Chapter 14 Areas of triangles and Quadrilaterals - Praadis Education

Chapter 14

Areas of triangles and Quadrilaterals

Exercise 14.1

Page No: 533

Question 1:

Find the area of the triangle whose base measures 24 cm and the

corresponding height measures 14.5 cm.

ANSWER:

We have:

Base = 24 cm

Height = 14.5 cm

Now,

Area of triangle= 1

2 ×Base×Height

= 1

2 ×24×14.5

=174 cm2

Question 2:

The base of a triangular field is three times its altitude. If the cost of

sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

ANSWER:

Let the height of the triangle be h m.

∴ Base = 3h m

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Now,

Area of the triangle = 𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡

𝑅𝑎𝑡𝑒 =

783

58

=13.5 ha

=135000 m2

We have:

Area of triangle = 135000 m2

⇒ 1

2 ×Base×Height =135000

⇒ 1

2×3h×h

⇒135000

⇒ ℎ2 = 135000×2

3

⇒ ℎ2 = 90000

⇒ ℎ = 300𝑚

Thus, we have:

Height = h = 300 m

Base = 3h = 900 m

Question 3:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in

length. Hence, find the height corresponding to the longest side.

ANSWER:

Let: a=42 cm, b = 34 cm and c=20 cm

∴s = 𝑎+𝑏+𝑐

2

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= 42+34+20

2

= 48 cm

By Heron's formula, we have:

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

= √48(48 − 42)(48 − 34)(48 − 20)

= √48 × 6 × 14 × 28

= √4 × 2 × 6 × 6 × 7 × 2 × 7 × 4

= 4 × 2 × 6 × 7

=336 cm2

We know that the longest side is 42 cm.

Thus, we can find out the height of the triangle corresponding to 42 cm.

We have:

Area of triangle = 336 cm2

⇒ 1

2 ×Base×Height = 336

⇒Height = 336×2

42 =16 cm

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30

cm in length. Also, find the length of the altitude corresponding to the

smallest side.

ANSWER:


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∴ 𝑠 = 𝑎 + 𝑏 + 𝑐

2

= 18+24+30

2

= 36𝑐𝑚.



= √36(36 − 18)(36 − 24)(36 − 30)

= √36 × 18 × 12 × 6

= √12 × 3 × 6 × 3 × 12 × 6

= 12 × 3 × 6

= 216𝑐𝑚2

We know that the smallest side is 18 cm.

Thus, we can find out the altitude of the triangle corresponding to 18 cm.

We have:

Area of triangle = 216𝑐𝑚2

= 1

2× 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔ℎ𝑡 = 216

⟹ 𝐻𝑒𝑖𝑔ℎ𝑡 = 216×2

18 = 24 cm.

Question 5:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105

m in length. Find the height corresponding to the longest side.

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ANSWER:

Let: a=91 m, b = 98 m and c=105 m

∴ 𝑠 = 𝑎+𝑏+𝑐

2

𝑠 = 91 + 98 + 105

2

= 𝑠 = 147𝑚


we have:


= √147(147 − 91)(147 − 98)(147 − 105)

= √147 × 56 × 49 × 42

= √7 × 3 × 7 × 2 × 2 × 2 × 7 × 7 × 7 × 7 × 3 × 2

= 7 × 7 × 7 × 2 × 3 × 2

= 4416𝑚2

We know that the longest side is 105 m.

Thus, we can find out the height of the triangle corresponding to 42 cm.

Area of triangle = 4116 m2

⟹ 1


⟹ 𝐻𝑒𝑖𝑔ℎ𝑡 = 4116×2

105 = 78.4m

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Question 6:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150

m. Find the area of the triangle.

ANSWER:

Let the sides of the triangle be 5x m, 12x m and 13x m.

We know:

Perimeter = Sum of all sides

or, 150 = 5x + 12x + 13x

or, 30x = 150

or, x = 5

Thus, we obtain the sides of the triangle.

5×5 = 25 m

12×5 = 60 m

13×5 = 65 m

Now,

Let: a=25 m, b = 60 m and c=65 m

∴s = 150

2

=75 m

By Heron's formula, we have


= √75(75 − 25)(75 − 60)(75 − 65)

= √75 × 50 × 15 × 10

= √15 × 5 × 5 × 10 × 15 × 10

= 15 × 5 × 10

= 750𝑚2

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Question 7: The perimeter of a triangular field is 540 m and its sides are

in the ratio 25: 17: 12. Find the area of the field. Also, find the cost of

ploughing the field at 5 per m2.

ANSWER:

Let the sides of the triangle be 25x m, 17x m and 12x m.

We know:


or, 540 = 25x + 17x + 12x

or, 54x = 540

or, x = 10

Thus, we obtain the sides of the triangle.

25××10 = 250 m

17××10 = 170 m

12××10 = 120 m

Now,

Let: 𝑎 = 250 𝑚, 𝑏 = 170 𝑚 and 𝑐 = 120 𝑚

∴ 𝑠 = 540

2= 270 𝑚



= √270(270 − 250)(270 − 170)(270 − 120)

= √270 × 20 × 100 × 150

= √30 × 3 × 3 × 20 × 20 × 5 × 30 × 5

= 30 × 3 × 20 × 5

= 9000𝑚2

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Cost of ploughing 1 m2 field = Rs 5

Cost of ploughing 9000 m2 field = 5×9000=Rs 450005×9000=Rs 45000.

