Mensuration 2D - Solved Questions and Answers

2D mensuration focuses on plane figures (flat shapes) like squares, rectangles, circles, triangles, etc.

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Mensuration - Formulas

Mensuration 2D questions and answers are provided below for you to learn and practice.

Question 1: Find the perimeter and area of an isosceles triangle whose equal sides are 5 cm and the height is 4 cm. 

Solution: 

Applying Pythagoras' theorem, 
(Hypotenuse)² = (Base)² + (Height)² 
(5)² = (0.5 x Base of isosceles triangle)² + (4)² 
0.5 x Base of isosceles triangle = 3 
Base of isosceles triangle = 6 cm 
Therefore, perimeter = sum of all sides = 5 + 5 + 6 = 16 cm 
Area of triangle = 0.5 x Base x Height
0.5 x 6 x 4 = 12 cm² 

Question 2: A rectangular piece of dimension 22 cm x 7 cm is used to make a circle of the largest possible radius. Find the area of the circle formed. 

Solution: 

In questions like this, the diameter of the circle is lesser in length and breadth. 
Here, the breadth Diameter of the circle = 7 cm 
Radius of the circle = 3.5 cm 
Therefore, area of the circle = π (Radius)²
π (3.5)² = 38.50 cm² 

Question 3: A pizza is to be divided into 8 identical pieces. What would be the angle subtended by each piece at the center of the circle? 

Solution: 

By identical pieces, we mean that area of each piece is the same. 
Area of each piece = (π x Radius² x θ) / 360 = (1/8) x Area of circular pizza 
(π x Radius² x θ) / 360 = (1/8) x (π x Radius²) 
θ / 360 = 1 / 8 
θ = 360 / 8 = 45
Therefore, the angle subtended by each piece at the center of the circle = 45° 

Question 4: Four cows are tied to each corner of a square field of side 7 cm. The cows are tied with a rope such that each cow grazes the maximum possible field, and all the cows graze in equal areas. Find the area of the ungrazed field. 

Solution: 

For maximum and equal grazing, the length of each rope has to be 3.5 cm. 
Area grazed by 1 cow = (π x Radius² x θ) / 360 
Area grazed by 1 cow = (π x 3.5² x 90) / 360 = (π x 3.5²) / 4 
Area grazed by 4 cows = 4 x [(π x 3.5²) / 4] = π x 3.5² 
Area grazed by 4 cows = 38.5 cm² 
Now, area of square field = Side² = 7² = 49 cm² 
Area ungrazed = Area of field - Area grazed by 4 cows 
Area ungrazed = 49 - 38.5 = 10.5 cm² 

Question 5: Find the area of the largest square that can be inscribed in a circle of radius 'r'. 

Solution: 

The largest square that can be inscribed in the circle will have the diameter of the circle as the diagonal of the square. 
Diagonal of the square = 2 r 
Side of the square = 2r / 2^1/2 
Side of the square = 2^1/2(r)
Therefore, area of the square = Side² = [2^1/2 r]² = 2r² 

Question 6: A contractor undertakes the job of fencing a rectangular field of length 100 m and breadth 50 m. The cost of fencing is Rs. 2 per meter, and the labor charges are Re. 1 per meter, both paid directly to the contractor. Find the total cost of fencing if 10 % of the amount paid to the contractor is paid as tax to the land authority. 

Solution: 

Total cost of fencing per meter = Rs. 2 + 1 = Rs. 3 
Length of fencing required = Perimeter of the rectangular field = 2 (Length + Breadth) 
Length of fencing required = 2 x (100 + 50) = 300 meter 
Amount paid to the contractor = Rs. 3 x 300 = 900 
Amount paid to the land authority = 10 % of Rs. 900 = Rs. 90 
Therefore, total cost of fencing = Rs. 900 + 90 = Rs. 990 

Question 7: A square has a circle inscribed in it. If the side of the square is 14 cm, find the area of the shaded region (square - circle).

Solution: 

Area of square = a²
= 14² = 196
Radius of circle = 7
Area of circle = πr²
=22/7 × 7 × 7 = 154
Shaded area = 196 - 154 = 42 cm²

Question 8: A garden is shaped like a rectangle with a semicircle on one end. The rectangle is 20 m long and 14 m wide. Find the total area.

Solution: 

Area of rectangle = 20 × 14 = 280 m²
Radius of semicircle = 7 m
Area of semicircle = 1/2 × πr² = 1/2 × 22/7 × 7 ²= 77 m²
Total area = 280 + 77 = 357 m²

Question 9: A rectangular garden is 20 m long and 15 m wide. A path 2 m wide runs along the inside edge of the garden. Find the area of the path.

Solution: 

Outer area = 20 × 15 = 300 m²
Inner area (excluding path): (20 - 4) × (15 - 4) = 16 × 11 = 176 m²
Path area = 300 - 176 = 124 m²

Question 10: A circular track has an outer radius of 21 m and an inner radius of 14 m. Find the area of the track and its width.

Solution: 

Track area = π(R² − r²) = 22/7(441−196) = 770 m²
Width = R−r = 7 m
Area = 770 m², Width = 7 m