SEBA Class 7 Mathematics Chapter 11 Perimeter and Area

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SEBA Class 7 Mathematics Chapter 11 Perimeter and Area

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 7 General Maths Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 7 Mathematics Chapter 11 Perimeter and Area Solutions for All Subject, You can practice these here.

Perimeter and Area

Chapter – 11

PART – II

Exercise – 11.1

1. Given below are any two values (base, height or area) of parallelograms. Fill up the gap by finding the value of other.

Ans: (i) Base of a parallelogram = 10 cm 

Height = 7cm 

Area = Base ×  Height 

(10 × 7 ) sq. cm = 70 sq. cm 

(ii) Base = 20 

Area = Base × Height 

400 sq. cm

Height = 400/20 cm 

= 20 cm

(iii) Base = 15 cm 

Height = 2.5 cm 

Area = Base × Height 

= 15 × 2.5 

= 37.50 sq.cm 

(iv) Height = 25 dm 

Area = 325 sq.dm 

Base = area/height 

= 325 sq.dm/25 dm 

= 13 dm

(v) Base = 15.6 cm 

Area = 16.38 sq.cm 

Height = Area/base 

= 16. 38 sq.cm/15.6cm 

= 1.05 cm 

(vi) Height = 8.4 dm 

Area = 48.72sq.dm

Base = Area/Height 

= 48.72/8.4 dm 

= 5.8 dm

2. Some structure any two values of [area (A), base (B) or height (H)] of triangle. Fill up the gaps.

(i) A = 64 sq.cm.

B = 8 cm.

H = ___________

Ans: A = 64 sq.cm

B = 8 cm.

Area = 1/2 × base × height

= 1/2 × 64 × 8 

H = 16 cm.

(ii) A = ___________

B = 3 m

H = 214 cm 

Ans: B = 3 m.

H = 214 cm = 214/100 = 2.14m

Area = 1/2 ​× base × height

= 1/2 × 3 × 2.14 

A = 3.21 sq. m 

(iii) A = 94 sq.cm

B = ___________

H = 7 m.

Ans: A = 94 sq.cm

H = 7 m.

Area = 1/2 ​× base × height

 94 = 1/2 ​× B × 7

=94 × 2 = B × 7 

= 188 = 7B

= 188/7 = B 

B = 26.86 cm 

(iv) A = 1256 sq.cm.

B = __________

H = 31.4 mm

Ans: A = 1256 sq.cm.

H = 31.4 mm 

Area = 1/2​ base × height

1256 = 1/2 × B × 31.4 

B = 1256 × 2/3.14 cm 

= 800 cm 

(v) A = 16.38 sq.cm.

B = 15.6 cm.

H = ___________

Ans: A = 16.38 sq.cm.

H = 4.2 cm.

Area = 1/2 base × height

16.38 = 1/2 × 15.6 × H

H = 16.38 × 2/15.6 cm 

= 2.1 cm

3. Find the area of each of the following trapeziums.

Ans: Area of a trapezium = 1/2 × distance between two parallel line × sum of length of two parallel lines

= 1/2 × 4 × (5 + 7.5) sq.cm

= 1/2 × 4 × 12.5 sq.cm

= (2 × 12.5) sq.cm.

= 25 sq.cm.

Ans: Area of trapezium

= 1/2 × (6 + 4) × 2.5 sq.cm 

= 1/2 × 10 × 2.5 sq.cm. 

= (5 × 2.5) sq.cm 

= 12.5 sq.cm

Ans: Area of trapezium, 

= 1/2 × 2 × (7.5 + 3.5) sq.cm. 

= 11 sq. cm.

Ans: Area of trapezium, 

= 1/2 × 2.5 × (5.5 + 2.5) 

= 1/2 × 2.5 × 8⁴ sq.cm.

= 10 sq.cm.

4. Give below are any two measures (base, height or diagonals) of thombus. Fill up the following table by using different formulas.

Ans: (i) Area of Rhombus 

= 1/2 × (Product of the diagonals) 

= 1/2 × 10 × 8.2 

= 41 sq.cm.

(ii) A = 56 sq. cm

B = 8cm 

⇒ Altitude = 56 sq.cm./8 cm.              

= 7 cm.

(iii) Area of a Rhombus = base × altitude 

= (20 × 7) sq.cm. 

= 140 sq.cm.

