Perimeter and Area Worksheet

Multiple Choice Questions

1. What happens to the area of a parallelogram if the height is increased 2 times and the base is halved?
   (a) Remains the same
   (b) Becomes twice the original area
   (c) Decreases to half the original area
   (d) Becomes thrice the original area 
   
   
2. The ratio of the area of semi-circle to the area of circle is
   (a) 1:2
   (b) 2:1
   (c) 3:1
   (d) 1:3
   
   
3. In a right-angled triangle LMN, the height of a triangle is twice its base. If its area is 169 cm2, find its base and height.
   
   (a) Base = 26 cm, Height = 13 cm
   (b) Base = 12 cm , Height = 24 cm
   (c) Base = 13 cm , Height = 26 cm
   (d) Base = 24 cm, Height = 12 cm 
   
   
4. The diameter of the below circular disc is 28 cm. Find the circumference of a quadrant-shaped disc. 
   
   (a) 116 cm
   (b) 50 cm
   (c) 85 cm
   (d) 100 cm
   
   
5. The area of the shaded region is
   
   (a) 600 cm²
   (b) 900 cm²
   (c) 580 cm²
   (d) 450 cm²
   
   
6. The area of a circle with diameter D is
   (a) πD²
   (b) 2πD
   (c) 
   (d) 
   
   
7. The adjacent sides of a parallelogram ABCD are 13 cm and 8 cm respectively. If the distance between its longer sides is 6 cm, find the distance between its shorter sides. 
   
   (a) 6 cm
   (b) 7 cm
   (c) 8.75 cm
   (d) 9.75 cm
   
   
8. If the ratio of the diameters of two circles is 5:7, the ratio of their circumferences is
   
   (a) 5:7
   (b) 7:5
   (c) 5:15
   (d) 15:5
   
   
9. If the area of the following rectangle is ‘b’ m², then the area of each triangle is
   
   (a) 2b m²
   (b)  
   (c)  
   (d) b² m²
   
   
10. A jogging ground is in the form of a circle whose diameter is 301 m. Ravi ran around the ground 3 times. Find the distance covered by Ravi. 
    (a) 946 m
    (b) 1892 m
    (c) 2838 m
    (d) 3784 m
    
    Subjective Questions
    
    
11. Find the height ‘h’ of a below parallelogram shape whose area is 391 cm².  
    
    
12. A flower garden of area 9856 m2 is in the form of a circle with circumference 352 m. A gardener wants to fence it at the rate of Rs. 30 per metre. Find the cost of fencing.
    
    
13. In ΔXYZ, XY = 6 cm, XZ = 12 cm and YT = 8 cm. Find the length of ZR.
    
    
14. How many times a tyre of radius 42 cm be rotated to travel 528 cm? 
    
    
15. Find the area of quadrilateral PQRS.
    
    
16. A parallelogram-shaped tile has base 9 cm and the corresponding height 6 cm. For covering a wall of area 648 cm², how many such tiles are required.
    
    
17. Find the distance covered by the tip of the second hand of a clock in 2 minutes, if the length of the second hand is 49 mm. 
    
    
18. PQRS is a square. Find the area of the shaded region.
    
    
19. Two parallelograms ABCD and PQRS of equal area are superimposed on one another as shown in the below figure. A square EBRF of side 3 cm is formed by their overlapping area. If BC = AD = 7 cm, find the area of parallelogram PQRS. 
     
20. Shramik wants to fence his circular farm with wire whose radius is 350 m. The cost of fencing is Rs. 13 per metre. He has Rs. 27000 with him. Is this money more or less than the required cost of fencing? By how much? 

Multiple Choice Questions

1. What happens to the area of a parallelogram if the height is increased 2 times and the base is halved?
   (a) Remains the same
   (b) Becomes twice the original area
   (c) Decreases to half the original area
   (d) Becomes thrice the original area
   Correct option: (a) Remains the same
   Solution:
   Let the original base and height of a parallelogram be ’b’ and ‘h’ respectively.
   Then, area of a parallelogram = b × h
    
   
   
2. The ratio of the area of semi-circle to the area of circle is
   (a) 1:2
   (b) 2:1
   (c) 3:1
   (d) 1:3
   Correct option: (a) 1:2
   Solution:
   Area of a circle 
   Area of a semi-circle 
   
   
   
3. In a right-angled triangle LMN, the height of a triangle is twice its base. If its area is 169 cm2, find its base and height.
   
   (a) Base = 26 cm, Height = 13 cm
   (b) Base = 12 cm , Height = 24 cm
   (c) Base = 13 cm , Height = 26 cm
   (d) Base = 24 cm, Height = 12 cm
   Correct option: (c) Base = 13 cm , Height = 26 cm
   Solution:
   Let the base of ΔLMN =  α cm
   Then, height of ΔLMN =  2α cm
   Given, area of ΔLMN = 169 cm²
   
   
   
4. The diameter of the below circular disc is 28 cm. Find the circumference of a quadrant-shaped disc. 
   
   (a) 116 cm
   (b) 50 cm
   (c) 85 cm
   (d) 100 cm
   Correct option: (b) 50 cm
   Solution:
   Diameter of a circular disc = 28 cm
    Radius of a circular disc = 14 cm
   Now, circumference of a quadrant-shaped disc 
   
   
   
   
5. The area of the shaded region is
   
   (a) 600 cm²
   (b) 900 cm²
   (c) 580 cm²
   (d) 450 cm²
   Correct option: (d) 450 cm²
   Solution:
   Area of a square EFGH = 30 cm × 30 cm = 900 cm²
   
   Therefore, area of shaded region = Area of square EFGH – Area of ΔPHG = 450 cm²
   
   
6. The area of a circle with diameter D is
   (a) πD²
   (b) 2πD
   (c) 
   (d) 
   Correct option: (c) 
   Solution:
   Diameter = D, Radius = R
   Now, D = 2 × R
   
   
   
   
7. The adjacent sides of a parallelogram ABCD are 13 cm and 8 cm respectively. If the distance between its longer sides is 6 cm, find the distance between its shorter sides. 
   