Question 8:

Two sides of a triangular field are 85 m and 154 m in length and its

perimeter is 324 m. Find (i) the area of the field and (ii) the length of the

perpendicular from the opposite vertex on the side measuring 154 m.

ANSWER:

(i) Let: a=85 m and b = 154 m

Given:Perimeter = 324 m

or,

a+b+c =324

⇒c =324−85−154

=85 m

∴s = 324

2

= 162m



= √162(162 − 85)(162 − 154)(162 − 85)

= √1296 × 77 × 77

= √36 × 77 × 77 × 36

= √3 × 6 × 77

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= 2772𝑚2

(ii) We can find out the height of the triangle corresponding to 154 m in

the following manner:

We have:

Area of triangle = 2772𝑚2

= 1


⟹ 𝐻𝑒𝑖𝑔ℎ𝑡 = 2772×2

154

= 36m

Question 9:

Find the area of an isosceles triangle each of whose equal sides measures

13 cm and whose base measures 20 cm.

ANSWER:

We have: a=13 cm and b=20 cm

∴Area of isosceles triangle=

𝑏

4√4𝑎2 − 𝑏2

= 20

4√4(13)2 − 202

= 5√676 − 400

= 5√276

= 5 × 16.6

= 83.06𝑐𝑚2

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Question 10:

The base of an isosceles triangle measures 80 cm and its area is 360 cm2.

Find the perimeter of the triangle.

Find the perimeter of the triangle.

ANSWER:

Let △PQR be an isosceles triangle and PX⊥QR.

Now,


⟹1

2× 𝑄𝑅 × 𝑃𝑋 = 360

⟹ ℎ = 720

80 = 9cm.

Now, QX = 12×80 = 40 cm

and PX = 9 cm

Area of triangle =360 cm2

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⇒1

2 ×QR×PX

= 360

⇒h = 720

80 =9 cm

Now,

QX = 1

2 ×80 = 40 cm and

PX = 9 cm

Also,

𝑃𝑄 = √𝑄𝑋2 + 𝑃𝑋2

𝑎 = √402 + 92

= √1600 + 81

= √1681

= 41cm

∴ Perimeter = 80 + 41 + 41 = 162 cm

Question 11:

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal

side to its base is 3 : 2. Find the area of the triangle.

ANSWER:

The ratio of the equal side to its base is 3 : 2.

⇒⇒Ratio of sides = 3 : 3 : 2.

Let the three sides of triangle be 3x, 3x, 2x.

The perimeter of isosceles triangle = 32 cm.

⇒3x+3x+2x=32 cm

⇒8x=32

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⇒x=4 cm

⇒3x+3x+2x=32 cm

⇒8x=32

⇒x=4 cm

Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.

Let S be the semi-perimeter of the triangle. Then,

𝑆 = 1

2(12 + 12 + 8)

= 32

2

= 16

Area of the triangle will be

= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √16(16 − 12)(16 − 12)(16 − 8)

= √16 × 4 × 4 × 8

= 4 × 4√8

= 4 × 4 × 2√2

= 32√2𝑐𝑚2

Disclaimer: The answer does not match with the answer given in the

book.

Question 12:

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The perimeter of triangle is 50 cm. One side of the triangle is 4 cm

longer than the smallest side and the third side is 6 cm less than twice

the smallest side. Find the area of the triangle.

ANSWER:

Let ABC be any triangle with perimeter 50 cm.

Let the smallest side of the triangle be x.

Then the other sides be x + 4 and 2x − 6.

Now,

x + x + 4 + 2x − 6 = 50 (∵ perimeter is 50 cm)

⇒ 4x − 2 = 50

⇒ 4x = 50 + 2

⇒ 4x = 52

⇒ x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.

∴ Semi-perimeter of the triangle is

𝑠 = 13 + 17 + 20

2

𝑠 = 50

2 = 25cm



= √25(25 − 13)(25 − 17)(25 − 20)

= √25(12)(8)(5)

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= 20√30 𝑐𝑚2

Hence, the area of the triangle is 20√30 𝑐𝑚2.

Question 13:

The triangular side walls of a flyover have been used for advertisements.

The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield

an earning of Rs 2000 per m2 a year. A company hired one of its walls

for 6 months. How much rent did it pay?

ANSWER:

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


𝑠 = 13 + 14 + 15

2

= 42

2

=21m



= √21(21 − 13)(21 − 14)(21 − 15)

= √21(8)(7)(6)

= 84𝑚2

Now,

The rent of advertisements per m2 per year = Rs 2000

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The rent of the wall with area 84 m2 per year = Rs 2000 × 84

= Rs 168000

The rent of the wall with area 84 m2 for 6 months = Rs 168000

2

= Rs 84000

Hence, the rent paid by the company is Rs 84000.

Question 14:

The perimeter of an isosceles triangle is 42 cm and its base is 11

2 times

each of the equal sides. Find (i) the length of each side of the triangle,

(ii) the area of the triangle, and (iii) the height of the triangle.

ANSWER:

Let the equal sides of the isosceles triangle be a cm each.

∴ Base of the triangle, b = 3

2a cm

(i) Perimeter = 42 cm

or, a + a + 3

2a = 42

or, 2a +3

2a= 42

⇒2a+ 3

2a = 42

⇒ 7𝑎

2 =42

⇒ 𝑎=12

⇒2a+ 3

2a = 42

⇒ 7𝑎

2=42

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⇒a=12

So, equal sides of the triangle are 12 cm each.