(iv) Area of a Rhombus 

= 1/2 × (18 × 14) 

= 1/2 × (252) sq.cm. 

= 125 sq.cm

(v) Area of a Rhombus, = base × altitude 

= 4.725 sq.cm. 

= 2.7 cm × altitude

Now,

Altitude = 4.725 sq.cm

= 1.75 cm

(vi) 120 sq. cm 

= 1/2 × 30 × other diagonal

⇒ Another diagonal, 

= 120 × 2/30 cm 

= 8 cm

5. The measure of base and height of a parallelogram are 1m 50 cm and height 75 cm respectively. Find the area in sq.cm.

Ans: Area of a parallelogram = base × height

= 150cm × 75cm

= 11250 sq.cm.

= 11250/1000 sq .m

= 1.125 sdq.m.

6. The measures of adjacent sides of a parallelogram are 12 cm and 9 cm respectively. The perpendicular distance of the long sides is 6 cm. Find the area of the parallelogram. (Draw the figure).

Ans: 

Area of a parallelogram

= base × height

= (12 × 6) sq .cm

= 72 sq.cm

7. The area of a right angle triangle is 30 sq.cm. One of the sides of the right angle is 6 cm, find out measure of other side of the right angle.

Ans:  

Area of a right angle, = 1/2 × base × height 

⇒ 30 sq.cm 1/2 × 6 × height 

⇒ height = 30 sq. cm/. 3 cm 

= 10 cm.

∴ Other side of the right angle = 10 cm.

8. Triangle ABC is an isosceles triangle where AB = AC= 7.5 cm and BC = 9 cm (see diagram). The height AD which is drawn from A to Bc measures 6 cm. Find the area of the triangle ∆ABC. Also find the measure of the height CE.

Ans: Area of a ∆ABC

= 1/2 × BC × AD 

= 1/2 × 9 × 6 sq.cm 

= 27 sq. cm 

Again, 

△ABC = 1/2 × AB × CE

⇒ 27 = 1/2 × 7.5 × CE 

⇒ CE = 2 × 27/7.5               

= 7.2 cm

Now,

∴ Area of ∆ABC = 27 sq.cm and CE = 7.2 cm

9. In the right angle triangle ABC, < A = 90° AD is perpendicular on BC. If AB = 12 cm, AC = 5 cm, BC = 13 cm, then find.

(i) Area of ∆ABC

Ans: Area of ∆ABC = 1/2 × AC× AB 

= 2 × 5 × 12 sq. cm 

= 30 sq. cm 

(ii) Length of AD

Ans: Again, 

△ABC = 1/2 × BC × AD

= 1/2 × 5 × 12 sq .cm

⇒ 30 sq. cm.  

= 1/2 × 13 cm × AD 

⇒ AD = 30 × 2/13 cm 

= 4.62 cm 

10. In a quadrilateral ABCD the diagonal AC is 12 m. If BL ⏊ AC, DM ⏊ AC and BL = 3 cm, DM = 7 cm then find the area of a quadrilateral ABCD?

Ans: Area of quadrilateral ABCD 

= 1/2 × AC × (BL + DM)

= 1/2 × 12 × (3 + 7) sq.cm 

= 60 sq.cm. 

11. EG is the diagonal of quadrilateral EFGH. FM and HN are perpendicular on EG. If EG = 28 cm, FM = 7 cm and HN = 5 cm, then find–

(i) Area of ∆EFG 

Ans: Area of ∆EFG 

= 1/2 × EG × FM

= 1/2 × 28 × 7 sq. cm 

= 98 sq. cm 

(ii) Area ∆EHG

Ans: Area of ∆EHG   

= 1/2 × EG × HN 

= 1/2 × 28 × 5 sq .cm 

= 70 sq. cm 

(iii) Area of quadrilateral EFGH.

Ans: Area of quadrilateral EFGH

= AEFG + ΔEHG

= (98 + 70) sq. cm 

= 168 sq. cm 

12. The area of a quadrilateral is 11 sq. cm. The measure of the sub perpendiculars which are drawn on the diagonal are 2.5 cm and 1.5 cm respectively. Find the measure of the diagonal.

Ans: Area of quadrilateral

= 1/2 × diagonal × (2.5 × 1.5)

⇒ 11 sq. cm 

= 1/2 × diagonal × (4) 

⇒ Diagonal = 11 × 2 /4  cm. 

