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# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 14 - Areas of Triangles and Quadrilaterals

Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 14 - Areas of Triangles and Quadrilaterals.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 14 - Areas of Triangles and Quadrilaterals Page/Excercise 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 ## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 14 - Areas of Triangles and Quadrilaterals Page/Excercise 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

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