Also,

Base = 3

2a =

3

2×12=18 cm32×12=18 cm

(ii)

Area of isosceles triangle = 𝑏

4√4𝑎2 − 𝑏2

= 18

4√4(12)2 − 182 (a = 12 cm and b = 18 cm)

= 4.5√576 − 324

= 4.5 × √253

= 4.5 × 15.87

= 71.42𝑐𝑚2

(iii)

Area of triangle = 71.42𝑐𝑚2

⟹ 1

2× 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔ℎ𝑡 = 71.42

⟹ 𝐻𝑒𝑖𝑔ℎ𝑡 =71.42×2

2

= 7.94cm

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Question 15:

If the area of an equilateral triangle is 36√3 𝑐𝑚2 find its perimeter.

ANSWER:

Area of equilateral triangle = √3

4× (𝑆𝑖𝑑𝑒)2

⟹ 36√3

⟹ (𝑆𝑖𝑑𝑒)2 = 144

⟹ 𝑆𝑖𝑑𝑒 = 12𝑐𝑚

Thus, we have:

Perimeter = 3 × Side = 3 × 12 = 36 cm

Question 16:

If the area of an equilateral triangle is 81√3 𝑐𝑚2, find its height.

ANSWER:

Area of equilateral triangle √3


⟹ 81√3

⟹ (𝑆𝑖𝑑𝑒)2 = 324

⟹ 𝑆𝑖𝑑𝑒 = 18𝑐𝑚

Thus, we have:

Perimeter = √3

2× Side

= √3

2 × 18

= 9√3 cm

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Question 17:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of

the triangle, correct to 2 places of decimal and (ii) the height of the

triangle, correct to 2 places of decimal. Take √3 =1.732.

ANSWER:

Side of the equilateral triangle = 8 cm

Area of equilateral triangle √3


⟹ √3

4× (8)2

⟹ 1.732×64

4

⟹ 27.71𝑐𝑚2

(ii) Height = √3

2× 𝑠𝑖𝑑𝑒

= √3

2× 8

= 1.732×8

2

= 6.93 𝑐𝑚.

Question 18:

The height of an equilateral triangle measures 9 cm. Find its area,

correct to 2 places of decimal. Take √3 =1.732.

ANSWER:

Height of the equilateral triangle = 9 cm

Thus, we have:

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Height = √3


⟹ 9 = √3


⟹ √3


⟹ 𝑆𝑖𝑑𝑒 = 18

√3 =

18

√3×

√3

√3 = 6√3 cm

Also,


4× (𝑠𝑖𝑑𝑒)2

= √3

4× (6√3)2

= 108

4√3

= 27√3

= 47.76𝑐𝑚2

Question 19:

The base of a right-angled triangle measures 48 cm and its hypotenuse

measures 50 cm. Find the area of the triangle.

ANSWER:

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Let △PQR be a right-angled triangle and PQ⊥QR.

Now,

PQ = √𝑃𝑅2 − 𝑄𝑅2

= √502 − 482

= √2500 − 2304

= √196

= 14cm.

Area of triangle = 1

2× 𝑄𝑅 × 𝑃𝑄

= 1

2× 48 × 14

= 336𝑐𝑚2

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Question 20:

Find the area of the shaded region in the figure given below.

ANSWER:

In right angled ∆ABD,

AB2 = AD2 + DB2 (Pythagoras Theorem)

⇒ AB2 = 122 + 162

⇒ AB2 = 144 + 256

⇒ AB2 = 400

⇒ AB = 20 cm

Area of ∆ADB = 1

2 ×DB×AD

= 1

2 × 16× 12

= 96𝑐𝑚2 ..(i)

In ∆ACB,

The sides of the triangle are of length 20 cm, 52 cm and 48 cm.


s = 20+52+48

2 =

120

2 = 60cm

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∴ By Heron's formula,

Area of ΔACB= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √60(60 − 20)(60 − 52)(60 − 48)

= √60(40)(8)(12)

= 480𝑐𝑚2 …(2)

Now,

Area of the shaded region = Area of ∆ACB − Area of ∆ADB

= 480 − 96

= 384 cm2

Hence, the area of the shaded region in the given figure is 384 cm2.

Question 21:

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm

and 14 cm respectively and the angle between the first two sides is a

right angle. Find its area. (Given, √6 =2.45).

ANSWER:

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8

cm, 12 cm and 14 cm respectively and the angle between the first two

sides is a right angle.

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Join AC.

In right angled ∆ABC,

𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 (Pythagoras Theorem)

⇒ 𝐴𝐶2 = 62 + 82

⇒ 𝐴𝐶2 = 36 + 64

⇒ 𝐴𝐶2 = 100

⇒AC = 10cm

Area of ∆ABC = 1

2 ×AB×BC

= 1

2×6×8

= 24 cm2 ....(1)

In ∆ACD,


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s = 10+12+14

2

= 36

2

=18 cm


Area of ΔACD = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √18(18 − 10)(18 − 12)(18 − 14)

= √18(8)(6)(4)

= 24√6 cm2

= 26(2.45) cm2

= 58.8 cm2

Thus,

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

= (24 + 58.8) cm2

= 82.8 cm2

Hence, the area of quadrilateral ABCD is 82.8 cm2.

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Question 22:

Find the perimeter and area of a quadrilateral ABCD in which BC = 12

cm, CD = 9 cm, BD = 15 cm, DA = 17 cm and ∠ABD = 90°.

ANSWER:

We know that △ABD is a right-angled triangle.