= 5.5 cm

13. The measures of parallel lines of a trapezium are 18 cm and 16 cm respectively. The distance between the parallel lines is 8 cm. Find the area of the trapezium.

Ans: Area of a trapezium 

= 1/2 × 8 × (18 + 16) sq. cm 

= 1/2 × (8 × 34) sq. cm 

= (1/2 × 272) sq. cm 

= 136 sq. cm 

14. The area of a trapezium shape field is 600 sq.cm. The length of the parallel bank are 20m and 30m respectively. Find the distance between the parallel bank.

Ans: Area of a trapezium shape field 

=1/2 × (20 + 30) × h 

⇒ 600 = 25h 

⇒ h = 600/25 m 

= 24m

∴ The distance between the parallel bank = 24m.

15. The area of a trapezium shaped paper is 11 sq. cm. The distance between the parallel side is 5.5 cm and the measure of a parallel side is 2.5 cm. Find the length of the other parallel side.

Ans: The area of a trapezium shaped paper = 1/2 × (5.5 + a) × 2.5

⇒ 11 = 1/2 × 5.5 (2.5+a) 

⇒ 5.5 × (2.5 + a) = 22 

⇒ 2.5 + a = 22/5.5 = 4 

⇒ 5.5 a = (4-2.5) cm 

= 1.5 cm

∴ The length of the other parallel side = 1.5 cm 

16. Find the area of a rhombus in which the length of the diagonals are 7m and 6m respectively. Find the area of the rhombus and express it in terms of sq.cm.

Ans: The length of the diagonals are 7m = 700 cm and 6m = 600 cm.

∴ Area of a rhombus = 1/2 × product of two diagonals 

= 1/2 × 700 × 600 sq. cm

= 2,10,000 sq. cm

17. Find the area of a rhombus, in which the length of one side is 6 cm and height is 8cm. If the length of one diagonal is 8 cm, find the length of the other diagonal.

Ans: 

Length of CD = 8cm, h = 6 cm. 

∴ Area of Rhombus ABCD = CD × h 

= (8 × 6) sq. cm 

= 48 sq. cm. 

Again,

48 sq. cm = 1/2 × AC × BD

⇒ 48 = 1/2 × 8 × BD

⇒ BD = 2 × 48/8

= 12 cm

∴ Area of Rhombus = 48 sq.cm. and the length of a other diagonal = 12 cm

18. The area of a rhombus is 56 sq. cm. Find the height of the rhombus if its perimeter is 32 cm.

Ans: Area of a rhombus = 56 sq. m. perimeter = 32 m 

∴ Measure of one side = 32/4 m = 8 m

∴ 56 sq.m. DC × AD

⇒ 56 = 8 × AD

⇒ 56 =  8 × AD

⇒ AD = 56/8 = 7 m

∴ Height = 7 m.

19. The diagonal of a rhombus is 6 cm., if the area of the rhombus is 24 sq. cm. then find the length of the other diagonal.

Ans: one diagonal of a rhombus = 6m and Area = 24 sq.m.

∴ Area = 1/2 × 6 × other diagonal 

⇒ 24 = 3 × other diagonal 

⇒ other diagonal = 24/3 

= 8m

20. If the area of a parallelogram is 15 sq. cm and base is 5 cm then find its height.

Ans: Area of a parallelogram = 15 sq. cm and base = 5 cm

∴ Area of a parallelogram = base × height

⇒ 15 sq.cm. = 5cm × height

⇒ height = 15 sq. cm /5 cm 

= 3 cm

1. Find out the diameter of the circle using the circumference given below: (π = 22/7)

(a) 28 cm

Ans: Diameter = 28 cm.

∴ Radius (r) = 28/2 cm 

= 14 cm.

∴ Circumference of the circle = 2πr

= 2 × 22/7 × 14 cm  

= 88 cm.

(b) 56 m.m.

Ans: Diameter (D) = 56 m.m.

r = 56/2mm. 

= 28 m.m.

Circumference = 2πr 

= 2 × 22/7 × 28mm.

= 176 mm.

(c) 42 cm.

Ans: Diameter (D) = 42 cm

∴ r = 42/2 

= 21 cm.

Circumference = 2πr 

= 2 × 22/7 × 21cm.

= 13.2 cm.

2. What length of fencing wire will be required to fence a garden of radius 14 m? What will be the expenditure if the cost of fencing wire is Rs. 55-00 per meter? (π = 22/7)

Ans: radius (r) = 14m.

