## Chapter 17 – Perimeter and Areas of Plane Figures Exercise Ex. 17A

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14. 5 cm.

_{}

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.

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

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 of the triangle.

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

The sides of a triangle are in the ratio 5 : 12 : 13, and it perimeter is 150 m. find the area of the triangle.

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}

The perimeter of a triangular field is 540 m, and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs. 18. 80 per _{}

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.

The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

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}

The difference between the sides at right angles in a right – angled triangle is 7 cm. The area of the triangle is_{}. Find its perimeter.

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

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is _{}, find the perimeter of the triangle.

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

Each side of an equilateral triangle is

10 cm. Find

(i) the area of a triangle and (ii) the height of the

triangle.

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take _{}

Let each side of the equilateral triangle be a cm

_{}

If the area of an equilateral triangle is _{}find its perimeter.

Let each side of the equilateral triangle be a cm

_{}

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

If the area of an equilateral triangle is _{}find its height.

Let each side of the equilateral triangle be a cm

_{}area of equilateral triangle =_{}

_{}

Height of equilateral triangle

_{}

The base of a right – angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Base of right angled triangle = 48 cm

Height of the right angled triangle =_{}

_{}

The hypotenuse of a right – angled triangle measure 6.5 m and its base measures 6 m. Find the length of perpendicular, and hence calculate the area of the 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}

Find the area of a right – angled triangle, the radius of whose circumcircle measure 8 cm and the altitude drawn to the hypotenuse measure 6 cm.

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}

Find the length of the hypotenuse of an isosceles right × angled triangle whose area is 200 _{}. Also, find the perimeter. Take _{}.

Let each equal side be a cm in length.

Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

The base of an isosceles triangle measure 80 cm and its area is _{}. Find the perimeter of the triangle.

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

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measure 12 cm. Find the area of the triangle.

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}.

In the given figure, ABC is an equilateral triangle the length of whose side is equal to 10 cm, and DBC is right angled triangle at D and BD = 8 cm. Find the area of the shaded region. _{}

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 _{}

Find the area and perimeter of an isosceles right – angled triangle, each of whose equal sides measures 10 cm. Take _{}

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 – Exercise

## Chapter 17 – Perimeter and Areas of Plane Figures Exercise FA

In the given figure ABCD is a quadrilateral in which ∠ABC = 90, ∠BDC = 90, AC = 17 cm, BC= 15 cm, BD = 12 cm and CD = 9cm. The area of quad. ABCD is

- 102 cm
^{2} - 114 cm
^{2} - 95 cm
^{2} - 57 cm
^{2}

In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CE ^ AB. Area of trap. ABCD is

- 306 m
^{2} - 316 m
^{2} - 296 m
^{2} - 284 m
^{2}

The sides of a triangle are in the ratio 12:14:25 and it perimeter is 25.5 cm. The largest side of the triangle is

- 7 cm
- 14 cm
- 12.5 cm
- 18 cm

The parallel sides of trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is

- 104 cm
^{2} - 78 cm
^{2} - 52 cm
^{2} - 65 cm
^{2}

Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall.

The length of the diagonal of a square is 24 cm. Find its area.

Find the area of a rhombus whose diagonals are 48 cm and 20 cm long.

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

A lawn is in the form of a rectangle whose sides are in the ratio 5:3 and its area is 3375 m^{2}. Find the cost of fencing the lawn at Rs.20 per metre.

Find the area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm.

Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.

The adjacent sides of a ‖ gm AECD measure 34 cm and 20 cm and the diagonal AC is 42 cm long. Find the area of the ‖ gm.

The cost of fencing a square lawn at Rs.14 per metre is Rs. 2800. Find the cost of mowing the lawn at Rs.54 per 100 m^{2}.

Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD =29 cm, DA = 39 cm and diag. BD = 20 cm.

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 ‖gm is 66m long, find its corresponding altitude.

The diagonal of a rhombus are 48 cm and 20cm long. Find the perimeter of the rhombus.

The adjacent sides of parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides.

In a four-sided field, the length of the longer diagonal is 128 m. The length of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field.

## Chapter 17 – Perimeter and Areas of Plane Figures Exercise MCQ

The

length of a rectangular hall is 5 m more than its

breadth. If the area

of the hall is 750 m^{2} then its length

is

- 15

m - 20

m - 25

m - 30

m

The length of a rectangular field is 23 m

more than its breadth. If the perimeter of the field is 206 m, then its area

is

- 2420

m^{2} - 2520

m^{2} - 2480

m^{2} - 2620

m^{2}

The length of a rectangular field is 12 m

and the length of its diagonal is 15 m. Then area of the field is

The cost of carpeting a room 15m long

with a carpet 75 cm wide, at Rs.70 per metre,

is Rs. 8400. Then

width of the room is

- 9

m - 8

m - 6

m - 12

m

On increasing the length of a rectangle

by 20% and decreasing its breadth by 20%, what is the change in its area?