∴ 𝐴𝐵2 = √𝐴𝐷2 − 𝐷𝐵2

= √172 − 152

= √289 − 225

= √64

= 8 cm.

Now,

Area of triangle ABD= 1

2 ×Base×Height

= 1

2 ×AB×BD

= 1

2 ×8×15

=60 cm2

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Let:a=9 cm, b = 15 cm and c=12 cm

s= 𝑎+𝑏+𝑐

2 =

9+15+12

2 = 18cm.

=18 cm



= √18(18 − 9)(18 − 15)(18 − 12)

= √18 × 9 × 3 × 6

= √6 × 3 × 3 × 3 × 3 × 6

= 6 × 3 × 3

= 54𝑐𝑚2

Now,

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

= (60 + 54) cm2 =114 cm2

And,

Perimeter of quadrilateral ABCD

= AB + BC + CD + AD

= 17 + 8 + 12 + 9 = 46 cm

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Question 23:

Find the perimeter and area of the quadrilateral ABCD in which AB = 21

cm, ∠BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

ANSWER:


BC2 = AB2 + AC2 (Pythagoras Theorem)

⇒ BC2 = 212 + 202

⇒ BC2 = 441 + 400

⇒ BC2 = 841

⇒ BC = 29 cm

Area of ∆ABC = 1

2×AB×AC

= 1

2×21×20

= 210 cm2 ....(1)

In ∆ACD,



s = 20+34+42

2

= 96

2

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= 48 𝑐𝑚.



= √48(48 − 20)(48 − 34)(48 − 42)

= √48(28)(14)(6)

= 336𝑐𝑚2 ..(2)

Thus,

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

= (210 + 336) cm2

= 546 cm2

Also,

Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm

= 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546

cm2, respectively.

Question 24:

Find the area of the quadrilateral ABCD in which BCD is an

equilateral triangle, each of whose sides is 26 cm, AD = 24 cm

and ∠BAD = 90°. Also, find the perimeter of the quadrilateral.

(Given: √3 = 1.73.)

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ANSWER:

We know that △BAD is a right-angled triangle.

∴ AB= √𝐵𝐷2 − 𝐴𝐷2

= √262 − 242

= √676 − 576

= √100

= 10cm.

Now, Area of triangle BAD= 1

2×Base×Height

= 1

2 ×AB×AD

= 1

2 ×10×24 =120 cm2

Also, we know that △BDC is an equilateral triangle.

∴Area of equilateral triangle = √3

4× (𝑠𝑖𝑑𝑒)2

= √3

4× (26)2

= √3

4× 676

= 169√3

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= 292.37𝑐𝑚2

Now,

Area of quadrilateral ABCD = Area of △△ABD + Area of △△BDC

= (120 + 292.37) cm2 = 412.37 cm2

Perimeter of ABCD = AB + BC + CD + DA

= 10 + 26+ 26 + 24

= 86 cm

Question 25:

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26

cm and diagonal AC = 30 cm.

ANSWER:

Let: a=26 cm, b =30 cm and c=28 cm

s= 𝑎+𝑏+𝑐

2

= 26+30+28

2

= 42cm.


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

= √42(42 − 36)(42 − 30)(42 − 28)

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= √42 × 16 × 12 × 14

= √14 × 3 × 4 × 4 × 2 × 2 × 3 × 14

=14×4×2×3

=336 cm2

We know that a diagonal divides a parallelogram into two triangles of

equal areas.

∴ Area of parallelogram ABCD

= 2(Area of triangle ABC) = 2×336

=672 cm2

Question 26:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10

cm and AC = 16 cm. [Given: √3 =1.73]

ANSWER:

Let: a=10 cm, b =16 cm and c=14 cm

s = 𝑎+𝑏+𝑐

2

= 10+16+14

2

= 20cm.

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= √20(20 − 10)(20 − 16)(20 − 14)

= √20 × 10 × 4 × 6

= √10 × 2 × 10 × 2 × 2 × 3 × 2

= 10 × 2 × 2√3

= 69.2 cm2

We know that a diagonal divides a parallelogram into two triangles of

equal areas.

∴ Area of parallelogram ABCD = 2(Area of triangle ABC)

= 2×69.2 cm2

=138.4 cm2

Question 27:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64

cm, AL ⊥ BD and CM ⊥ BD such that AL = 16.8 cm and CM = 13.2 cm.

Calculate the area of quadrilateral ABCD.

ANSWER:

Area of ABCD=Area of △ABD + Area of △BDC

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= 1

2 ×BD×AL+

1

2×BD×CM

= 1

2×BD(AL+CM)

= 1

2×64(16.8+13.2)

=32×30

=960 cm2

Question 28:

The area of a trapezium is 475 cm2 and its height is 19 cm. Find the

lengths of its two parallel sides if one side is 4 cm greater than the other.

ANSWER:

In the given figure, ABCD is a trapezium with parallel sides AB and CD.

Let the length of CD be x.

Then, the length of AB be x + 4.

Area of trapezium = 1

2 ×sum of parallel sides×height

⇒475 = 1

2 ×(x+x+4)×19

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⇒475×2 = 19(2x+4)

⇒950 =38x+76

⇒ 38x =950−76

⇒38x =874

⇒x = 874

38

⇒x =23

∴ The length of CD is 23 cm and the length of AB is 27 cm.

Hence, the lengths of two parallel sides is 23 cm and 27 cm.

