Important Questions Class 9 Maths Chapter 12

Important Questions Class 9 Mathematics Chapter 12 – Heron’s formula

CBSE Class 9 Mathematics syllabus covers Heron’s formula, a foundational idea with application in a wide range of fields. The chapter assists us in finding the area of a triangle with three different side lengths. In addition to the formula, Heron made other contributions, the most noteworthy of which was creating the Aeolipile, the world’s first steam engine. The Heron was unable to use it for anything useful; rather, the ancient Greeks thus treated it as a toy and a source of interest. 

Extramarks is the preferred learning platform for a lot of students, from Class 1 to Class 12. We provide an online learning module and live classes to help students with their learning. We recommend students regularly practise solving a lot of questions so that they are able to self-assess their preparation and understanding of the subject. Using our question bank of Important Questions Class 9 Mathematics Chapter 12 students will be able to get access to questions from different sources in one place. The solutions to all important questions are prepared in accordance with the most recent CBSE term-wise syllabus for 2021–2022 by our experienced Mathematics teachers. The goal of the Important Questions Class 9 Mathematics Chapter 12 is to provide students with thorough, step-by-step explanations for each response to the problems provided in this chapter’s activities.

The study materials provided by Extramarks can be used by students to get a clearer understanding of Heron’s Formula. Along with the important questions and solutions, students will find many other study resources on the Extramarks platform

Important Questions Class 9 Mathematics Chapter 12 – With Solutions

The questions in our study aid for Class 9 Mathematics Chapter 12 Important Questions have been compiled by our team from various sources, including NCERT textbooks and exemplars, past year’s question papers, and other Mathematics related resources. 

Below are a few questions and answers from our question bank of Chapter 12 Class 9 Mathematics Important Questions. Students can register on the Extramarks website to get full access to our question bank along with all solutions. Regularly solving questions will definitely help students improve their understanding of the topic and score good marks on exams.

Question 1: An isosceles right triangle having an area of 8 cm2. The length of its hypotenuse is:

(A) √32 cm

(B) √24 cm

(C) √48 cm

(D) √16 cm.

Solution 1: (A) √32 cm

Explanation:

Let the height of the triangle be h

As the triangle is isosceles,

Let base = height = h

According to the question,

Area of the triangle = 8 cm2

⇒ ½ × Base × Height = 8

⇒ ½ × h × h = 8

⇒ h2 = 16

⇒ h = 4cm

Base = Height = 4cm

the triangle being right-angled,

Hypotenuse2 = Base2 + Height2

⇒ Hypotenuse2 = 42 + 42

⇒ Hypotenuse2 = 32

⇒ Hypotenuse = √32

Question 2: The perimeter of any equilateral triangle is 60 m. The area is:

(A) 20√3 m2

(B) 15√3 m2

(C)  10√3 m2

(D) 100√3 m2

Solution 2: (D) 100√3 m2

Explanation:

Area of an equilateral triangle of side a = √3/4 a2

As per the question,

The perimeter of the triangle = 60m

⇒ a + a + a = 60

⇒ 3a = 60

⇒ a = 20m

Area of triangle = (√3/4) a2

= (√3/4) (20)2

= (√3/4) (400)

= 100√3

Question 3: The sides of the triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is:

(A) 1322 cm2

(B) 1311 cm2

(C) 1344 cm2

(D) 1392 cm2

Solution 3: (C) 1344 cm2

Explanation:

As per the question,

Sides of the triangle,

a = 56, b = 60, c = 52

s = (a + b + c)/2

⇒ s = (56 + 60 + 52)/2

= 168/2 = 84.

Area of the triangle = √s(s-a)(s-b)(s-c)

= √84(84-56)(84-60)(84-52)

= √84×28×24×32

= 1344cm2

Question 4: The area of the equilateral triangle with sides 2√3 cm is:

(A) 5.196 cm2

(B) 1.732 cm2

(C) 3.496 cm2

(D)  0.866 cm2

Solution 4: (A) 5.196 cm2

Explanation:

Area of the equilateral triangle of side a = √3/4 a2

As per the question,

a = 2√3

Area of triangle = (√3/4) a2

= (√3/4) (2√3)2

= (√3/4)(12)

= 3√3

= 5.196

Question 5: An isosceles right triangle has an estimated area 8cm2. The length of its hypotenuse is:

(A). 16 cm

(B).48 cm

(C).32 cm

(D).24 cm

Solution 5:(C)32 cm

Explanation:

Height of triangle=h

As the triangle is isosceles,

Let base=height=h

Area of triangle=8cm2

⇒ 12 X Base X Height =8

⇒ 12 X h X h =8

⇒12×h×h=8

⇒h2 =  16

⇒h=4cm

Base =Height =4cm

Since the triangle is right-angled,

Hypotenuse2 = Base2 + Height2

⇒Hypotenuse2 = 42  + 42

⇒Hypotenuse2 =32

⇒Hypotenuse= 32

Question 6: The sides of a given triangle are 35cm,54cm and 61cm, respectively. The length of the longest altitude is:

1. 265cm
2. 28cm

C.105cm

1. 245 cm

Solution 6: (D) 245 cm

Explanation:

Semi-perimeter of a triangle,

s= a+b+c2

S =35+54+612

=75cm

Area, A

A= s(s−a)(s−b)(s−c)

= 75(75−35)(75−54)(75−61)

=4205  cm2

The area of the triangle is given as

A= 12 ×a×h

A=12×a×h

Where h is the longest altitude. 

Therefore,

12×a×h=4205

⇒h= 420×2×5a

⇒h= 420×2×535

Hence, the length of the altitude, h=245cm

Question 7: The sides of a given triangle are 56cm, 60cm and 52cm long. The area of the triangle is:

A.4311 cm2

B.4322cm2

1. 2392 cm2
2. None of these

Solution 7: (D) None of this 

Explanation:

The three sides of a given triangle are a=56cm, b=60cm and c=52cm. Then, semi-perimeter of a triangle,

s= a+b+c2

= 56+60+522

= 1682

=84cm

Area of a triangle 

= s(s−a)(s−b)(s−c)

= 84(84−56)(84−60)(84−52)

= 4×7×3×4×7×4×2×3×4×4×2

= (4)6 × (7)2 × (3)2

= (4) x 7 x 3

= 1344 cm2

The area of the triangle is 1344 cm2

Question 8: The area of an equilateral triangle is 163 m2. Its perimeter is:

1. 24m

B.12m

C.306m

D.48m

Solution 8: (A) 24m

Explanation:

 Let the length of the side of the equilateral triangle be am 

Now, area of equilateral △= 34(side)2

⇒163 = 34(a)2

⇒ (a)2 = 163 X 43 = 64

⇒a= 64

⇒a = 8 m

Substitute the value of a,

Perimeter of equilateral △= 3a= 3 × 8

=24m

Question 9: The perimeter of a given triangle is 30cm. The sides are in the ratio 1: 3: 2, then its smallest side is:

1. 15cm

B.5cm

C.1cm

D.10cm

Solution 9: (B) 5cm

Explanation:

The perimeter of the triangle =30cm

The ratio of the sides is =1:3:2

sides are x,3x,2x

⇒x+3x+2x=30cm

⇒6x=30

⇒x=5cm

Therefore the smallest side is 5cm.

