291.84 + 84

= 375.8 m2

Q3. The sides o f a quadrilateral, taken in order as 5 m, 12m, 14m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area.

Given that the sides of the quadrilateral are

AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m

Join AC

Now area of triangle ABC = | x AB x BC= \ x 5 x 12

= 30 m2

In triangle ABC, By applying Pythagoras theorem

AC2 = AB2 + BC2

AC = \ / 5 2 + 122

AC = 13 m

Now in triangle ADC,

Perimeter = 2s = AD + DC + AC

2s = 15 m +14 m +13 m

s = 21 m


Area of the triangle PSR = ^ /s x (s — a) x (s — b) x (s — c)

= y/21 x (21 - 15) x (21 - 14) x (21 - 13)

= 84m2

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

= (30 + 84) m 2

= 114 m 2Q4. A park in the shape o f a quadrilateral ABCD, has angle C = 9(fi, AB = 9 m,BC= 12 m,CD = 5 m, AD = 8m. How much area does it occupy?

Solution:

Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.

Let us join BD

In triangle BCD, apply Pythagoras theorem

B D 2 = B C 2 + C D 2

B D 2 = 122 + 52

BD = 13 m

Area of triangle BCD = ^ x B C x C D

= \ x 12 x 5

= 30 m2

Now, in triangle ABD

Perimeter = 2s = 9m + 8m + 13m

s = 15 m


Area of the triangle ABD = y /s x (s — a) x (s — b) x (s — c)

= ^/15 x (15 - 9) x (15 - 8) x (15 - 13)

= 35.49m2

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= (35.496 + 30) m 2

= 65.5m2.

Q5. Two parallel sides o f a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area o f the trapezium?

Solution:

Two parallel sides of trapezium are AB = 77 m and CD = 60 m

The other two parallel sides of trapezium are BC = 26 m, AD = 25m

Join AE and CF

DE is perpendicular to AB and also, CF is perpendicular to AB

Therefore, DC = EF = 60 m

Let AE = x

So, BF = 77 - 60 - x

BF = 17 - x

In triangle ADE,

By using Pythagoras theorem,

DE2 = AD2 - AE2 DE2 = 252 - a;2

In triangle BCF,

By using Pythagoras theorem,

CF2 = BC2 - BF2

CF2 = 262 - (17 - a;)2

Here, DE = CF

So, DE2 = CF2252 - a:2 = 262 - (17 - a;)2

252 - a:2 = 262 - (172 - 34a: + a:2)

252 - a:2 = 262 - 172 + 34a: + a:2

252 = 262 - 172 + 34a:

x = 7

DE2 = 252 - a:2

DE = -y/625^-49

DE = 24 m

Area of trapezium = ^ x (60 + 77) x 24

Area of trapezium = 1644 m 2

Q6. Find the area o f a rhombus whose perimeter is 80 m and one o f whose diagonal is 24 m.

Solution:

Given,

Perimeter of a rhombus = 80 m

As we know.

Perimeter of a rhombus = 4 x side = 4 x a

4 x a = 80 m

a = 20 m

There fore0A= | x AC OA = 12 m

In triangle AOB

OB2 = AB2 - OA2 OB2 = 202 - 122

OB = 16 m

Also, OB = OD because diagonal of rhombus bisect each other at 90°

Therefore, BD = 2 OB = 2x16 = 32 m

Area of rhombus = x BD x AC Area of rhombus = -| x 32 x 24

Area of rhombus = 384 m 2

Q7. A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate o f Rs 5 per meter square. Find the cost o f painting.

Solution:

Given that.

Perimeter of a rhombus = 32 m

We know that.

Perimeter of a rhombus = 4 x side

4 x side = 32 m

4 x a = 32 m

a = 8 m

Let AC = 10 m

0A = } x ACOA = | x 10

OA = 5 m

By using Pythagoras theorem

OB2 = AB2- 0 A2 OB2 = 82-52

OB =v /39m

BD = 2 x OB

BD = 2-\/39 m

Area of the sheet = | x BD x AC Area of the sheet = j x 2y/39 x 10

Therefore, cost of printing on both sides at the rate of Rs. 5 per m2

= Rs 2x10^39x5

= Rs. 625

Q8. Find the area o f the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. [Take \ /3 = 1.73]

Solution:/p>

Given that, in a quadrilateral ABCD in which AD = 24 cm.

Angle BAD = 90°

BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm

In triangle BAD, by applying Pythagoras theorem,

BA2 = BD2 - AD2 BA2 = 262 + 242

BA = •x/lOO

BA = 10 cm

Area of the triangle BAD = ^ x BA x AD Area of the triangle BAD = -| x 10 x 24

Area of the triangle BAD = 120 cm 2

Area of the equilateral triangle = — x side/n

Area of the equilateral triangle ORS = - j - x 26

Area of the equilateral triangle BCD = 292.37 cm 2

Therefore, the area of quadrilateral ABCD = area of triangle BAD + area of the triangle BCD

The area of quadrilateral ABCD = 120 + 292.37

= 412.37 cm2

Q9. Find the area o f quadrilateral ABCD in which AB = 42 cm, BC=21 cm, CD =29 cm, DA =34 cm and the diagonal BD =20 cm.