Question 29:

In the given figure, a ∆ABC has been given in which AB = 7.5 cm, AC =

6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same

area as that of ∆ABC is constructed. Find the height DL of the

parallelogram.

ANSWER:

In ∆ABC,

The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.

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𝑠 = 𝑎+𝑏+𝐶

2

= 7.5+6.5+7

2

= 21

2 = 10.5cm



= √10.5(10.5 − 7.5)(10.5 − 6.5)(10.5 − 7)

= √10.5(3)(4)(3.5)

= 21𝑐𝑚2 ..(2)

Now,

Area of parallelogram DBCE = Area of ∆ABC

= 21 cm2

Also,

Area of parallelogram DBCE = base × height

⇒21=BC×DL

⇒21=7×DL

⇒DL= 21

7

=3 cm

Hence, the height DL of the parallelogram is 3 cm.

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Question 30:

A field is in the shape of a trapezium having parallel sides 90 m and 30

m. These sides meet the third side at right angles. The length of the

fourth side is 100 m. If it costs Rs 5 to plough 1 m2 of the field, find the

total cost of ploughing the field.

ANSWER:

In the given figure, ABCD is a trapezium having parallel sides 90 m and

30 m.

Draw DE perpendicular to AB, such that DE = BC.

In right angled ∆ADE,

AD2 = AE2 + ED2 (Pythagoras Theorem)

⇒ 1002 = (90 − 30)2 + ED2

⇒ 10000 = 3600 + ED2

⇒ ED2 = 10000 − 3600

⇒ ED2 = 6400

⇒ ED = 80 m

Thus, the height of the trapezium = 80 m ...(1)

Now,

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2×sum of parallel sides×height

= 1

2 (90+30)×8012×90+30×80

= 4800 m2

The cost to plough per m2 = Rs 5

The cost to plough 4800 m2 = Rs 5 × 4800

= Rs 24000

Hence, the total cost of ploughing the field is Rs 24000.

Question 31:

A rectangular plot is given for constructing a house, having a

measurement of 40 m long and 15 m in the front. According to the laws,

a minimum of 3-m-wide space should be left in the front and back each

and 2 m wide space on each of the other sides. Find the largest area

where house can be constructed.

ANSWER:

Let ABCD be a rectangular plot is given for constructing a house, having

a measurement of 40 m long and 15 m in the front.

According to the laws, the length of the inner rectangle = 40 − 3 − 3 =

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34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.

∴ Area of the inner rectangle PQRS = Length × Breath

= 34 × 11

= 374 m2

Hence, the largest area where house can be constructed is 374 m2.

Question 32:

A rhombus-shaped sheet with perimeter 40 cm and one diagonal 12 cm,

is painted on both sides at the rate of Rs 5 per cm2. Find the cost of

painting.

ANSWER:

Let the sides of rhombus be of length x cm.

Perimeter of rhombus = 4x

⇒ 40 = 4x

⇒ x = 10 cm

Now,

In ∆ABC,

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s= 𝑎+𝑏+𝐶

2

= 10+10+12

2

= 32

2 = 16cm.


Area of ΔABC= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √16(16 − 10)(16 − 10)(16 − 12)

= √16(6)(6)(4)

= 48𝑐𝑚2 ..(1)

In ∆ADC,



𝑠 = 𝑎 + 𝑏 + 𝑐

2

= 32

2 = 16cm.

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Area of ΔADC= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √16(16 − 10)(16 − 10)(16 − 12)

=√16(6)(6)(4)

=48𝑐𝑚2 ..(2)

∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC

= 48 + 48

= 96 cm2

The cost to paint per cm2 = Rs 5

The cost to paint 96 cm2 = Rs 5 × 96

= Rs 480

The cost to paint both sides of the sheet = Rs 2 × 480

= Rs 960

Hence, the total cost of painting is Rs 960.

Question 33:

The difference between the semi perimeter and the sides of a ∆ABC are

8 cm, 7 cm and 5 cm respectively. Find the area of the triangle.

ANSWER:

Let the semi-perimeter of the triangle be s.

Let the sides of the triangle be a, b and c.

Given: s − a = 8, s − b = 7 and s − c = 5 ....(1)

Adding all three equations, we get

3s − (a + b + c) = 8 + 7 + 5

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⇒ 3s − (a + b + c) =

⇒ 3s − 2s = 20

(∵ 𝑠 = 𝑎 + 𝑏 + 𝑐

2)

⇒ s = 20 cm ...(2)



= √20(8)(7)(5) from (1) and (2)

= 20√14 𝑐𝑚2

Hence, the area of the triangle is 20√14 𝑐𝑚2.

Question 34:

A floral design on a floor is made up of 16 tiles, each triangular in shape

having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles

at Re 1 per sq. cm.

ANSWER:

Area of one triangular-shaped tile can be found in the following manner:


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s = 𝑎+𝑏+𝑐

2

= 16+12+20

2

= 24cm.



= √24(24 − 16)(24 − 12)(24 − 20)

= √24 × 8 × 12 × 4

= √6 × 4 × 4 × 4 × 4 × 6

= 96𝑐𝑚2

Now,

Area of 16 triangular-shaped tiles = 16×96=1536𝑐𝑚2

Cost of polishing tiles of area 1 cm2 = Rs 1

Cost of polishing tiles of area 1536 cm2 = 1×1536 =Rs 1536

Question 35:

An umbrella is made by stitching 12 triangular pieces of cloth, each

measuring (50 cm × 20 cm × 50 cm). Find the area of the cloth used in

it.