Question 10: Find the area of a given triangular garden whose sides are 40m.,90m and 70m. (use 5 =2.24)

Solution 10: Let a=40m,b=90m and c=70m

The half perimeter,

s= (a+b+c)2

⇒ (40+90+70)2

⇒ 2002

s=100

By Heron’s formula of area of triangle = s(s−a)(s−b)(s−c)

⇒ 100(100−40)(100−90)(100−70)

⇒ 100×60×10×30

⇒10 18000

⇒10×60− 5

=10×134.4

=1344 m2

The area of the triangular garden= 1344 m2

Question 11: Find the total cost of levelling the ground in the form of a triangle with sides of 16m, 12m and 20m at Rs. 4 per sq. meter.

Solution 11: Let the sides be a=16,b=12,c=20.

By using the Herons formula 

s= a+b+c2

= 16+12+202

= 482

=24 

The area of the triangle,

⇒A= s(s−a)(s−b)(s−c)

⇒A= (24−16)(24−12)(24−20)

⇒A= 24×8×12×4

⇒A= (2×2×3×2)(2×2×2)(2×3×2)(2×2)

⇒A=2×2×2×2×2×3

⇒A=96m sq

Cost per metre =4 

Cost for 96m=4×96

=384 Rs.

Question 12: Find the area of a given triangle, two sides of which are 8cm and 11cm, and the perimeter is 32cm.

Solution 12: Let a,b, and c be the sides of the triangle and 2s be its perimeter so that

a=8cm,b=11cm and 2s=32cm

Now,

a+b+c=2s

8+11+c=32

c=13

Therefore,

s−a=16−8=8,

s−b=16−11=5,

s−c=16−13=3

Hence, the area of the given triangle

= s(s−a)(s−b)(s−c)

= 16×8×5×3

= 830 cm2

Question 13: The area of a given isosceles triangle is 12cm2. If one of the equal sides is 5cm. Find its base.

Solution 13: Let equal sides be (a)=5 cm and base (b) =?

Area of an isosceles triangle =12 sq. cm

Area of an isosceles triangle 

= b44a2 – b2

12= b44×(5)2 – b2

48=b 100- b2

Squaring both sides, we get

2304 = b2( 100 – b2)

b4 = 100 b2 + 2304 = 0

b2 – 64b2 – 36b2 + 2304 = 0

b2 (b2 -64) – 36 ( b2 – 64) = 0

(b2 -64)(b2 -36)=0

Either  ( b2 – 64) = 0

⇒  b2 = 64 ⇒  b = 8

Or, (b2 -36)=0

⇒  b2 = 36 ⇒  b = 6

Hence base =8 cm or 6 cm.

Question 14: Find the area of a triangle if ∠B= 900

Solution 14: Since ∠B= 900, ABC is a right-angle triangle.

Pythagoras Theorem,

⇒AB2 +BC2 =AC2

⇒AB2+ 42=52

⇒AB2+42=52

⇒AB2=25−16

⇒AB2=25−16

⇒AB2=9

⇒AB=3cm

⇒AB=3cm

Area(△ABC) =  12 ×AB×BC

= 12×3×4

=6cm2

Question 15: The diagonals of a given rhombus are 24 cm and 10 cm. Find the area and perimeter.

Solution 15: To find the area, 

Area =12×24×10

=120cm2

Perimeter s2= (242) 2  + (102)2

= 122 + 52

=169

s=13

The perimeter of the rhombus = 4×13= 52cm

Question 16: Two sides of the parallelogram are 10 cm and 7cm. One of its diagonals is 13 cm. Find the area.

Solution 16:

ABCD is parallelogram

AB=CD=10cm

AD=CB=7cm

Diagonal BD=13cm

The diagonal divides the parallelogram into two equal triangles

The area of triangle ABD

Area = s(s−a)(s−b)(s−c)

s= a+b+c2

a=10

b=7

c=13

Substitute the values in the formula:

s= 10+7+132

Area = 15(15−10)(15−7)(15−13)

Area = 34.6410161514

Area of parallelogram =2× Area of triangle =2×34.6410161514=69.2820 cm2

Hence the area of the parallelogram is 69.2820 sq. cm.

Question 17: A rhombus-shaped sheet with a perimeter of 40 cm and one diagonal of 12 cm is painted on both sides at the rate of 5 per m2. Find the cost of painting.

Solution 17: Let ABCD be a rhombus, then AB=BC=CD=DA=x

The perimeter of the rhombus =40cm

⇒4x=40cm⇒x=10cm

∴AB=BC=CD=DA=10cm

In △ABC, S= a+b+c2 = 10+10+122 =16cm

Ar △ABC=  16(16−10)(16−10)(16−12) = 16×6×6×4 = 48cm2

ar.ABCD=2×48=96 cm2

Cost of painting the sheet =Rs(5×96×2)= Rs 960 on both sides

Question 18:The sides of a given quadrilateral ABCD are 6cm,8cm,12cm, and 14cm (taken in order), respectively, and the angle between the first two sides is a right angle. Find its area.

Solution 18: Applying Pythagoras theorem in △ABC, we get

AC= AB2 + BC2  = 62 + 82  = 36 + 64 = 100 = 10 cm

So, the area of △ABC= 12× base × height

=12×AB×BC

=12×6×8=24

Now, in △ACD, we have,

AC=10cm,   CD=12cm, AD=14cm

As per Heron’s formula, the area of the triangle

(A)= [s(s−a)(s−b)(s−c)]

Where,  2s=(a+b+c)

Here, a=10cm,b=12cm,c=14cm

s= (10+12+14)2

= 362 =18

Area of △ACD= [18×(18−10)(18−12)(18−14)]

= (18×8×6×4)

= (2×3×3×2×2×2×2×3×2×2)

= [(2×2×2×2×2×2×3×3)×2×3]

= 2x2x2x3x6

=246

So, total area of quadrilateral ABCD=△ABC+△ACD

=24+246

=24( 6 + 1)

Question 19: The perimeter of a given isosceles triangle is 32cm. The ratio of the equal side to its base is 32. Find the area of the triangle.

Solution 19: The value of the ratio of the equal side to the base is 32.

We suppose the sides be 3x,2x. Let the third =3x

Given, perimeter=32

We know the perimeter is equal to the sum of the sides. Thus, ⇒3x+2x+3x=32

⇒8x=32

⇒x=4

⇒ 322=16

Thus, the sides are 12cm,8cm,12cm

Thus, the Area of the triangle = 322(16 – 12)(16 -8)(16 -12)

= 16×4×8×4

=32 2 cm2.

Question 20: The sides of a given triangular field are 41m,40m and 9m. Find the total number of flower beds that can be prepared in the field if each flower bed needs 900cm2 space.

Solution 20: By Heron’s formula. Area of a triangular= s×(s−a)(s−b)(s−c), where a,b, and c are sides of the triangle and s is the semi perimeter. so, area of the field = [45×(45−41)(45−40)(45−9)]

=(45×4×5×36)

= 32400

=180 m2

= 1800000 cm2

now, the space needed for a flower bed =900cm2

so, number of flower beds = 1800000900 =2000. 

Question 21: The perimeter of a given triangular ground is 420m, and its sides are in the ratio 6:7:8. Find the area of the triangular ground.

Solution 21: The perimeter of the triangular field =420m.