Length of fencing wire (2πr) = (2 × 22/7 × 14) m = 88 m.

The expenditure Rs. (55.00 × 88) 

= Rs. 4,840

3. Find the circumference of the half circle shaped region given in the adjacent figure.

Ans: 

The circumference of the half circle shaped region given in the adjacent figure

= 2πr/2 + 10 cm

= πr + 10 cm

= (22/7 × 10/2)cm + 10 cm

= (22/7 × 5)cm + 10 cm

= 15.71 cm + 10cm

= 25.71 cm

4. The diameter of a wheel of a vehicle is 70cm. How many times will the wheel be needed to rotate to complete the distance of 33 km? (π = 22/7)

Ans: Diameter of a wheel (D) = 70 cm

Perimeter of the wheel = 2πr 

= (2r) π

= (70 × 22/7) cm

= 220 cm 

= 2.2m 

Distance covered = 33 km 

= 33 × 1000m

∴ Number of rotation = 33000/(2.2) 

= 330000 / 22 

= 30000 / 2 

= 15000

5. The radius of a circle is 84 cm and the radius of another circle is 91 cm. By how much is the circumference of the second circle more than the first circle?

Ans: Circumference of first circle = 2 × 22/7 × 85 cm

= 528 cm 

Circumference of 2nd circle 

= 2 × 22/7 × 91 cm

= 572 cm

∴ The circumference of second circle (572- 528) cm 

= 44 cm more than the first circle.

6. The length of the side of a square is 3 m and the radius of a circle is 7m. Find the difference of the perimeter of the square and the circumference

Circle.

Ans: Perimeter of square (4 × 3)m

= 12m and Circumference of circle = 2 × 22/7 × 7 m 

= 44m

∴  The difference of the perimeter of the square and the circumference of the circle = (44-12) m = 32m.

7. Runima made a circle with a 44 cm string. Find out the diameter of the circle? If a square is made with the same string, then what will be the length of each side of the square?

Ans: If a circle is made with a 44 cm string, then circumference will be 44 cm.

∴  2πr = 44 ⇒ r = 44/(2π) 

= (44×7)/(2×22) 

= 7cm.

∴ Diameter (2 × 7) cm 

= 14 cm.

If a square is made with the same string, then the perimeter of the square will be 44 cm.

∴  Perimeter of square = 4 × one side. 

⇒ 4 × one side = 44

⇒ one side = 44/4

= 11 cm  

∴ 14 cm, 11 cm.

8. The diameter of a wheel of a vehicle is 98 cm. What distance the vehicle will cover when the wheel rotates 300 times?

Ans: Diameter = 98 cm

∴ radius (r) = 98/2 cm 

= 49 cm

∴ Circumference of the wheel = 2 × 22/7 × 49 = 308 cm

When the wheel rotates 1 time, it covers 308 cm.

When the wheel rotates 300 times, it cover (308 × 300) cm

= 92400 cm

9. Rs. 2640-00 has been spent to fence the circular garden. What will be the circumference of the garden if the cost of fencing is Rs. 28.00 per meter?

Ans: Rs. 28.00 has been spent for 1 m to fence

∴ Rs. 1.00 has been spent for 1/28 m to fence

∴ Rs. 2640 has been spent for 1/28 × 2640 m 

= 94.29m 

10. A circular portion of radius 4 cm is taken away from a circular paper of radius 10 cm. By what length the circumference of the removed piece of paper is smaller than the circumference of the first piece of paper

Ans: Circumference of a piece of paper of radius 10 cm

= 2 × 22/7 × 10 cm 

= 440/7 cm

= 62.86 cm

and, Circumference of a piece of paper of radius 4 cm

= 2 × 22/7 × 4 cm

= 25-15 cm (Approx)

∴ The length of Circumference of the removed piece of paper is (62.86 – 25.14) cm = 37.72 cm (Approx) smaller than the circumference of the first piece of paper.

1. Write the correct answer from the given options:

(i) The area of the circle of radius 10.5 cm will be.

(a) 346.5 sq.cm.

(b) 340.5 sq.cm.

(c) 34.65 sq.cm.

(d) 34.05 sq.cm.

Ans: (a) 346.5 sq.cm.

(ii) The radius of a circular paper of area 616 sq. cm is-

(a) 7 cm.

(b) 28 cm.

(c) 14 cm.

(d) 3.5 cm.