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ANSWER:

We know that the triangle is an isosceles triangle.

Thus, we can find out the area of one triangular piece of cloth.


4√4𝑎2 − 𝑏2

= 20

4 × √4(50)2 − (20)2 (a = 50 cm and b = 20 cm)

= 5× √20000 − 400

= 5 × √9600

= 5 × 40√6

= 200√6

= 490𝑐𝑚2

Now,

Area of 1 triangular piece of cloth = 490 cm2

Area of 12 triangular pieces of cloth = 12×490 =5880 cm2

Question 36: In the given figure, ABCD is a square with diagonal 44

cm. How much paper of each shade is needed to make a kite given in the

figure?

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ANSWER:

In the given figure, ABCD is a square with diagonal 44 cm.

∴ AB = BC = CD = DA. ....(1)


AC2 = AB2 + BC2 (Pythagoras Theorem)

⇒ 442 = 2AB2

⇒ 1936 = 2AB2

⇒ AB2 = 1936

2

⇒ AB2 = 968

⇒ AB = 22√2 cm ...(2)

∴ Sides of square = AB = BC = CD = DA = 22√2 cm

Area of square ABCD = (side)2

= ( 22√2)2

= 968 cm2 ...(3)

Area of red portion = 968

4 = 242 cm2

Area of yellow portion = 968

2 = 484 cm2

Area of green portion = 968

4= 242 cm2

Now, in ∆AEF,



s= 20+24+12

2 =

54

2 = 27cm

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= √27(27 − 20)(27 − 20)(27 − 14)

= √27(7)(7)(13)

= 21√39

= 131.04 𝑐𝑚2 ….(4)

Total area of the green portion = 242 + 131.04 = 373.04 cm2

Hence, the paper required of each shade to make a kite is red paper 242

cm2, yellow paper 484 cm2 and green paper 373.04 cm2.

Question 37:

A rectangular lawn, 75 m by 60 m, has two roads, each road 4 m wide,

running through the middle of the lawn, one parallel to length and the

other parallel to breadth, as shown in the figure. Find the cost of

gravelling the roads at Rs 50 per m2.

ANSWER:

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Area of rectangle ABCD = Length × Breath

= 75 × 4

= 300 m2

Area of rectangle PQRS = Length × Breath

= 60 × 4

= 240 m2

Area of square EFGH = (side)2

= (4)2

= 16 m2

∴ Area of the footpath = Area of rectangle ABCD + Area of

rectangle PQRS − Area of square EFGH

= 300 + 240 − 16

= 524 m2

The cost of gravelling the road per m2 = Rs 50

The cost of gravelling the roads 524 m2 = Rs 50 × 524

= Rs 26200

Hence, the total cost of gravelling the roads at Rs 50 per m2 is Rs 26200.

Question 38:

The shape of the cross section of a canal is a trapezium. If the canal is 10

m wide at the top, 6 m wide at the bottom and the area of its cross

section is 640 m2, find the depth of the canal.

ANSWER:

The top and the bottom of the canal are parallel to each other.

Let the height of the trapezium be h.

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⇒ 640 = 1

2 ×(10+6)×h

⇒ 640 = 8×h

⇒ h = 640

8

⇒ h = 80 m

Hence, the depth of the canal is 80 m.

Question 39:

Find the area of a trapezium whose parallel sides are 11 m and 25 m

long, and the nonparallel sides are 15 m and 13 m long.

ANSWER:

In the given figure, ABCD is the trapezium.

Draw a line BE parallel to AD.

In ∆BCE,

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


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𝑠 = 𝑎 + 𝑏 + 𝑐

2

= 42

2 = 21m



= √21(21 − 15)(21 − 13)(21 − 14)

= √1(6)(8)(7)

= 84𝑚2 ….(1)

Also,

Area of ∆BCE

= 1

2 ×Base×Height

⇒84= 1

2 ×14×Height

⇒84=7×Height

⇒Height= 84

7

⇒Height=12 m

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,

Area of the parallelogram ABED = Base × Height

= 11 × 12

= 132 m2 ...(2)

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∴ Area of the trapezium = Area of the parallelogram ABED + Area of

the triangle BCE

= 132 + 84

= 216 m2

Hence, the area of a trapezium is 216 m2.

Question 40:

The difference between the lengths of the parallel sides of a trapezium is

8 cm, the perpendicular distance between these sides is 24 cm and the

area of the trapezium is 312 cm2. Find the length of each of the parallel

sides.

ANSWER:

Let the length of the parallel sides be x and x − 8.

The height of the trapezium = 24 cm



⇒ 312 = 1

2 ×(x+x−8)×24

⇒ 312 = 1

2 (2x − 8)

⇒ 2x − 8 = 312

12

⇒ 2x − 8 = 26

⇒ 2x = 26 + 8

⇒ 2x = 34

⇒ x = 17 cm

Hence, the lengths of the parallel sides are 17 cm and 9 cm.

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Question 41:

A parallelogram and a rhombus are equal in area. The diagonals of the

rhombus measure 120 m and 44 m. If one of the sides of the

parallelogram measures 66 m, find its corresponding altitude.

ANSWER:

Diagonals d1 and d2 of the rhombus measure 120 m and 44 m,

respectively.

Base of the parallelogram = 66 m

Now,

Area of the rhombus = Area of the parallelogram

⇒ 1

2 ×d1×d2=Base × Height

⇒ 1

2×120×44=66×Height

⇒ 60×44=66×Height

⇒2640 = 66×Height

⇒ Height = 2640

44

⇒ Height=40 m

Hence, the measure of the altitude of the parallelogram is 40 m.