Given the ratios of the sides are 6:7:8

Sum of the ratios =6+7+8=21

Length of the first side of the field = 621×420

=6×20

=120m

Length of the second side of the field = 721 ×420=7×20=140m

Length of the third side of the field = 821×420=8×20=160m

As per Heron’s formula, the area (A) of the triangle with sides a, b & c is given as,

A= 2s(s−a)(s−b)(s−c) where 2s=(a+b+c)

Here, a=120m,b=140m,c=160m,

s= (120+140+160)2

= 4202

=210

Area of triangular field= 210×(210−120)(210−140)(210−160)

= (210×90×70×50)

= (3×7×3×3×7×5×10000)

=[(7×3×100×7×3×100)×3×5]

=210015

=210015.

Question 22: If each side of a given triangle is doubled, then find the ratio of the area of the new triangle thus formed from the given triangle.

Solution 22: Let a,b and c denote the length of the sides of the triangle.

Area of the triangle, A1 = s(s−a)(s−b)(s−c)

A1=s(s−a)(s−b)(s−c), where s is the semi-perimeter of the triangle. So, semi perimeter s= a+b+c2

When the sides of a triangle are doubled, we get s’ = 2a+2b+2c2 =a+b+c = 2s, where s’ is the semi-perimeter of the new triangle

Area of the new triangle, A2 = s′(s′−2a)(s′−2b)(s′−2c)

= 2s(2s−2a)(2s−2b)(2s−2c)

= 16s(s−a)(s−b)(s−c)

=4s(s−a)(s−b)(s−c)

=4A1

Thus, the ratio of the area of the new triangle to the given triangle =  A2A1=4A1A1 =4:1

Question 23: A field which is in the shape of a trapezium in which the parallel sides are 25m and 10m. If the non-parallel sides are 14m and 13m, find their area.

Solution 23: 

Let ABCD be a trapezium with,

AB=25m

CD=10m

BC=14m

AD=13m

Draw CE || DA. So, ADCE is a parallelogram with, CD=AE=10m

CE=AD=13m

BE=AB−AE=25−10=15m

In ΔBCE, the semi perimeter will be,

s= a+b+c2

s= 14+13+152

s=21m

Area of ΔBCE, A=s(s−a)(s−b)(s−c)

= 21(21−14)(21−13)(21−15)

= 21(7)(8)(6)

= 7056

=84m2

Also, the area of ΔBCE is,

A= 12× base × height

84= 12 ×15×CL

84×215 = CL

CL= 565 m

The area of the trapezium is A= 12 (sum of parallel sides) (height)

A= 12 ×(25+10)( 565)

A=196 m2

Thus, the area of the trapezium is 196 m2

Question 24: An umbrella was made by stitching 10 triangular pieces of cloth of 5 different colours, each piece measuring 20cm, 50cm and 50cm. How much cloth is required for each colour of one umbrella? 

(6=2.45)

Solution 24: Area of a triangle = 12bh

Here, b=20

h= 502− 102

= 2500−100

=2400

= 6×400

Therefore, Area = 122 x 20⋅206

= 10 x 206 = 2006

= 200 x245 = 490 cm2

Each colour cloth is used 2 times.

∴ The area of each coloured cloth required for one umbrella =490×2 =980cm2

Question 25: A triangle and a parallelogram have the same base and some area. When the sides of the triangle are 26cm,28cm and 30cm and the base of the parallelogram is 28cm. Find the height of the parallelogram.

Solution 25: To find Perimeter of Triangle, 2S=26+28+30=84

⇒S=42cm

Use Heron’s formula,

Area s(s−a)(s−b)(s−c)

Area = 42(42−26)(42−28)(42−30)

= 42×16×14×12 

Area =336 cm2

Area of parallelogram = Area of triangle

⇒h×28=336

⇒h=12cm

Height of parallelogram =12cm.

Question 26: Write True/False and justify the answers:

(i). The area of the triangle with a base of 4 cm and height of 6 cm is 24 cm 2

Solution 26(i):

False

Justification:

Area of the triangle = ½ × Base × Altitude

= ½ × 4 × 6

= 12cm2

Hence, the given statement “the area of the triangle with base 4 cm and height 6 cm is 24 cm2” is False.

(ii). The area of the △ ABC is 8 cm2 where AB = AC = 4 cm and ∠A = 90º.

Solution 26(ii):

It is true

Justification:

Area of the triangle = ½ × Base × Altitude

= ½ × 4 × 4

= 8cm2

Thus, the statement is “area of the △ABC is 8 cm2 where AB = AC = 4 cm and ∠A = 90º” is True.

(iii). The area of a given isosceles triangle is 5/4 √11cm2 if the perimeter is 11 cm and the base is 5 cm.

Solution 26(iii):

True

Justification:

According to the question,

Perimeter = 11cm

And base, a = 5

As the triangle is isosceles, b = c

Perimeter = 11

⇒ a + b + c = 11

⇒ 5 + b + b = 11

⇒ 5 + 2b = 11

⇒ 2b = 6

⇒ b = 3

So, we have,

a = 5, b = 3, c = 3

s = (a + b + c)/2

⇒ s = (5 + 3 + 3)/2 = 11/2

Area of triangle = √s(s-a)(s-b)(s-c)

=112(112 -5)(112 -3)(112-3)

=112(12)(52)(52)

⇒ Area of the triangle = (5√11)/4cm2

Hence, the statement “The area of an isosceles triangle is 5/4 √11cm2 when the perimeter is 11 cm, and the base is 5 cm” is True.

(iv). The area of an equilateral triangle is 20√3 cm2, and each side is 8 cm.

Solution 26 (iv):

It is false

Justification:

Area of the given equilateral triangle of side a = √3/4 a2

As per the question,

Area of the triangle = 20√3 cm2

⇒ √3/4 a2 = 20√3

⇒ a2 = 20× 4

⇒ a2 = 80

⇒ a = 4√5 cm

Hence, the statement “the area of an equilateral triangle is 20√3 cm2 having each side is 8 cm” is False.

Question 27:  Find the total cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m2.

Solution 27:

As per the question,

The sides of the given triangular field are 50 m, 65 m and 65 m.

Cost of laying grass all around the triangular field = Rs 7 per m2

Let a = 50, b = 65, c = 65

s = (a + b + c)/2

⇒ s = (50 + 65 + 65)/2

= 180/2

= 90.

Area of the triangle = √s(s-a)(s-b)(s-c)

= √90(90-50)(90-65)(90-65)

= √90×40×25×25

= 1500m2

Cost of laying grass all around= Area of the triangle ×Cost per m2

= 1500×7

= Rs.10500

Question 28: The triangular side walls of a given flyover have been used for advertisements. The measurements of the sides of the walls are 13 m, 14 m, and 15 m. The advertisements bring in Rs 2000 per m2 per year.  A company hired one of the walls for a term of 6 months. How much rent did it pay for it?

Solution 28:

As per the question,

The sides of the given triangle are 13 m, 14 m and 15 m

Let a = 13, b = 14, c = 15

s = (a + b + c)/2

⇒ s = (13 + 14 + 15)/2

= 42/2

= 21.

Area of the triangle = √s(s-a)(s-b)(s-c)

= √21(21-13)(21-14)(21-15)

= √21×8×7×6

= 84m2

Cost of the advertisements incurred for a year = Area of triangle × Cost per m2

= 84×2000

= Rs. 168000

We know the board is rented for only 6 months:

Cost of the advertisements incurred  for 6 months = (6/12) × 168000

= Rs. 84000

Question 29: From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm, and 6 cm. Find the area of the triangle.

