R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 14 - Areas of Triangles and Quadrilaterals

Page / Exercise

Chapter 14 - Areas of Triangles and Quadrilaterals MCQ

Solution 1

  

Solution 2

  

Solution 3

  

Solution 4

  

Solution 5

  

Solution 6

  

Solution 7

  

Solution 8

  

Solution 9

  

Solution 10

  

Solution 11

  

Solution 12

 

  

Solution 13

  

Solution 14

  

Solution 15

  

Solution 16

  

Chapter 14 - Areas of Triangles and Quadrilaterals Ex. 14

Solution 1

Solution 2

Solution 3


Solution 4

Solution 5


Solution 6

Solution 7

It is given that the sides a, b, c of the triangle are in the ratio 25 : 17 : 12,

i.e. a : b : c = 25 : 17 : 12

a = 25x, b = 17x and c = 12x

Given, perimeter = 540 m

25x + 17x + 12x = 540

54x = 540

x = 10

So, the sides of the triangle are

a = 25x = 25(10) = 250 m

b = 17x = 17(10) = 170 m

c = 12x = 12(10) = 120 m

Cost of ploughing the field = Rs. 5/m2

Cost of ploughing 9000 m2 = Rs. (5 × 9000) = Rs. 45,000

Solution 8

Solution 9

Solution 10

Solution 11

It is given that the ratio of equal side to its base is 3 : 2.

Ratio of sides of isosceles triangle = 3 : 3 : 2

i.e. a : b : c = 3 : 3 : 2

a = 3x, b = 3x and c = 2x

Given, perimeter = 32 cm

3x + 3x + 2x = 32

8x = 32

x = 4

So, the sides of the triangle are

a = 3x = 3(4) = 12 cm

b = 3x = 3(4) = 12 cm

c = 2x = 2(4) = 8 cm

Solution 12

Let the three sides of a triangle be a, b and c respectively such that c is the smallest side.

Then, we have

a = c + 4

And, b = 2c - 6

Given, perimeter = 50 cm

a + b + c = 50

(c + 4) + (2c - 6) + c = 50

4c - 2 = 50

4c = 52

c = 13

So, the sides of the triangle are

a = c + 4 = 13 + 4 = 17 cm

b = 2c - 6 = 2(13) - 6 = 20 cm

c = 13 cm

  

Solution 13

Three sides of a wall are 13 m, 14 m and 15 m respectively.

i.e.

a = 13 m, b = 14 m and c = 15 m

Rent for a year = Rs. 2000/m2

Rent for 6 months = Rs. 1000/m2

Thus, total rent paid for 6 months = Rs. (1000 × 84) = Rs. 84,000 

Solution 14

Solution 15

Solution 16

Solution 17

(i) Area of an equilateral triangle=

Where a is the side of the equilateral triangle

Solution 18

 


Solution 19

Solution 20

In right triangle ADB, by Pythagoras theorem,

AB2 = AD2 + BD2 = 122 + 162 = 144 + 256 = 400

AB = 20 cm

For ΔABC, 

Thus, area of shaded region

= Area of ΔABC - Area of ΔABD

= (480 - 96) cm2

= 384 cm2

Solution 21

  

Let ABCD be the given quadrilateral such that ABC = 90° and AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm.

In ΔABC, by Pythagoras theorem,

AC2 = AB2 + BC2 = 62 + 82 = 36 + 64 = 100

AC = 10 cm

In ΔACD, AC = 10 cm, CD = 12 cm and AD = 14 cm

Let a = 10 cm, b = 12 cm and c = 14 cm

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (24 + 58.8) cm2

= 82.8 cm2

Solution 22

  

In ΔABD, by Pythagoras theorem,

AB2 = AD2 - BD2 = 172 - 152 = 289 - 225 = 64

AB = 8 cm

Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 8 + 12 + 9 + 17

= 46 cm

In ΔBCD, BC = 12 cm, CD = 9 cm and BD = 15 cm

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

Thus, area of quadrilateral ABCD

= A(ΔABD) + A(ΔBCD)

= (60 + 54) cm2

= 114 cm2 

Solution 23

  

In ΔBAC, by Pythagoras theorem,

BC2 = AC2 + AB2 = 202 + 212 = 400 + 441 = 841

BC = 29 cm

Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 21 + 29 + 42 + 34

= 126 cm

In ΔACD, AC = 20 cm, CD = 42 cm and AD = 34 cm

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

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (210 + 336) cm2

= 546 cm2 

Solution 24

Perimeter of quad. ABCD = AB + BC + CD + DA = 10 + 26 + 26 + 24 = 86 cm

Solution 25

Solution 26

Solution 27

Solution 28

Let the smaller parallel side of trapezium = x cm

Then, larger parallel side = (x + 4) cm

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

Solution 29

  