Question 42:

A parallelogram and a square have the same area. If the sides of the

square measure 40 m and altitude of the parallelogram measures 25 m,

find the length of the corresponding base of the parallelogram.

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ANSWER:

It is given that,

Sides of the square = 40 m

Altitude of the parallelogram = 25 m

Now,

Area of the parallelogram = Area of the square

⇒Base × Height=(side)2

⇒Base×25=(40)2

⇒Base×25=1600

⇒Base= 1600

25

⇒Base=64 m

Hence, the length of the corresponding base of the parallelogram is 64

m.

Question 43:

Find the area of a rhombus one side of which measures 20 cm and one of

whose diagonals is 24 cm.

ANSWER:

It is given that,

The sides of rhombus = 20 cm.

One of the diagonal = 24 cm.

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In ∆ABC,



s = 20+20+24

2

s = 64

2

s= 32



= √32(32 − 20)(32 − 20)(32 − 24)

= √32(12)(12)(8)

= 192𝑐𝑚2 ..(1)

In ∆ACD,



s = 20+20+24

2

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s = 64

2

s= 32


Area of ΔACD= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √32(32 − 20)(32 − 20)(32 − 24)

= √32(12)(12)(8)

= 192𝑐𝑚2 ..(2)

∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD

= 192 + 192

= 384 cm2

Hence, the area of a rhombus is 384 cm2.

Question 44:

The area of a rhombus is 480 cm2, and one of its diagonals measures 48

cm. Find (i) the length of the other diagonal, (ii) the length of each of its

sides, and (iii) its perimeter.

ANSWER:

It is given that,

Area of rhombus = 480 cm2.

One of the diagonal = 48 cm.

(i) Area of the rhombus = 1

2 ×d1×d2

⇒480= 1

2×48×d2

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⇒480=24×d2

⇒d2= 480

24

⇒d2=20 cm

Hence, the length of the other diagonal is 20 cm.

(ii) We know that the diagonals of the rhombus bisect each other at right

angles.

In right angled ∆ABO,

AB2 = AO2 + OB2 (Pythagoras Theorem)

⇒ AB2 = 242 + 102

⇒ AB2 = 576 + 100

⇒ AB2 = 676

⇒ AB = 26 cm

Hence, the length of each of the sides of the rhombus is 26 cm.

(iii) Perimeter of the rhombus = 4 × side

= 4 × 26

= 104 cm

Hence, the perimeter of the rhombus is 104 cm.

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CCE Test Paper – 14

Question 1:

In a ∆ABC it is given that base = 1

2 cm and height = 5 cm. Its area is

(a) 60 cm2

(b) 30 cm2

(c) 15√3 𝑐𝑚2

(d) 45 cm2

ANSWER:

(b) 30 cm2


2 ×Base×Height

Area of ΔABC= 1

2 ×12×5=30 cm2

Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The

area of the triangle is

(a) 96 cm2

(b) 120 cm2

(c) 144 cm2

(d) 160 cm2

ANSWER:

(a) 96 cm2


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s = 𝑎+𝑏+𝑐

2

=20+16+12

2

=24 cm



= √24(24 − 20)(24 − 16)(24 − 12)

= √24(4)(8)(12)

= √6 × 4 × 4 × 4 × 4 × 6

= 6 × 4 × 4

= 96𝑐𝑚2 …..(2)

Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the

triangle is

(a) 8√3 𝑐𝑚2

(b) 16√3 𝑐𝑚2

(c) 32 √3 𝑐𝑚2

(d) 48 cm2

ANSWER:

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(b) 6√3 𝑐𝑚2


4 × (𝑠𝑖𝑑𝑒)2

= √3

4 × (8)2

= √3

4 × 64

= 64√3 𝑐𝑚2

Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides

measures 6 cm. The area of the triangle is

(a) 16√5 𝑐𝑚2

(b) 8√5 𝑐𝑚2

(c) 16√3 𝑐𝑚2

(d) 8√3 𝑐𝑚2

ANSWER:

(b) 8√5 𝑐𝑚2


4√4𝑎2 − 𝑏2

Here,

a= 6 cm and b=8 cmThus, we have:

= 8

4 × √4(6)2 − 82

= 8

4 × √144 − 64

= 8

4 × √80

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= 8

4× 4√5

=8√5 𝑐𝑚2

Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5

cm. The height of the triangle is

(a) 8 cm

(b) √30 cm

(c) 4 cm

(d) √11 cm

ANSWER:

(c) 4 cm

Height of isosceles triangle = 1

2√4𝑎2 − 𝑏2

= 1

2√4(5)2 − 62 (a=5 cm and b=6 cm)

= 1

2× √100 − 36

= 1

2× √64

= 1

2× 8

= 4 cm

Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long.

Its area is

(a) 5√10 𝑐𝑚2

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(b) 50 cm2

(c) 10√3 𝑐𝑚2

(d) 75 cm2

ANSWER:

(b) 50 cm2

Here, the base and height of the triangle are 10 cm and 10 cm,

respectively.