Solution 29:

As per the question,

The lengths of the perpendiculars are given as 14 cm, 10 cm and 6 cm.

We know,

Area of the equilateral triangle of side a = √3/4 a2

We divide the triangle into three triangles,

Area of the triangle = (1/2 × a × 14) + (1/2 × a × 10) + (1/2 × a × 6)

√3/4 a2 = ½ × a × (14 + 10 + 6)

√3/4 a2 = ½ × a × 30

a = 60/√3

= 20√3

Area of triangle = √3/4 a2

=√3/4 (20√3)2

= 300√3 cm2

Question 30: The perimeter of a given isosceles triangle is 32 cm. The ratio of the equal side to its base is 3: 2. Find its area.

Solution 30:

As per the question,

The perimeter of the given isosceles triangle = 32 cm

The ratio of equal side to base = 3: 2

Let the equal side be 3x

Thus, base = 2x

The perimeter of the triangle is= 32

⇒ 3x + 3x + 2x = 32

⇒ 8x = 32

⇒ x = 4.

Equal side will be= 3x = 3×4 = 12

Base will be= 2x = 2×4 = 8

The sides of the triangle given are 12cm, 12cm and 8cm.

Let a = 12, b = 12, c = 8

s = (a + b + c)/2

⇒ s = (12 + 12 + 8)/2

= 32/2

= 16.

Area of the triangle = √s(s-a)(s-b)(s-c)

= √16(16-12)(16-12)(16-8)

= √16×4×4×8

= 32√2 cm2

Question 31: Find the area of a given parallelogram given in the figure given below. Find the length of the altitude from vertex A on the side DC.

Solution 31:

We know,

Area of the given parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)

For Area (ΔBCD),

We have,

a = 12, b = 17, c = 25

s = (a + b + c)/2

⇒ s = (12 + 17 + 25)/2 = 54/2 = 27.

Area of the (ΔBCD) = √s(s-a)(s-b)(s-c)

= √27(27-12)(27-17)(27-25)

= √27×15×10×2

= 90 cm2

Given, ABCD is a parallelogram,

Therefore, Area(ΔBCD) = Area(ΔABD)

Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)

= 90 + 90

= 180 cm2

Let altitude from A be = x

Also, Area of parallelogram(ABCD) = CD × (Altitude from A)

⇒ 180 = 12 × x

⇒ x = 15 cm

Question 32: Find the area of a given triangle in which the two sides are 18 cm and 10 cm, and the perimeter is 42cm.

Solution 32:

Assuming the third side of the given triangle is “x”.

Now, the three sides of the triangle become 18 cm, 10 cm, and “x” cm

Given the perimeter of the triangle = 42cm

So, x = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle,

= √[s (s-a) (s-b) (s-c)]

= √[21(21 – 18) (21 – 10) (21 – 14)] cm2

= √[21 × 3 × 11 × 7] m2

= 21√11 cm2

Question 33: A field which is in the shape of a trapezium in which the parallel sides are 25 m and 10 m. The non-parallel sides of which are 14 m and 13 m. Find the area of the given field.

Solution 33:

First, We draw a line segment BE that is parallel to the line AD. Now, from B, draw a perpendicular on the line segment CD.

Now, it is seen that the given quadrilateral ABED is also a parallelogram. Thus,

AB = ED = 10 m

AD = BE = 13 m

EC = 25 – ED = 25 – 10 = 15 m

Now, considering the triangle BEC,

Its semi perimeter (s) would be = (13+ 14 + 15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

=s(s-a)(s-b)(s-c)

=21 x (21-13) x(21-14)x(21-15) m2

=21 x8x7x6 m2

= 84 m2

We know the area of the ΔBEC = (½) × CE × BF

84 cm2 = (½) × 15 × BF

=> BF = (168/15) cm = 11.2 cm

So, the total area of the figure ABED will be BF × DE, i.e. 11.2 × 10 = 112 m2

∴ Area of the given field = 84 + 112 = 196 m2

Question 34: A rhombus-shaped field has green grass for 18 cows to graze. If each of the sides of the rhombus is 30 m and the longer diagonal is 48 m, how much area of the grass field will each cow get to graze?

Solution 34:

We draw a rhombus-shaped field first with its vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles, which have equal areas. The diagram is as follows.

Considering the triangle BCD,

Its semi-perimeter would be= (48 + 30 + 30)/2 m = 54 m

Using Heron’s formula,

Area of the ΔBCD =

=s(s-a)(s-b)(s-c)

=54(54-48)(54-30)(54-30) m2

= 54 x 6 x24 x 24 m2

= 432 m2

∴ Area of the given field = 2 × area of the ΔBCD = (2 × 432) m2 = 864 m2

Thus, the area of the grass field which each cow will be getting = (864/18) m2 = 48 m2

Question 35: Find the total cost of laying grass within a triangular field having sides of 50 m, 65 m, and 65 m at the rate of Rs 7 per m2.

Solution 35:

As per the question,

The sides of the given triangular field are 50 m, 65 m and 65 m.

Cost of laying grass in the given triangular field = Rs 7 per m2

Let a = 50, b = 65, c = 65

s = (a + b + c)/2

⇒ s = (50 + 65 + 65)/2

= 180/2

= 90.

Area of the given triangle = √(s(s-a)(s-b)(s-c))

= √(90(90-50)(90-65)(90-65))

= √(90×40×25×25)

= 1500m2

Cost of laying grass all over = Area of the triangle ×Cost per m2

= 1500×7

= Rs.10500

Question 36: The perimeter of a given isosceles triangle is 32 cm. Given the ratio of the equal side to its base is 3: 2. Find the area of the given triangle.

Solution 36:

As per the question,

The perimeter of the given isosceles triangle = 32 cm

Also,

The ratio of the equal side to base = 3: 2

We suppose the equal side be 3x

So, the base will be= 2x

The perimeter of the given triangle = 32

⇒ 3x + 3x + 2x = 32

⇒ 8x = 32

⇒ x = 4.

Equal side will be= 3x = 3×4 = 12

Base = 2x = 2×4 = 8

The sides of the given triangle = 12cm, 12cm and 8cm.

Let a = 12, b = 12, c = 8

s = (a + b + c)/2

⇒ s = (12 + 12 + 8)/2

= 32/2

= 16.

Area of the given triangle = √(s(s-a)(s-b)(s-c))

= √(16(16-12)(16-12)(16-8))

= √(16×4×4×8)

= 32√2 cm2

Question 37: A rectangular plot which is given for constructing a house has a measurement of 40 m in length and 15 m in width. The laws require a minimum of 3 m of space in the front and back, and 2 m of space on each of the other sides. Find the largest area where a house can be constructed.

Solution 37:

Let the rectangle be PQRS,

As per the question,

PQ = 40m and QR = 15m

As 3m is left from both the front and back,

AB = PQ -3 -3

⇒ AB = 40 -6

⇒ AB = 34m

Again,

Given that 2m has to be left from both sides,

BC = QR -2 – 2

⇒ BC = 15 -4

⇒ BC = 11m

Now, Area left for the construction of the house construction is the area of ABCD.

Hence,

Area(ABCD) = AB × CD

= 34 × 11

= 374 m2

Question 38: How much paper of each shade is required to make a kite given in the figure below, where ABCD is a square with a diagonal of 44 cm?