- 20%

increase - 20%

decrease - No

change - 4%

decrease

A

rectangular ground 80 m x 50 m has a path 1 m wide

outside around it. The

area of the path is

- 264

m^{2} - 284

m^{2} - 400

m^{2} - 464

m^{2}

The

area of a square field is 6050 m^{2}. The length of its

diagonal

is

- 135

m - 120

m - 112

m - 110

m

The area of a square field is 0.5

hectare. The length of its diagonal is

Each

side of an equilateral triangle is 8 cm. Its area is

The base and height of a triangle are in

the ratio 3:4 and its area is 216 cm^{2}. The height of the triangle

is

- 18

cm - 24

cm - 21

cm - 28

cm

The lengths of the sides of a triangular

field are 20 m, 21m and 29 m. The cost of cultivating the field at Rs. 9 per m^{2} is

- Rs. 2610
- Rs. 3780
- Rs. 1890
- Rs. 1800

The side of a square is equal to the side

of an equilateral triangle. The ratio of their areas is

The side of an equilateral triangle is

equal to the radius of a circle whose area is 154 cm^{2}. The area of

the triangle is

The area of a rhombus is 480 cm^{2}

and the length of one of its diagonals is 20 cm. The length of each side of

the rhombus is

- 24

cm - 30

cm - 26

cm - 28

cm

One side of a rhombus is 20 cm long and

one of its diagonals measures 24 cm. The area of the rhombus is

- 192

cm^{2} - 480

cm^{2} - 240

cm^{2} - 384

cm^{2}

## Chapter 17 – Perimeter and Areas of Plane Figures Exercise Ex. 17B

The length of a rectangular park is twice

its breadth and its perimeter is 840 m. Find the area of the park.

A lawn is in the form of a rectangle

whose sides are in the ratio 5 : 3. The area of the

lawn is 3375 m^{2}. Find the cost of fencing the lawn at Rs. 65 per metre.

A room is 16 m long and 13.5 m broad.

Find the cost of covering its floor with 0.75-m-wide carpet at Rs. 60 per metre.

The floor of a rectangular hall is 24 m long and 18 m wide. How many carpets, each of length 2.5 m and breadth 80 cm, will be required to cover the floor of the hall?

Area of floor = Length Breadth

_{}

Area of carpet = Length Breadth

_{= }

Number of carpets = _{}

= 216

Hence the number of carpet pieces required = 216

A 36 m-long, 15 m-broad verandah is to be paved with stones, each measuring 6 dm by 5 dm. How many stones will be required?

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.

The area of a rectangle is 192 _{}and its perimeter is 56 cm. Find the dimensions of the rectangle.

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

A rectangular park 35 m long and 18 m wide is to be covered with grass, leaving 2.5 m uncovered all around it. Find the area to be laid with grass.

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}

A rectangular plot measures 125 m by 78 m. It has a gravel path 3 m wide all around on the outside. Find the area of the path and the cost of gravelling it at Rs. 75 per _{}.

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

A footpath of uniform width runs all around the inside of a rectangular field 54m long and 35 m wide. If the area of the path is 420 _{}, find the width of the path.

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

_{}

The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911_{}. Find the dimensions of the garden.

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

A room 4.9 m long and 3.5 m broad is

covered with carpet, leaving an uncovered margin of 25 cm all around the

room. If the breadth of the carpet is 80 cm, find its cost at Rs.80 per metre.

A carpet is laid on the floor of a room 8m by 5m. There is a border of constant width all around the carpet. If the area of the border is _{}, find its width.

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 – _{})_{}

_{}

A 80 m by 64 m rectangular lawn

has two roads, each 5 m wide, running through its middle, one parallel to its

length and the other parallel to its breadth. Find the cost of gravelling the

roads at Rs.40 per m^{2}.

The dimensions of a room are 14 m x 10 m

x 6.5 m. There are two doors and 4 windows in the room. Each door measures

2.5 m x 1.2 m and each window was of the measure 1.5 m x 1 m. Find the cost

of painting the four walls of the room at Rs.35 per m^{2}.

The cost of painting the four walls of a

room 12 m long at Rs.30 per m^{2} is Rs.7560 and the cost of covering the floor

with mat at Rs.25 per m^{2} is Rs.2700. Find the dimensions of the room.

Find the area and perimeter of a square

plot of land whose diagonal is 24 m long. []

Find the length of the diagonal of a square of area 128 _{}. Also find the perimeter of the square, correct to two decimal places.

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

The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km per hour?

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.

The cost of harvesting a square field at Rs.900 per hectare is Rs.8100. Find the cost of putting a fence

around it at Rs.18 per metre.

The cost of fencing a square at Rs. 14 per meter is Rs. 28000. Find the cost of mowing the lawn at Rs. 54 per 100 _{}.

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 =_{}

In the given figure, ABCD is a

quadrilateral in which diagonal BD = 24 cm, AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 12

cm. Calculate the area of the quadrilateral.

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29cm, DA = 34 cm and diagonal BD = 20 cm.

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

_{}

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ACB = 90^{o} and AC = 15 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

Find the area of the quadrilateral ABCD in which AD = 24cm, _{}BAD = 90^{o} and ΔBCD forms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take _{}

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

Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.

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

= (25 16.8) _{}

The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find the distance between the shorter sides.

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

The area of a parallelogram is 392_{}. If its altitude is twice the corresponding base, determine the base and the altitude.

_{}

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

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

_{}

_{}

Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also find the perimeter of the rhombus.

_{}

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

_{}

The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long find (i) the length of the other diagonal and (ii) the area of the rhombus.

(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 = _{}

_{}

The area of a rhombus is 480_{} and one of its diagonals measures 48 cm. Find (i) the length of the other diagonals, (ii) the length of each of its sides, and (iii) its perimeter.

(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

The parallel sides of a trapezium are 12

cm and 9 cm and the distance between them is 8 cm. Find the area of the

trapezium.

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 _{}, find the depth of the canal.

Areaof cross section = _{}

_{}

Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the non parallel sides are 15 m and 13 m long.

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

_{}