Thus, we have:


2×Base×Height

= 1

2×10×10

=50 cm2

Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the

triangle is

(a) 10 √3 cm

(b) 5√3 cm

(c) 10√2 cm

(d) 5 cm

ANSWER:

(b) 5√3 cm

Height of equilateral triangle

= √3

2 ×Side

= √3

2 ×10

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=5√3 cm

Question 8:

The height of an equilateral triangle is 6 cm. Its area is

(a) 12√3 𝑐𝑚2

(b) 6√3 𝑐𝑚2

(c) 12√2 𝑐𝑚2

(d) 18 cm2

ANSWER:

(a) 12√3 𝑐𝑚2

Height of equilateral triangle = √3

2 ×Side

⇒6= √3

2 ×Side

⇒ Side= 12

√3×

√3

√3 =

12

3× √3 = 4√3 𝑐𝑚.

Now,Area of equilateral triangle

= √3

4 ×(Side)2

= √3

4× 48

= 12√3 𝑐𝑚2

Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32

m respectively. The area of the triangle is

(a) 480 m2

(b) 320 m2

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(c) 384 m2

(d) 360 m2

ANSWER:

(c) 384 m2

Let: a=40 m, b = 24 m and c=32 ms=

s = 𝑎+𝑏+𝑐

2

= 40+24+32

2

= 48m



= √48(48 − 40)(48 − 24)(48 − 32)

= √48 × 8 × 24 × 16

= √24 × 2 × 8 × 24 × 8 × 2

= 24 × 8 × 2

= 384𝑚2

Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150

cm. The area of the triangle is

(a) 375 cm2

(b) 750 cm2

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(c) 250 cm2

(d) 500 cm2

ANSWER:

(b) 750 cm2

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.


or, 150 = 5x + 12x + 13x

or, 30x = 150

or, x = 5

Thus, the sides of the triangle are 5×5 cm, 12×5 cm and 13×5 cm, i.e.,

25 cm, 60 cm and 65 cm.

Now,


s = 150

2 =75 cm



= √75(75 − 25)(75 − 60)(75 − 65)

= √75 × 50 × 15 × 10

= √15 × 5 × 5 × 10 × 15 × 10

= 15 × 5 × 10

= 750 𝑐𝑚2

Question 11:

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The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm

respectively. The length of the altitude of the triangle corresponding to

the smallest side is

(a) 24 cm

(b) 18 cm

(c) 30 cm

(d) 12 cm

ANSWER:

(a) 24 cm

Let:a=30 cm, b = 24 cm and c=18 cm

s = 𝑎+𝑏+𝑐

2

s= 30+24+18

2

s =36cm.

On applying Heron's formula, we get:

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

= √36(36 − 30)(36 − 24)(36 − 18)

= √36 × 6 × 12 × 18

= √12 × 3 × 12 × 6 × 3

= 12 × 3 × 6

= 216 𝑐𝑚2

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The smallest side is 18 cm.

Hence, the altitude of the triangle corresponding to 18 cm is given by:


⇒ 1

2×Base×Height

= 216×2

18 = 24cm

Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm2. The

perimeter of the triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cm

ANSWER:

(b) 36 cm

Let △PQR be an isosceles triangle and PX⊥QR.

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Now,

Area of triangle =48 cm2

⇒ 1

2 ×QR×PX

= 48

⇒h =96

16=6 cm

Also, QX = 1

2 ×24 = 12 cm

and PX = 12 cm

PQ= √𝑄𝑋2 + 𝑃𝑋2

𝑎 = √82 + 62

𝑎 = √64 + 36

𝑎 = √100

𝑎 = 10𝑐𝑚

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

Question 13:

The area of an equilateral triangle is 36√3 cm2. Its perimeter is

(a) 36 cm

(b) 12 √3 cm

(c) 24 cm

(d) 30 cm

ANSWER:

(a) 36 cm


4 × (𝑆𝑖𝑑𝑒)2

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= 36√3’

⇒ (𝑆𝑖𝑑𝑒)2 = 144

⇒ Side = 12cm

Now,

Perimeter = 3 × Side = 3 × 12 = 36 cm

Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base

is 24 cm. The area of the triangle is

(a) 156 cm2

(b) 78 cm2

(c) 60 cm2

(d) 120 cm2

ANSWER:

(c) 60 cm2


4√4𝑎2 − 𝑏2

Here, a= 13 cm and b=24 cm

Thus, we have:

= 24

4√4(13)2 − 242

= 6 × √676 − 576

= 6 × √100

=6 × 10

= 60 𝑐𝑚2

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Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm

long. The area of the triangle is

(a) 168 cm2

(b) 252 cm2

(c) 336 cm2

(d) 504 cm2

ANSWER:

(c) 336 cm2

Let △PQR be a right-angled triangle and PQ⊥QR.

Now,

PQ= √𝑃𝑅2 − 𝑄𝑅2

= √502 − 482

= √2500 − 2304

= √196

= 14cm

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∴Area of triangle = 1

2 × 𝑄𝑅 × 𝑃𝑄

= 1

2× 48 × 14

= 336𝑐𝑚2

Question 16:

The area of an equilateral triangle is 81√3 𝑐𝑚2 . Its height is

(a) 9√3cm

(b) 6√3cm

(c) 18√3 cm

(d) 9 cm

ANSWER:

(a) 9√3 cm

Area of equilateral triangle = 81√3𝑐𝑚2

⇒ √3

4 × (𝑠𝑖𝑑𝑒)2=81√3

⇒(𝑠𝑖𝑑𝑒)2=81×4

⇒(𝑠𝑖𝑑𝑒)2= 324

⇒Side=18 cm

Now, Height = √3

2 ×Side

=√3

2 ×18

=9√3 cm.

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Chapter 14 Areas of triangles and Quadrilaterals - Praadis Education

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