Ans: (c) 14 cm.

2. Draw the circles with following radius and the area of the circles – Take [π=22/7]

(a) 5 cm.

Ans: Area of a circle = π × r²

= 22/7 ×(5)² cm²

= 550/7 cm²

= 78.57 sq. cm 

(b) 46 cm

Ans: Area of a circle with radius 4.6 cm

= 22/7 × (4.6)² sq.cm

= 465.52 / 7 sq. cm.

= 66.07sq.cm.

(c) 5.5 m

Ans: Area of a circle with radius 5.5m

= 22/7× (5.5)²  sq.m. 

= 665. 5 / 7 sq.m.

= 95.07 sq.m.

3. Find the radius and diameter of the circles whose area is given below

(a) 154 cm²

Ans: Area of a circle

= 154 sq. cm. 

⇒ πr²= 154

⇒ r² = 154/π

⇒ r² = (154 × 7)/22

⇒ r² = 7 × 7

⇒ r = √(7)² 

⇒ r = 7

∴ Radius = 7 cm 

Diameter 

= (2×7) cm 

= 14 cm 

∴ 7 cm, 14 cm.

(b) 550/7 sq. cm.

Ans: 550/7 sq. cm

∴ πr²= 550/7 

⇒ r² = (550×7)/(7×22) = 25 

⇒ r² = 25

⇒ r = √5

∴ Radius = 5 cm

= 5 cm 

Diameter

= (2×5) cm 

= 10 cm. 

∴ 10 cm 

4. Find the cost of polishing a circular table of radius 3 meter if the cost of expenditure is Rs. 30-00 per sq. meter. (π= 3.14).

Ans: Area of a circular table 

= 22/7 × (2/3)2sq. cm. 

= 22/7 × 9/4 sq. m.

=  99/14 sq. m.

= 7.07 sq. m.

The cost of expenditure per sq. meter = Rs. 30 

∴ The cost of expenditure of 7.07 sq. m 

= Rs. (30×7.07) 

= Rs. 212.10

5. The radius of a circle is 84 cm and the radius of another circle is 91cm. By how much is amount the area of the second circle more than the first circle.(π= 22/7).

Ans: Area of first circle = 22/7 × 84)² sq. cm

= 155232/7 sq. cm.

= 22.176 sq. cm.

Area of the second circle, = 22/7 × (91)² sq. cm.

= 182182/7 sq. cm

= 26,026 sq. cm

∴ Area of the second circle is (26,026- 22,176) sq. cm. 

= 3850 sq. cm more than the first circle.

6. A square is made by folding a string from a circle of circumference 28 cm.

(a) Find the area of square and circle

Ans: 2πr = 28 

⇒ π = 28/(2π) 

= 49/11 = 4.45cm 

∴ Area of circle = 22/7 × (4.45)² sq. cm.

= 435.66/7 

= 62.24 sq. cm.

Again,

4l = 28 

l = 28/4 = 7 cm.

Area = (7cm)²

= 49 sq. cm.

(b) Which area will be greater and how much?

Ans: Area of circle is (62.24 – 49) sq. cm 

= 13.24 sq cm more than the area of the square.

7. Find the area of the shaded part of the figures given

Ans: From the first figure:

Area of the shaded part = Area of square – Area of circle 

= {(8)²– 22/7 × (4)²} sq. cm 

= {64-50.29} sq. cm. 

= 13.71 sq. cm (Approx)  

From the second figure:

Area of the shaded part

8. There is a path of breadth 1 meter around a circular flower garden. The diameter of the garden is 66 meters. Find out the area of the path. Take (π = 3.14)

Ans: Diameter (D) = 66m 

∴ radius (r) = 66/2 m = 33m 

Area of circle with road = π × (34)² sq. m. 

Area of circle without road = π × (33)² sq. m.

∴ The area of the path = π (34²– 33²) sq. m.

= 22/7 × (1156 – 1089) sq. m.

= 22/7 × 67 sq. m. 

= 1474/7 sq. m 

= 210.57m² (Approx)

9. The circumference of a circle is 31.4 cm. Find out the radius and area of the circle (π= 3.14)

Ans: 2πr = 31.4

⇒ r = 31 . 4 ×7/2×22 cm

⇒ r = 219.8/44

⇒ r=  4.995 cm  

= 5 cm Approx 

∴ Area = 22/7 × (5)² sq. cm 

= 22/7 × 25 sq. cm.