In ΔABC, AB = 7.5 cm, BC = 7 cm and AC = 6.5 cm

Let a = 7.5 cm, b = 7 cm and c = 6.5 cm

  

Solution 30

  

Construction: Draw BT CD

In ΔBTC, by Pythagoras theorem,

BT2 = BC2 - CT2 = 1002 - 602 = 10000 - 3600 = 6400

BT = 80 m

AD = BT = 80 m

Cost of ploughing 1 m2 field = Rs. 5

Cost of ploughing 4800 m2 field = Rs. (5 × 4800) = Rs. 24,000

Solution 31

Length of rectangular plot = 40 m

Width of rectangular plot = 15 m

Keeping 3 m wide space in the front and back,

length of rectangular plot = 40 - 3 - 3 = 34 m

Keeping 2 m wide space on both the sides,

width of rectangular plot = 15 - 2 - 2 = 11 m

Thus, largest area where house can be constructed

= 34 m × 11 m

= 374 m2

Solution 32

  

Let ABCD be the rhombus-shaped sheet.

Perimeter = 40 cm

4 × Side = 40 cm

Side = 10 cm

AB = BC = CD = AD = 10 cm

Let diagonal AC = 12 cm

Since diagonals of a rhombus bisect each other at right angles,

AO = OC = 6 cm

In right ΔAOD, by Pythagoras theorem,

OD2 = AD2 - AO2 = 102 - 62 = 100 - 36 = 64

OD = 8 cm

BD = 2 × OD = 2 × 8 = 16 cm

Now,

Cost of painting = Rs. 5/cm2

Cost of painting rhombus on both sides = Rs. 5 × (96 + 96)

= Rs. 5 × 192

= Rs. 960

Solution 33

Let the sides of a triangle be a, b, c respectively and 's' be its semi-perimeter.

Then, we have

s - a = 8 cm

s - b = 7 cm

s - c = 5 cm

Now, (s - a) + (s - b) + (s - c) = 8 + 7 + 5

3s - (a + b + c) = 20

3s - 2s = 20

s = 20

Thus, we have

a = s - 8 = 20 - 8 = 12 cm

b = s - 7 = 20 - 7 = 13 cm

c = s - 5 = 20 - 5 = 15 cm

Solution 34

Solution 35

Solution 36

  In ΔAEF, AE = 20 cm, EF = 14 cm and AF = 20 cm

Let a = 20 cm, b = 14 cm and c = 20 cm

Solution 37

For road ABCD, i.e. for rectangle ABCD,

Length = 75 m

Breadth = 4 m

Area of road ABCD = Length × Breadth = 75 m × 4m = 300 m2

For road PQRS, i.e. for rectangle PQRS,

Length = 60 m

Breadth = 4 m

Area of road PQRS = Length × Breadth = 60 m × 4 m = 240 m2

 

For road EFGH, i.e. for square EFGH,

Side = 4 m

Area of road EFGH = (Side)2 = (4)2 = 16 m2

 

Total area of road for gravelling

= Area of road ABCD + Area of road PQRS - Area of road EFGH

= 300 + 240 - 16

= 524 m2

 

Cost of gravelling the road = Rs. 50 per m2

Cost of gravelling 524 m2 road = Rs. (50 × 524) = Rs. 26,200

Solution 38

Area of cross section = Area of trapezium = 640 m2

Length of top + Length of bottom

= sum of parallel sides

= 10 m + 6 m

= 16 m

 Thus, the depth of the canal is 80 m.

Solution 39

  

From C, draw CE DA.

Clearly, ADCE is a parallelogram having AD EC and AE DC such that AD = 13 m and D = 11 m.

AE = DC = 11 m and EC = AD = 13 m

BE = AB - AE = 25 - 11 = 14 m

Thus, in ΔBCE, we have

BC = 15 m, CE = 13 m and BE = 14 m

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

Solution 40

Let the smaller parallel side = x cm

Then, longer parallel side = (x + 8) cm

Height = 24 cm

Area of trapezium = 312 cm2

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

Solution 41

Area of parallelogram = Area of rhombus

Solution 42

Area of parallelogram = Area of square

  

Solution 43

Let ABCD be a rhombus and let diagonals AC and BD intersect each other at point O.

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

OD2 = AD2 - OA2 = 202 - 122 = 400 - 144 = 256

OD = 16 cm

BD = 2(OD) = 2(16) = 32 cm

Solution 44

(i) Area of a rhombus = 480 cm2

  

(ii) Let diagonal AC = 48 cm and diagonal BD = 20 cm

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

AD2 = OA2 + OD2 = 242 + 102 = 576 + 100 = 676

AD = 26 cm

AD = BC = CD = AD = 26 cm

Thus, the length of each side of rhombus is 26 cm.

 

(iii) Perimeter of a rhombus = 4 × side = 4 × 26 = 104 cm