Solution 38:

As shown in the figure,

AC = BD = 44cm

AO = 44/2 = 22cm

BO = 44/2 = 22cm

From ΔAOB,

AB2 = AO2 + BO2

⇒ AB2 = 222 + 222

⇒ AB2 = 2 × 222

⇒ AB = 22√ 2 cm

Area of the square = (Side)2

= (22√2)2

= 968 cm2

Area of each of the triangles (I, II, III, IV) = Area of the square /4

= 968 /4

= 242 cm2

To find the area of the lower triangle,

Suppose a = 28, b = 28, c = 14

s = (a + b + c)/2

⇒ s = (28 + 28 + 14)/2 = 70/2 = 35.

Area of triangle = √s(s-a)(s-b)(s-c)

= √35(35-28)(35-28)(35-14)

= √35×7×7×21

= 49√15 = 189.77cm2

Thus,

We get,

Area of the Red = Area of the IV

= 242 cm2

Area of the Yellow = Area of the I + Area of the II

= 242 + 242

= 484 cm2

Area of the Green = Area of the III + Area of the lower triangle

= 242 + 189.77

= 431.77 cm2

Question 39: The perimeter of a triangle is 50 cm. The length of one side of the triangle is 4 cm longer than the smaller side, and the third side is 6 cm less than twice the smaller side. Find the area of a triangle.

Solution 39:

Let the smaller side be = x cm

Then, larger side = (x + 4) cm

And, third side = (2x-6) cm

Given,

Perimeter = 50 cm

⇒ x + (x + 4) + (2x-6) = 50

⇒ 4x-2 = 50

⇒ 4x = 52

⇒ x = 13

Thus, the smaller side = 13cm

Larger side will be = x + 4 = 13 + 4 = 17cm

Third side = 2x-6 = 2×13 – 6 = 26-6 = 20cm

To find the area of the triangle,

Let a = 13, b = 17, c = 20

s = (a + b + c)/2

⇒ s = (13 + 17 + 20)/2 = 50/2 = 25.

Area of triangle = √s(s-a)(s-b)(s-c)

= √25(25-13)(25-17)(25-20)

= √25×12×8×5

= 20√30 cm2

Question 40: The area of the trapezium is 475 cm2, and the height is 19 cm. Find the lengths of its two parallel sides when one side is 4 cm greater than the other.

Solution 40:

Let PQRS be the given trapezium, shown in the figure.

According to the question,

PQ = 19cm

Suppose RQ = x cm

Thus,

PS = (x + 4)cm

Construction:

Drawing a perpendicular from R on PS such that it will also be parallel to PQ.

Now,

We find,

PQRT is a rectangle,

Area of the rectangle PQRT = PQ × QR

⇒ Area(PQRT) = 19×x = 19x

Now,

PS = PT + TS

Since PT = QR = x cm

(x + 4) = x + TS

⇒ TS = 4cm

Area of triangle RST = ½ × RT × ST

Since RT = PQ = 19cm

⇒ Area(ΔRST) = ½ × 19 × 4

= 38cm2

Area(PQRS) = Area(PQRT) + Area(ΔRST)

⇒ 475 = 19x + 38

⇒ 19x = 475 -38

⇒ 19x = 437

⇒ x = 23 cm

(x + 4) = 23 + 4 = 27cm

Thus, the lengths of the parallel sides are 23cm and 27cm.

Question 41: A traffic signal board, which indicates ‘SCHOOL AHEAD’, is an equilateral triangle with a side ‘a’. Find the area of the signal board using Heron’s formula. What will be the area of the signal board when its perimeter is 180 cm?

Solution 41:

Given,

Side of signal board = a

Perimeter of signal board = 3a = 180 cm

∴ a = 60 cm

Semi perimeter of signal board (s) = 3a/2

Using Heron’s formula,

The area of the triangular signal board will be =

=s(s-a)(s-b)(s-c)

=(3a2)(3a2-a)(3a2-a)(3a2-a)

=3a2 x a2 xa2 xa2

=3a416

=3a24

=34 x 60 x 60 = 9003 cm

Question 42: The triangular side walls of any flyover have been used for advertisements. The lengths of the walls are 122 m, 22 m, and 120 m (see Fig. 12.9). The advertisements gain an average of 5000 per m2 per year. A company hired one of its employees for 3 months. How much rent did it pay?

Solution 42:

The sides of triangle ABC are 122 m, 22 m and 120 m, respectively.

Now, the perimeter would be (122+22+120) = 264 m

And, the semi perimeter (s) = 264/2 = 132 m

Using Heron’s formula,

Area of the triangle =

s(s-a)(s-b)(s-c)

=132)132-122)(132-22)(132-120) m2

=132 x 10x 110 x 12 m2

=1320 m2

Given the rent of the advertising per year = ₹ 5000 per m2

∴ The rent for one wall in 3 months would be = Rs. (1320×5000×3)/12 = Rs. 1650000

Question 43: There is a slide in a park. One of the side walls has been painted in some colour, indicating the message “KEEP THE PARK GREEN AND CLEAN” . If the three sides of the wall are 15 m, 11 m, and 6 m, find the area painted in colour.

Solution 43:

It is given that the three sides of the wall are 15 m, 11 m and 6 m.

Thus, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Using Heron’s formula,

Area of the message =

s(s-a)(s-b)(s-c)

= √[16(16-15)(16-11) (16-6)] m2

= √[16×1×5×10] m2 = √800 m2

= 20√2 m2

Question 44: Find the area of a given triangle, two sides of which are 18 cm and 10 cm, and the perimeter is 42cm.

Solution 44:

Assuming the third side of the triangle to be “x”.

Now, the three sides of the given triangle are 18 cm, 10 cm, and “x” cm

The perimeter of the triangle = 42cm

Thus, x = 42-(18+10) cm = 14 cm

∴ The semi perimeter of the triangle = 42/2 = 21 cm

By using Heron’s formula,

Area of the triangle,

= s(s-a)(s-b)(s-c)

= √[21(21-18)(21-10)(21-14)] cm2

= √[21×3×11×7] m2

= 21√11 cm2

Question 45: Sides of a given triangle are in the ratio of 12: 17: 25, and the perimeter is 540cm. Find the area.

Solution 45:

The ratio of the length of the sides of the triangle is given as 12: 17: 25

Now, suppose the common ratio between all the sides of the given triangle is “x.”

∴ The sides are 12x, 17x and 25x

The perimeter of the triangle = 540 cm

12x+17x+25x = 540 cm

54x = 540cm

Thus, x = 10

Now, the sides of the given triangle are 120 cm, 170 cm, and 250 cm.

Then, the semi perimeter of the triangle (s) will be= 540/2 = 270 cm

Using Heron’s formula,

Area of the triangle

s(s-a)(s-b)(s-c)

=[270(270-120)(270-170)(270-250)] cm2

=[270 x 150 x 100 x 20] cm2

= 9000 cm2

Question 46: An isosceles triangle has a perimeter of 30 cm, and the equal sides are 12 cm each. Find the area of the triangle.

Solution 46:

First, let the third side be x.

The length of the equal sides is 12 cm, and the perimeter is 30 cm.

