# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 15 - Perimeter and Areas of Plane Figures

## Chapter -

## Chapter 17 - Perimeter and Areas of Plane Figures MCQ

## Chapter 17 - Perimeter and Areas of Plane Figures FA

## Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15A

_{}

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

_{}

(i)Area of triangle = _{}

_{}

(ii)Let base = 42 cm and corresponding height = h cm

Then area of triangle = _{}

_{}

Hence, the height corresponding to the longest side = 16 cm

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

Then,2s = (18 + 24 + 30) cm = 72 cm

s = 36 cm

(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(i)Area of triangle = _{}

_{}

(ii)Let base = 18 cm and altitude = x cm

Then, area of triangle = _{}

_{}

Hence, altitude corresponding to the smallest side = 24 cm

On dividing 150 m in the ratio 5 : 12 : 13, we get

Length of one side = _{}

Length of the second side =

Length of third side = _{}

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

_{}

(s a) = 50 cm, (s b) = 15 cm, and (s c) = 10 cm

_{}

Hence, area of the triangle = 750 m^{2}

On dividing 540 m in ratio 25 : 17 : 12, we get

Length of one side = _{}

Length of second side = _{}

Length of third side = =120 m

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

_{}

Then, (s a) = 29 m, (s b) = 100 m, and (s c) = 150m

_{}

The cost of ploughing 100 _{}area is = Rs. 18. 80

The cost of ploughing 1 _{}is = _{}

The cost of ploughing 9000 _{}area = _{}

= Rs. 1692

Hence, cost of ploughing = Rs 1692.

Let the length of one side be x cm

Then the length of other side = {40 (17 + x)} cm = (23 - x) cm

Hypotenuse = 17 cm

Applying Pythagoras theorem, we get

_{}

_{}

Hence, area of the triangle = 60 cm^{2}

Let the sides containing the right - angle be x cm and (x - 7) cm

_{}

One side = 15 cm and other = (15 - 7) cm = 8 cm

_{}

_{}perimeter of triangle (15 + 8 + 17) cm = 40 cm

Let the sides containing the right angle be x and (x 2) cm

_{}

_{}

One side = 8 cm, and other (8 2) cm = 6 cm

_{}

= 10 cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Let each side of the equilateral triangle be a cm

_{}

Let each side of the equilateral triangle be a cm

_{}

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm

Let each side of the equilateral triangle be a cm

_{}area of equilateral triangle =_{}

_{}

Height of equilateral triangle

_{}

Base of right angled triangle = 48 cm

Height of the right angled triangle =_{}

_{}

Let the hypotenuse of right - angle triangle = 6.5 m

Base = 6 cm

_{}

_{}

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm^{2}

The circumcentre of a right - triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)

= (2 × 8) cm = 16 cm

Base = 16 cm, height = 6 cm

Area of right angled triangle

_{}

Hence, area of the triangle= 48 cm^{2}

Let each equal side be a cm in length.

Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Let each equal side be a cm and base = 80 cm

_{}

_{}

_{}perimeter of triangle = (2a + b) cm

= (2 41 + 80) cm

= (82 + 80) cm = 162 cm

Hence, perimeter of the triangle = 162 cm

Let the height be h cm, then a= (h + 2) cm and b = 12 cm

_{}

Squaring both sides,

_{}

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

_{}

Hence, area of the triangle = 48 cm^{2}.

Area of shaded region = Area of _{}ABC – Area of _{}DBC

First we find area of _{}ABC

_{}

Second we find area of _{DBC which }is right angled

_{}

Area of shaded region = Area of _{}ABC – Area of _{}DBC

= (43.30 - 24) _{}= 19. 30 _{}

Area of shaded region = 19.3 _{}

Let _{}ABC is a isosceles triangle. Let AC, BC be the equal sides

Then AC = BC = 10cm. Let AB be the base of _{}ABC right angle at C.

_{}

Area of right isosceles triangle ABC

_{}

Hence, area = 50 cm^{2} and perimeter = 34.14 cm

## Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15B

Area of floor = Length Breadth

_{}

Area of carpet = Length Breadth

_{= }

Number of carpets = _{}

= 216

Hence the number of carpet pieces required = 216

Area of verandah = (36 × 15) _{}= 540 _{}

Area of stone = (0.6 × 0.5) _{}[10 dm = 1 m]

Number of stones required =

Hence, 1800 stones are required to pave the verandah.

Perimeter of rectangle = 2(l + b)

_{}2(l + b) = 56 Þ l + b = 28 cm

b = (28 l) cm

Area of rectangle = 192_{}

l (28 l) = 192

28l - _{}= 192

_{}- 28l + 192 = 0

_{}- 16l 12l + 192 = 0

l(l 16) 12(l 16) = 0

(l 16) (l 12) = 0

l = 16 or l = 12

Therefore, length = 16 cm and breadth = 12 cm

Length of the park = 35 m

Breadth of the park = 18 m

Area of the park = (35 18) _{=} 630 _{}

Length of the park with grass =(35 5) = 30 m

Breadth of the park with grass = (18- 5) m = 13 m

Area of park with grass = (30 13) _{}= 390 _{}

Area of path without grass = Area of the whole park area of park with grass

= 630 _{}- 390 _{}= 240 _{}

Hence, area of the park to be laid with grass = 240 m^{2}

Length of the plot = 125 m

Breadth of the plot = 78 m

Area of plot ABCD = (125 78) _{}= 9750 _{}

Length of the plot including the path= (125 + 3 + 3) m = 131 m

Breadth of the plot including the path = (78 + 3 + 3) m = 84 m

Area of plot PQRS including the path

= (131 84) _{}= 11004 _{}

Area of path = Area of plot PQRS Area of plot ABCD

= (11004 9750) _{}

= 1254 _{}

Cost of gravelling = Rs 75 per m^{2}

_{}Cost of gravelling the whole path = Rs. (1254 75) = Rs. 94050

Hence, cost of gravelling the path = Rs 94050

Area of rectangular field including the foot path = (54 35) _{}

Let the width of the path be x m

Then, area of rectangle plot excluding the path = (54 2x) (35 2x)