   (a) 6 cm
   (b) 7 cm
   (c) 8.75 cm
   (d) 9.75 cm
   Correct option: (d) 9.75 cm
   Solution:
    
   Let the distance between shorter sides = ‘h’ cm
   Now, if we choose longer side as base = CD = 13 cm, distance between the longer sides is height, that is CF = 6 cm
   Now, area of a parallelogram ABCD = base × height
   = 13 cm × 6 cm
   = 78 cm²
   Also, area of parallelogram ABCD = AD × h
   78 = 8 × h
    h = 9.75 cm
   
   
8. If the ratio of the diameters of two circles is 5:7, the ratio of their circumferences is
   (a) 5:7
   (b) 7:5
   (c) 5:15
   (d) 15:5
   Correct option: (a) 5:7
   Solution:
   
   
   
9. If the area of the following rectangle is ‘b’ m², then the area of each triangle is
   
   (a) 2b m²
   (b)  
   (c)  
   (d) b² m²
   Correct option: (b)   
   Solution:
   Diagonal of a rectangle divides it into two triangles of equal area.
   
   
10. A jogging ground is in the form of a circle whose diameter is 301 m. Ravi ran around the ground 3 times. Find the distance covered by Ravi. 
    (a) 946 m
    (b) 1892 m
    (c) 2838 m
    (d) 3784 m
    Correct option: (c) 2838 m
    Solution:
    Diameter of a circular jogging ground, d = 301 m
    Then, circumference of a circular jogging ground 
    Number of rounds made by Raavi = 3
    Therefore, total distance covered by Raavi = 3 × Circumference = 3 × 946 = 2838m
    
    Subjective Questions
    
    
11. Find the height ‘h’ of a below parallelogram shape whose area is 391 cm². 
    Solution:
    Area of a parallelogram = base × height
    
    Therefore, the height of the given parallelogram is 17 cm.
    
    
12. A flower garden of area 9856 m2 is in the form of a circle with circumference 352 m. A gardener wants to fence it at the rate of Rs. 30 per metre. Find the cost of fencing.
    Solution:
    Total cost of fencing = Circumference × Rate of fencing per metre
    = Rs. (352 × 30)
    = Rs. 10,560
    Therefore, the total cost of fencing is Rs. 10,560.
    
    
13. In ΔXYZ, XY = 6 cm, XZ = 12 cm and YT = 8 cm. Find the length of ZR.
    Solution:
    
    Also,
    
    
    Therefore, the length of ZR is 16 cm.
    
    
14. How many times a tyre of radius 42 cm be rotated to travel 528 cm? 
    Solution:
    Circumference of a tyre    
    Number of rotations of tyre 
    Therefore, a tyre should be rotated 2 times to travel 528 cm.
    
    
15. Find the area of quadrilateral PQRS.
    Solution:
    
    Area of quadrilateral PQRS = Area of ΔQSR + Area of ΔQPS
                                           = 750 m² +300 m²
                                           = 1050 m²
    
    
16. A parallelogram-shaped tile has base 9 cm and the corresponding height 6 cm. For covering a wall of area 648 cm², how many such tiles are required.
    Solution:
    Area of a parallelogram-shaped tile = base × height
    = 9 cm × 6 cm
    = 54 cm2
    Area of a wall = 648 cm²
    Number of tiles required = 
    
    
17. Find the distance covered by the tip of the second hand of a clock in 2 minutes, if the length of the second hand is 49 mm. 
    Solution:
    Distance covered by second hand in 1 minute is the circumference of a circle and length of the second hand is the radius of the circle.
    Then, circumference of a circle   = distance covered by second hand in 1 minute
    Therefore, distance covered by second hand in 2 minutes = 2 × Circumference = 2 × 308 = 616 mm
    
    
18. PQRS is a square. Find the area of the shaded region.
    Solution:
    Area of a square PQRS = (Side)² = (10 cm)² = 100 cm².
    
    
    Area of shaded region = Area of a square – (Area of ΔQRX + Area of ΔPMX + Area of ΔSMR)
    = 100 – (25 + 17.5 + 15)
    = 100 – 57.5
    = 42.5 cm²
    
    
19. Two parallelograms ABCD and PQRS of equal area are superimposed on one another as shown in the below figure. A square EBRF of side 3 cm is formed by their overlapping area. If BC = AD = 7 cm, find the area of parallelogram PQRS. 
    Solution:
    Area of parallelogram PQRS = Area of parallelogram ABCD
    = BC × BE
    = 7 cm × 3 cm
    = 21 cm²
    
    
20. Shramik wants to fence his circular farm with wire whose radius is 350 m. The cost of fencing is Rs. 13 per metre. He has Rs. 27000 with him. Is this money more or less than the required cost of fencing? By how much? 
    Solution:
    Radius of a circular farm = 350 m
    Then, circumference of a circular farm
    Cost of fencing = Rs. 13 per metre
    Then,
    Total cost of fencing = Circumference of a circular farm × Cost of fencing per metre
    = Rs. (2200 × 13)
    = Rs. 28600
    Amount with Shramik = Rs. 27000, which is less than total cost of fencingand, difference = Rs. 28600 – Rs. 27000 = Rs. 1600
    Hence, the amount with Shramik is less than the total cost of fencing by Rs. 1600.