So, 30 = 12+12+x

∴ The length of the third side is 6 cm

The semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm

Using Heron’s formula,

Area of the triangle

=s(s-a)(s-b)(s-c)

= √[15(15-12)(15-12)(15-6)] cm2

= √[15×3×3×9] cm2

= 9√15 cm2

Question 47: A park which in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area will it occupy?

Solution 47:

First, we construct a quadrilateral ABCD and then join BD.

We know,

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The figure is shown below:

Now, applying the Pythagoras theorem in ΔBCD

BD2 = BC2 +CD2

⇒ BD2 = 122+52

⇒ BD2 = 169

⇒ BD = 13 m

Now, the area of the figure ΔBCD = (½ ×12×5) = 30 m2

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Using Heron’s formula,

Area of ΔABD

s(s-a)(s-b)(s-c)

=15(15-13)(15-9)(15-8) m2

= 15 x 2 x 6 x 7 m2

= 6√35 m2 = 35.5 m2 (approximately)

∴ The area of the given quadrilateral ABCD = Area of the ΔBCD+Area of the ΔABD

= 30 m2+35.5m2 = 65.5 m2

Question 48: Find the area of a given quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution 48:

First, we construct a diagram with the given parameter.

Now, applying the Pythagorean theorem in ΔABC,

AC2 = AB2+BC2

⇒ 52 = 32+42

⇒ 25 = 25

Thus, it is concluded that ΔABC is right-angled at B.

So, area of the figure ΔBCD = (½ ×3×4) = 6 cm2

The semi perimeter of the figure ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Using Heron’s formula,

Area of ΔACD

s(s-a)(s-b)(s-c)

= 7(7-5)(7-5)(7-4) cm2

=(7 x 2 x 2 x3)cm2

= 2√21 cm2 = 9.17 cm2 (approximately)

Area of the quadrilateral ABCD = Area of the ΔABC + Area of theΔACD = 6 cm2 +9.17 cm2 = 15.17 cm2

Question 49: Both a triangle and a parallelogram have the same base and area. If the sides of the given triangle are 26 cm, 28 cm, and 30 cm, and the base of the parallelogram is 28 cm, find the height of the parallelogram.

Solution 49:

Given,

As per the question, the parallelogram and triangle have equal areas.

The sides of the given triangle are 26 cm, 28 cm and 30 cm.

So, the perimeter will be = 26+28+30 = 84 cm

And its semi perimeter will be= 84/2 cm = 42 cm

Using Heron’s formula, the area of the triangle =

s(s-a)(s-b)(s-c)

= √[42(42-26)(42-28)(42-30)] cm2

= √[42×16×14×12] cm2

= 336 cm2

Now, suppose the height of the parallelogram is h.

As the area of the parallelogram = area of the triangle,

28 cm× h = 336 cm2

∴ h = 336/28 cm

So, the height of the given parallelogram is 12 cm

Question 50: If a rhombus-shaped field has green grass for 18 cows to graze upon. If each of the sides of the rhombus is 30 m and the longer diagonal is 48 m, how much total area of grass field will each cow get to graze?

Solution 50:

We first draw a rhombus-shaped field with vertices as ABCD. The diagonal AC thus divides the rhombus into two congruent triangles, which have equal areas. The figure is as follows.

Considering the triangle BCD,

Its semi-perimeter will be= (48 + 30 + 30)/2 m = 54 m

By using Heron’s formula,

Area of the ΔBCD =\

s(s-a)(s-b)(s-c)

=54(54-48)(54-30)(54-30) m2

=54 x 6 x 24 x 24 m2

= 432 m2

∴ Area of the field = 2 × area of ΔBCD = (2 × 432) m2 = 864 m2

Thus, the area of grass field that each cow will  get = (864/18) m2 = 48 m2

Question 51: An umbrella is supposedly made by stitching 10 triangular pieces of cloth from two different colours (see figure given below), each piece of measurement 20 cm, 50 cm and 50 cm. What measurement of cloth of each colour will be required for the umbrella?

Solution 51:

For each triangular piece, semi perimeter would be

s = (50+50+20)/2 cm = 120/2 cm = 60cm

By using Heron’s formula,

Area of the triangular piece

= s(s-a)(s-b)(s-c)

= √[60(60-50)(60-50)(60-20)] cm2

= √[60×10×10×40] cm2

= 200√6 cm2

∴ The area of the triangular pieces = 5 × 200√6 cm2 = 1000√6 cm2

Question 52: A kite in the shape of a square with its diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each shall be made of three different shades, as shown in the figure given below. How much will paper for each shade be used in it?

Solution 52:

As the kite is the shape of a square, the area will be

A = (½)×(diagonal)2

Area of kite = (½)×32×32 = 512 cm2.

The area of the shade I = Area of the shade II

512/2 cm2 = 256 cm2

So, the total area of the paper required in each shade is 256 cm2

For triangle section (III),

The sides are 6 cm, 6 cm and 8 cm

Now, the semi perimeter of the isosceles triangle = (6+6+8)/2 cm = 10 cm

Using Heron’s formula, the area of III triangular pieces will  be

s(s-a)(s-b)(s-c)

= √[10(10-6)(10-6)(10-8)] cm2

= √(10×4 ×4×2) cm2

= 8√5 cm2 = 17.92 cm2 (approx.)

Question 53: A floral design on the floor is made up of 16 tiles that are triangular; the sides of the triangle are 9 cm, 28 cm, and 35 cm (see figure given below). Find the total cost of polishing the tiles at the rate of 50p per cm2.

Solution 53:

The semi perimeter of each triangular shape = (28+9+35)/2 cm = 36 cm

Using Heron’s formula,

The area of each triangular shape will be

s(s-a)(s-b)(s-c)

= 36 x (36-35) x(36-28) x(36 -9)

=36 x 1 x 8 x 27 cm2

= 36√6 cm2 = 88.2 cm2

That is, the total area of 16 tiles = 16×88.2 cm2 = 1411.2 cm2

Given that the polishing cost of tiles is = 50 paise/cm2

∴ The total polishing cost of the tiles is = Rs. (1411.2×0.5) = Rs. 705.6

Question 54: A field which is in the shape of a trapezium with parallel sides is 25 m and 10 m. The non-parallel sides of which are 14 m and 13 m. Find the area of the given field.

Solution 54:

Firstly, we draw a line segment BE that is parallel to the line AD. Then, from B, we draw a perpendicular line segment CD.

Now, as shown that the quadrilateral ABED is a parallelogram. Thus,

AB = ED = 10 m

AD = BE = 13 m

EC = 25-ED = 25-10 = 15 m

Now, considering the triangle BEC,

Its semi perimeter (s) will be = (13+14+15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

s(s-a)(s-b)(s-c)

=21 x (21-13) x(21-14) x( 21-15) m2

= 21 x 8 x 7 x 6 m2

= 84 m2

We know the area of ΔBEC = (½)×CE×BF

84 cm2 = (½)×15×BF

BF = (168/15) cm = 11.2 cm

So, the total area of the figure ABED will be BF×DE, i.e. 11.2×10 = 112 m2

∴ Area of the given field = 84+112 = 196 m2

Question 55: The perimeter of a given triangle is 450m, and the sides are in the ratio of 13:12:5.

Find the area of the triangle.

Solution 55: Suppose the sides of the triangle be 13x,12x and 5x

The perimeter of the triangle =450m

∴13x+12x+5x=450m

or

30x=450

∴x=15

∴ The sides are 13×15,12×15, and 5×15

I.e.