Solution:

Given

AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and the diagonal

BD = 20 cm.

Perimeter of triangle ABD 2s = AB + BD + DA

2s = 34 cm + 42 cm + 20 cm

s = 48 cm/p>


Area of the triangle ABD = \ / s x (s — a) x (s — b) x (s — c)

= y^48 x (48 - 42) x (48 - 20) x (48 - 34)

= 336cm2

Now, for the area of triangle BCD

Perimeter of triangle BCD 2s = BC + CD + BD

2s = 29cm + 21 cm + 20cm

s = 35 cm


Area of the triangle BCD = y /s x (s — a) x (s — b) x (s — c)

= ^/35 x (14) x (6) x (15)

= 210cm2Therefore, Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Area of quadrilateral ABCD = 336 + 210

Area of quadrilateral ABCD = 546 cm 2

Q10. Find the perimeter and the area o f the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB =90°.

Solution:

Given are the sides of the quadrilateral ABCD in which

AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and an angle ACB = 90°

By using Pythagoras theorem

BC2 = AB2 - AC2BC2 = 172 - 152

BC = 8 cm

Now, area of triangle ABC = x AC x BC area of triangle ABC = -| x 8 x 15

area of triangle ABC = 60 cm2

Now, for the area of triangle ACD

Perimeter of triangle ACD 2s = AC + CD + AD

2s = 15+ 12+9

s = 18 cm


Area of the triangle ACD = y 's x (s — a) x (s — b) x (s — c)

= 0 8 x (3) x (6) x (9)

= 54cm2

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of quadrilateral ABCD = 60 cm2 + 54cm2

Area of quadrilateral ABCD = 114 cm 2

Q77. The adjacent sides o f a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal A C measures 42 cm. Find the area o f parallelogram.

Solution:

Given,

The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm.

Area of the parallelogram = Area of triangle ADC + Area of triangle ABC

Note: Diagonal of a parallelogram divides into two congruent triangles

Therefore,

Area of the parallelogram = 2 x (Area of triangle ABC)

Now, for area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 34 cm + 20 cm + 42 cm

s = 48 cm


Area of the triangle ABC = ^ /s x (s — a) x (s — b) x (s — c)

= ^/48 x (14) x (28) x (6)

= 336cm2

Therefore, area of parallelogram ABCD = 2 x (Area of triangle ABC)

Area of parallelogram = 2 x 336cm2

Area of parallelogram ABCD = 672 cm2

Q12. Find the area o f the blades o f the magnetic compass shown in figure given below.

Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB

Now, for the area of triangle ADB

Perimeter = 2s = AD + DB + BA

2s = 5 cm + 1 cm + 5 cm

s = 5.5 cm


Area of the triangle DEF = y 's x (s — a) x (s — b) x (s — c)

= V 5 .5 x (0.5) x (4.5) x (0.5)

= 2.49cm2

Also, area of triangle ADB = Area of triangle CDB

Therefore area of the blades of the magnetic compass = 2 x area of triangle ADB

Area of the blades of the magnetic compass = 2 X2.49

Area of the blades of the magnetic compass = 4.98 cm 2

Q13.A hand fan is made by sticking 10 equal size triangular strips o f two different types o f paper as shown in the figure. The dimensions o f equal strips are 25 cm, 25 cm and 14 cm. Find the area o f each type o f paper needed to make the hand fan.

Given that.

The sides of AOB

AO = 25 cm

OB = 25 cm

BA = 14 cm

Area of each strip = Area of triangle AOB

Now, for the area of triangle AOB

Perimeter = AO + OB + BA

2s = 25 cm +25 cm + 14 cm

s = 32 cm


Area of the triangle AOB = y /s x (s — a) x (s — b) x (s — c)

= v/32V(7yiT(4y^r(i8)

= 168cm2

Also, area of each type of paper needed to make a fan = 5 x Area of triangle AOB

Area of each type of paper needed to make a fan = 5 x 168 cm 2

Area of each type of paper needed to make a fan = 840cm2

Q14. A triangle and a parallelogram have the same base and the same area. I f the sides o f the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height o f a parallelogram.

Solution:

The sides of the triangle DCE are

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the h be the height of parallelogram ABCD

Now, for the area of triangle DCE

Perimeter = DC + CE + ED

2s = 15 cm + 13 cm + 14 cm

s = 21 cm


Area of the triangle AOB = ^ s x (s — a) x (s — b) x (s — c)= V/2TlT(7)V(8)^r{6)= 84cm2

Also, area of triangle DCE = Area of parallelogram ABCD=>84cm2

24 x h = 84cm2

h = 6 cm.

rd sharma solutions class 9 herons formula ex …...rd sharma solutions class 9 herons formula ex...

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