= 550/7 sq. cm.

= 78.57 sq. cm.

10. From a square sheet of aluminium having a side of 6 cm, a circular portion of radius 2 cm is cut out. What is the area of the aluminium sheet that will be left over? (π = 3.14)

Ans: Area of a square sheet aluminium having side of 6 cm = (6cm)²

= 36 sq cm.

Again, area of a circle of radius 2cm = 22/7 × (2)² sq cm.

= 88/7 sq  cm.

= 12.57 sq cm. 

∴ The area of aluminium sheet that will be left over 

= (36 – 12.57) sq. cm. 

= 23.43 sq. cm. (Approx)

11. Find out the circumference and area of a half circle of diameter 21 cm. (π=22/7)

Ans: Diameter = 21 cm 

∴ Radius (r) = 21/2 cm.  

∴ Area of a half circle = πr + 21 cm 

= (2 × 2) + 21 cm (সে মি.)

= {22/7 × 21/2} + 21 cm

= 33 cm  + 21 cm 

= 54cm 

and area = πr²/2 cm 

= (22/(7 × 2) × (21/2)²sq. cm.

= 4851/28 sq. cm.

= 173.25 sq. cm.

12. The area of a circular disc is 38.5 sq. cm. What is the circumference of the disc? (π = 22/7)

Ans: Area of a circular disc 

= 38.5 sq.cm. 

∴ circumference  = 2 × 22/7 × 3.5 cm

= 154/7 cm

= 22 cm

13. A path of 1 meter breadth is constructed inside a square shaped garden of side measuring 30 meters. Find the area of the path?

Ans:

Area a square shaped garden of side measuring 30m = (30m)²

= 900 sq. m.

Without path, area of garden = (28m)² = 784 sq. m.

∴ Area of the path  = (900-784) sq. m.

= 116 sq. m.

14. Two pedestrian path ways of breadth 1 meter each is constructed mutually perpendicular through the middle of a square garden of side measuring 30 me ter.

(i) Find out the area of pedestrian path ways.

Ans: 

Area captured by road Area of PORS rectangle + Area of EFGH rectangle –

Area of square KLMN = (30×1+30×1-1×1) sq. m. 

= (60-1) sq. m.

= 59 sq. m.

(ii) What will be the total amount of expenditure to plant carpet grass in the remaining area if the cost of per sq. meter of carpet grass is Rs. 40?

Ans: Without road, area of garden = (30m)² – 59 sq.m.

= 900 sq. m. – 59 sq. m.

= 841 sq. m.

∴ The cost of per sq. meter of carpet grass is Rs. 40 

∴ The cost of 841 sq. m of carpet grass = ( 841 × 40 ) 

= Rs . 33 . 640

15. There is a path of 3 meter breadth out side of a park of length 125 meter and breadth 65 meter. Find out the area of the path.

Ans: Without road, area of garden 

= 125 × 65m 

= 8, 125 sq .m 

Area of garden with road 

= 131m × 71m 

= 9301 sq m. 

∴ Area of road 

= (9,301 – 8,125) sq .m 

= 1 ,176 sq. m.

16. From a rectangular aluminium sheet of 10 meter length and 5 meter breadth, two circles of 2 meter radius are removed. Find the area of the remaining part of the rectangular sheet.

Ans: Area of remaining part of the rectangular sheet

= 10m × 5m – ( 22/7×2²) sq. m. 

= 50sq.m – 88/7 sq. m. 

= 50 sq. m.-12.57 sq. m. 

= 37-43 sq. m. 

17. A veranda of breadth 2.25 meter is constructed surrounding a rectangular room of length 5.5 meter and breadth 4 meter-

(i) Find the area of the veranda বাৰাণ্ডাৰ কালি উলিওৱা.

Ans: Area of the room = 5.5m × 4m. 

= 22 sq m

Area of the room with veranda

= (5 .50 + 4.50) × (4 +4.50) sq.m. 

= (10×8.50) sq. m.

= 85 sq .m. 

∴ (i) Area of the veranda = (85  – 22) sq.m. 

= 63 sq .m

(ii) Find out the cost of construction of the veranda if the cost of per sq. meter of construction is Rs. 200.

Ans: The cost of construction of the veranda if the cost of per sq. meter of construction is Rs. 200, then the total cost

= Rs. (200× 63)

= Rs. 12.600

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