195m,180m and 75m

∴S= a+b+c2

= 4502

=225m

∴ Area of triangle = s(s−a)(s−b)(s−c) sq m

= 225(225−195)(225−180)(225−75) sq m

= 225×30×45×150 sq m

= (15×15×2×3×5)sq m

=6750 sqm

Question 56: The sides of the triangle are 39cm,42cm, and 45cm. A parallelogram is standing on the greatest side of the triangle and has the same area as that of the triangle. Find the height of the parallelogram.

Solution 56: To find the area of ΔABC

S= 45+42+392 cm

S= 63cm

Therefore, the Area of ΔABC= 63(63−45)(63−42)(63−39)sq cm

= 63×18×21×24 sq cm

=9×7×2×3×2sqcm

=756sqcm

Suppose h be the height of the parallelogram

Now,

Area of the parallelogram BCDE= Area of ΔABC

∴h×BC=756

or

45h=756

h= 75645

h=16.8cm

Hence, the height of the parallelogram =16.8cm

Question 57: The students of a particular school staged a rally for the purpose of a cleanliness campaign. They walked through the lanes in a group of two. One of the groups walked through lanes AB, BC, and CA, whereas the other group went through lanes AC, CD, and DA. Then they cleaned the total area within their lanes. What if AB=9m, BC=40m, CD=15m, DA=28m and ∠B= 90∘? Which of the groups cleaned more areas and by how much more? Find the total area cleaned by the students.

Solution 57: We have right angle ΔAB

AC2 = AB2 + BC2

AC2 = 92 + 402

AC2 = 92 + 402

AC2 = 1681

∴AC=41

The first group is supposed to clean the area of ΔABC, which is the right-angled triangle

Now,

Area of ΔABC= 12 ×40m×9m

=180sqm

The second group is supposed to clean the area of ΔACD, which has AD=28m,

DC=15m and AC=41

Hence,

S= 28+15+412

=42m

∴ Area of ΔACD= 42(42−28)(42−15)(42−41) sq m

= 42×14×27×1 sq m

= 7×3×2×7×2×9×3 sq m

=7×3×2×7×2×9×3sqm

=126sqm

∴ The first group cleaned more =(180−126)sqm =54sqm.

Therefore, the Total area which was cleaned by students =(180+126)sqm =306sq m.

Question 58: A traffic signal board which indicates ‘school ahead’ is an equilateral triangle with side ‘a’. Find the area of the signal board using the Herons formula. Its perimeter is 180cm. What will be its area?

Solution 58:  To find the area of the single board,

S= a+a+a2 units

S= 3a2 units 

∴ Area of triangle = 3a2 x (3a2 -a)(3a2 -a)(3a2 -a)

= 3a2 xa2 xa2xa2

= a243 sq units

Perimeter =180cm

Thus, each side = 1803 = 60 cm

Area of signal board = 34(60)2 sq cm

= 9003 sq cm

Question 59: A parallelogram in which the length of the sides is 80m and 40m has one diagonal 75m long. Find the area of the parallelogram.

Solution 59: As according to the question,

AB = DC = 80cm

BC =  AD = 40cm and

AC = 75cm

In △ABC, S= 80+40+752 =97.5cm

Area of triangle ABC= s(s−a)(s−b)(s−c)

= 97.5(97.5−80)(97.5−40)(97.5−75) sq m

= 97.5×17.5×57.5×22.5 sq m

= 2207460.94

=1485.75sqm

Area of parallelogram ABCD=2× Area of △ABC

=2×1485.7

=2971.4sq m

Question 60: The side of a given triangular field is 52m,56m, and 60m. Find the cost of levelling the field Rs18/m  when the space of 4 cm is to be left for the entry gate.

Solution 60: The side of the triangular field is 52 m,56 m, and 60 m

The cost of levelling is Rs. 18/m.

To find:

The total cost of levelling.

1) Leveling when done at the boundary of the field so we will find the perimeter of the field first

The perimeter of the field:

52 + 56 + 60

168 m.

The length needed to be levelled 168−4=164 m (space of 4 m is left for the entry gate)

2) Cost of levelling is 164×18 =Rs. 2952

The total cost of the levelling is Rs. 2952

Question 61: A floral design of the floor is made up of 16 tiles which are triangular. The side of the triangle being 9 cm, 28 cm, and 35 cm. Find the cost of polishing the tiles at 50 paisa/ sq cm

Solution 61: For each triangular tile, we have

S= 35+28+92 cm=36cm

∴ Area of Each tile = s(s−a)(s−b)(s−c)

= 36(36−35)(36−28)(36−9) sq cm

= 366  sq cm

Area of 16 tile =16×366 sq cm

Therefore, cost of polishing =Rs[ 12×16×366] = Rs 2886

=Rs(288×2.45)

=Rs705.60

Question 62: The measure of one side of the right triangle is 42m. If the difference between the lengths of its hypotenuse and the other side is 14cm, find the measurement of two on the unknown side.

Solution 62: Suppose AB=y and AC=x and BC=42cm

Therefore, By the given condition,

x−y=14(i)

By Pythagoras theorem,

x2−y2 =1764

(x+y)(x−y)=1764

∴14(x+y)=1764 using (ii)

∴x+y= 176414 = 126 (iii)

Adding (ii) and (iii), we get

2x=140

I.e. x=70

∴y=126−x

∴y=126−x

y=126−70

=56

Question 63: The perimeter of the rhombus ABCD is 80cm. Find the area of the rhombus if its diagonal BD measures 12cm.

Solution 63: Given that,

The perimeter of the rhombus =80 m

The perimeter of the rhombus =4× side

⇒4a=80

⇒a=20 m

Now in△ABD,

∴S= 20+20+122 = 26

So, Area of ΔABD= 26×6×6×14 sq cm =114.4sqcm

Area of rhombus =2× area of △ABD

=2×114.4sqcm

=228.8sqcm

Question 64: Find area of a quadrilateral ABCD in which AB=5cm,BC=6cm,CD = 6cm,DA=7cm

AB=5cm,BC=6cm,CD = 6cm,DA=7cm, and AC=7cm

Solution 64: Area of the quadrilateral ABCD = Area of the ΔABC+ Area of ΔACD (i)

In ΔABC,

S= 5+6+72 =9cm

Area of ΔABC= 9(9−5)(9−6)(9−7) sq cm

ΔABC=9(9−5)(9−6)(9−7)sqcm

= 9×4×3×2 sq cm

=66 sq cm

=66sqcm

=14.4sqcm

In ΔACD,

S= 7+7+62 =10cm

∴ Area of ΔACD= 10(10−7)(10−7)(10−6) sq cm

ΔACD=10(10−7)(10−7)(10−6)sqcm

= 10×3×3×4 sq cm

=18.9sqcm

Area of the quadrilateral ABCD=(14.4+18.9)sq cm

=33.3sqcm

Question 65: Shashi Kant owns a vegetable garden, which is in the rhombus. The length of each side of the garden is 35m, and its diagonal is 42m long. After growing the vegetables, he wanted to divide them into seven equal parts and decided to look after each part once a week. Find the area of the garden that he has to look after daily.

Solution 65: Let ABCD be garden

∴DC=35m

DB=42m

Draw CE⊥DB

The diagonals of any rhombus bisect each other at right angles.