Area of path = (54 35) (54 2x) (35 2x)

(54 35) (54 2x) (35 2x) = 420

1890 1890 + 108x + 70x - 4_{} = 420

178x - 4_{} = 420

4_{} - 178x + 420 = 0

2_{} - 89x + 210 = 0

2_{} - 84x 5x + 210 = 0

2x(x 42) 5(x 42) = 0

(x 42) (2x 5) = 0

_{}

Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45_{}

Length of park excluding the path = (9x 7) m

Breadth of the park excluding the path = (5x 7) m

Area of the park excluding the path = (9x 7)(5x 7) _{}

_{}Area of the path = _{}

_{}

_{}(98x 49) = 1911

98x = 1911 + 49

_{}

Length = 9x = 9 20 = 180 m

Breadth = 5x = 5 20 = 100 m

Hence, length = 180 m and breadth = 100 m

Let the width of the carpet = x meter

Area of floor ABCD = (8 5) _{}

Area of floor PQRS without border

= (8 2x)(5 2x)

= 40 16x 10x + _{}

= 40 26x + _{}

Area of border = Area of floor ABCD Area of floor PQRS

= [40 (40 26x + _{})] _{}

=[40 40 + 26x - _{}] _{}

= (26x - _{})_{}

_{}

Area of the square = _{}

Let diagonal of square be x

_{}

Length of diagonal = 16 cm

Side of square = _{}

Perimeter of square = [4 side] sq. units

=[ 4 11.31] cm = 45.24 cm

Let d meter be the length of diagonal

Area of square field = _{}

_{ }

Time taken to cross the field along the diagonal

_{}

Hence, man will take 6 min to cross the field diagonally.

Rs. 14 is the cost of fencing a length = 1m

Rs. 28000 is the cost of fencing the length= _{}

Perimeter = 4 side = 2000

_{}side = 500 m

Area of a square = _{}

= 250000 _{}

Cost of mowing the lawn =_{}

Area of quad. ABCD = Area of ABD + Area ofDBC

For area of ABD

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

_{}

For area of DBC

a = 29 cm, b = 21 cm, c = 20 cm

_{}

Area of quad. ABCD = Area of _{}ABC + Area of _{}ACD

_{}

Now, we find area of a _{}ACD

_{}

_{}

Area of quad. ABCD = Area of _{}ABC + Area of _{}ACD

_{}

Perimeter of quad. ABCD = AB + BC + CD + AD

=(17 + 8 + 12 + 9) cm

= 46 cm

_{}Perimeter of quad. ABCD = 46 cm

ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm

By Pythagoras theorem

_{}

For area of equilateral _{}DBC, we have

a = 26 cm

_{}

Area of quad. ABCD = Area of _{}ABD + Area of _{}DBC

= (120 + 292.37) _{}= 412.37 _{}

Perimeter ABCD = AD + AB + BC + CD

= 24 cm + 10 cm + 26 cm + 26 cm

= 86 cm

Area of the ||gm = (base height) sq. unit

= (25 16.8) _{}

Longer side = 32 cm, shorter side = 24 cm

Distance between longer sides = 17.4 cm

Let the distance between the shorter sides be x cm

Area of ||gm = (longer side distance between longer sides)

= (shorter side distance between the short sides)

_{}

_{}distance between the shorter side = 23.2 cm

_{}

Area ofparallelogram = 2 area of DABC

Opposite sides of parallelogram are equal

AD = BC = 20 cm

And AB = DC = 34 cm

In _{}ABC we have

a = AC = 42 cm

b = AB = 34 cm

c = BC = 20 cm

_{}

Then, (s a) = 6 cm, (s b) = 14 cm and (s c) = 28 cm

_{}

_{}

_{}

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

OA = OC = 15 cm,

And OB = OD = 8 cm

And _{}AOB = 90

_{}By Pythagoras theorem, we have

_{}

(i)Perimeter of rhombus = 4 side

_{}4 side = 60 cm

_{}

By Pythagoras theorem

_{}

OB = 12 cm

OB = OD = 12 cm

_{}BD = OB + OD = 12 cm + 12 cm = 24 cm

Length of second diagonal is 24 cm

(ii) Area of rhombus = _{}

_{}

(i)Area of rhombus = 480 _{}

One of its diagonals = 48 cm

Let the second diagonal =x cm

_{}

Hence the length of second diagonal 20 cm

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

_{}AC = 48, BD = 20 cm

_{}OA = OC = 24 cm and OB = OD = 10 cm

By Pythagoras theorem , we have

_{}

(iii)_{}Perimeter of the rhombus = (4 26) cm = 104 cm

Areaof cross section = _{}

_{}

Let ABCD be a given trapezium in which

AB = 25, CD = 11

BC = 15, AD = 13

Draw CE || AD

In ||gm ADCE, AD || CE and AE || CD

AE = CD = 11 cm,

And BE = AB BE

= 25 11 = 14 cm

In BEC,_{}

Area of BEC = _{}

Let height of _{}BEC is h

Area of _{}BEC =

From (1) and (2), we get

7h = 84 _{}h = 12 m

Area of trapezium ABCD

_{}

### Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability### R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects

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