∴DE= 12, DB= 12×42  or 21m

Now

CE2=CD2−DE2

= 352 − 212

=784

CE=28m

Area of ΔDBC=  12×DB×CE

= 12 ×42×28

=588sqcm

∴ Area of the garden ABCD=2×588 sqm

=1176 sqm

Area of the piece of garden which he has to be looked after daily = 11767 sq m

=168 sqm

Question 66: The perimeter of the triangle is 480 meters, and its sides are in the ratio of 1:2:3.

Find the area of the triangle.

Solution 66: Suppose the sides of the triangle be x,2x,3x

The perimeter of the triangle =480m

∴x+2x+3x=480m

6x=480m

x=80m

Therefore, the sides are 80m,160m,240m

So, S= 80+160+2402

= 4802

=240m

And,

∴ Area of triangle = s(s−a)(s−b)(s−c) sq m

= 240(240−80)(240−160)(240−240) sq m

=0 sq m

Therefore, any triangle doesn’t exist with its ratio of 1:2:3, whose perimeter is 480m.

Question 67: Find the total cost of levelling the ground, which is in the form of an equilateral triangle having a side of 12m at Rs 5 per square meter.

Solution 67: Here, the sides are 12m,12m,12m

∴S= 12+12+122 =18cm

And,

Area of equilateral triangle= s(s−a)(s−b)(s−c) sq m

= 18(18−12)(18−12)(18−12) sq m

=18(18−12)(18−12)(18−12)sqm

= 18×6×6×6 sq m

= 6×3×6×6×6 sq m

=363 sq m

∴ Cost of leveling ground =5×36×1.73  = Rs311.4\m

Question 68: A kite which is in the shape of a square with a diagonal of 32cm and an isosceles triangle of base 8cm and side 6cm each of which is to be made of three different shades. How much of the paper of each shade has been used in it? ( use 5=2.24)

Solution 68: Suppose ABCD be the square and ΔCEF an isosceles triangle.

Let diagonals bisect each other at O.

Thus, AO= 12 ×32cm =16cm

Area of shaded portion, I= 12 ×16×32 sq cm =256 sq cm

And,

Area of portion III = a44b2  –  a2 = 844 x (6)2 -8

=17.92sqcm

Thus, the papers of all three shades required are 256 sq cm,256sq cm and 17.92sqcm.

Question 69: The sides of a quadrangular field are 29m,36m,7m, and 24m, respectively. The angle, as contained by the last two sides, is a right angle. Find its area.

Solution 69: The sides are given in order. Thus, the length of the last two sides is 7 m and 24 m, respectively.

We know in a right-angled triangle using Pythagoras theorem, 

( Hypotenuse )2=( Base )2+( Height )2

Let us assume:

The diagonal of the field is d.

Substituting the values,

(d)2=(24)2 +(7)2

(d)2=576+49

(d)2=625

d=25

Hence, the Diagonal of the park =25 m

Semi-perimeter =Perimeter/2

=(29+36+25)m/2

=90 m/2

=45 m

Substituting the values,

 Area = 45(45−29)(45−36)(45−25)

= 45×16×9×20

= 129600

=360

 

Question 70: Finding the area of the 2nd triangle whose measurements are given below:

1st side =7 m

2nd side=24 m

3rd side=25 m

Solution 70: 

Finding the semi-perimeter of the triangle:

Semi-perimeter = Perimeter/2 

=(7+24+25)m/2

=56 m/2

=28 m

Finding the area of the 2nd triangle using Heron’s formula:

 Area =28(28−7)(28−24)(28−25)

= 28×21×4×3

= 7056

=84

Thus, the area of the 2nd triangle is 84m2

Finding the area of the quadrangular field:

Area of the field =( Area of 1 stΔ) + ( Ar. of 2nd Δ )

=(360+84)m2

=444 m2

=444 m2

Hence, the area of the field=444 m2

Question 71: A field which is in the shape of a trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 14m and 13m. Find the area of the field.

Solution 71: Let ABCD be a trapezium with,

AB∥CD

AB=25 m

CD=10 m

BC=14 m

AD=13 m

Draw CE | DA. So, ADCE is a parallelogram with,

CD=AE=10 m

CE=AD=13 m

BE=AB−AE=25−10=15 m

In ΔBCE, the semi perimeter will be,

s= a+b+c2

s= 14+13+152

s=21 m

Area of △BCE,

A= s(s−a)(s−b)(s−c)

= 21(21−14)(21−13)(21−15)

= 21(7)(8)(6)

= 7056

=84 m2

Also, the area of ΔBCE is,

A= 12 × base × height 

84= 12 ×15×CL

84×215=CL

CL= 565 m

Now, the area of the trapezium is,

A= 12 (sum of parallel sides) (height) 

A= 12×(25+10)(565)

A=196  m2

Thus, the area of the given trapezium is 196 m2

Question 72: The perimeter of the right triangle is 24cm. If its hypotenuse is 10cm, find the length of the other two sides. Find the area by using the formula area of a right triangle. Verify your result by using Heron’s formula.

Solution 72: Let the sides of the right Δ be ‘a ‘ cm and ‘b ‘cm.

Then,

a+b+c=24

⇒a+b+10=24

⇒a+b=24−10

⇒a+b=14…………(1)

a2+b2=(10)2

Also, a2+b2=100

⇒a2+b2=100……(2)

We know that

(a+b)2=a2+2ab+b2

(14)2=100+2ab

−2ab=100−196

−2ab=−96

ab= 962=48

ab=962=48

=>ab=48……

Also,

(a−b)2=a2−2ab+b2

(a−b)2=100−2×48

(a−b)2=100−2×48

(a−b)2=100−96

(a−b)2=4

(a−b)= 4 =2

⇒(a−b)=2

Solving (1) and (4), we get;

∴a=8 and b=6

Now,

a+b+c2

S= 242=12

Area of Δ= s(s−a)(s−b)(s−c)

⇒ 12(12−8)(12−6)(12−10)

⇒ 12×4×6×2

=> 2×2×3×2×2×2×3×2

=> 22× 22× 2232

=>2×2×2×3

=>24

Hence, The area of Δ is 24 cm.

Question 73: Radha drew a picture of an aeroplane using coloured paper, as shown in figure given below. find the total area of the paper used.

Solution 73: Area (1) = area of the isosceles triangle with a=1cm and b=5cm

= a44b2–a2

= 14100-1 = 994 sq cm(approx)

Area (ii) = area of a rectangle with

L=6.5cm

L=6.5cm and b=1cm

=6.5×1sqcm

=6.5sqcm

Area (iii) = Area of trapezium

=3× Area of equilateral Δ with side=1cm

=3× 34×(1)2 sq cm

= 3×1.7324 or 5.1964sqcm

=1.3sqcm (approx.)

Area of(IV+V)=2× 12×6×1.5sqcm=9sqcm

Total used area of the paper = Area (I+II+III+IV+V)

=(2.5+6.5+1.3+9) sq cm

=19.3 sqcm

Benefits of Solving Mathematics Class 9 Chapter 12 Important Questions

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All the topics are covered from the chapter Heron’s formula in our question bank of Important Questions Class 9 Mathematics Chapter 12. These topics include the area of a triangle based on its base and heights, area of triangle based on Heron’s Formula, and the application of Heron’s Formula in calculating areas of Quadrilaterals.

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