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Q1. The area of a triangle is 50 cm². If its base is 25 cm, then the corresponding height is

• 1) 3 cm
• 2) 4 cm
• 3) 5 cm
• 4) 6 cm

Solution

Area of triangle = Thus,

Q2. Anu wants to put a lace to decorate the edge of a circular cake having diameter 12 cm. Find the length of lace required and also the cost, if 1 cm of the lace costs Rs. 2.

Solution

Since the lace has to be put on the edge of the circular cake, Length of lace required = circumference of the cake Diameter of the cake = 12 cm So, radius (r) = = 6 cm Now, circumference of the cake = 2r = 2 x 3.14 x 6 = 37.68 cm Hence, the length of the lace required is 37.68 cm. Cost of 1 cm lace = Rs. 2 Thus, the cost of 37.68 cm of lace = Rs. (37.68 x 2) = Rs. 75.36

Q3. A circle and a square are equal in area. The side of the square will be _______.

• 1)
• 2)
• 3)
• 4)

Solution

Area of square = Area of circle i.e. (side)² = r² i.e. side = 

Q4. What is the circumference of a circle with radius 14 cm?

Solution

The formula for finding the circumference of a circle is: Circumference of a circle = 2r Here, radius (r) = 14 cm Thus, Circumference = 2 x x 14   = 2 x 22 x 2 = 88 cm So, the circumference of the circle = 88 cm

Q5. Find the area of triangle XYZ with base XY = 9 cm and the length of height drawn from Z is 5 cm.

• 1) 24.5 cm²
• 2) 22.5 cm²
• 3) 23.5 cm²
• 4) 25.5 cm²

Solution

Thus, area = 22.5 cm²

Q6. A wire is in the shape of a square of side 22 cm. If the wire is re-bent into a circle, find its radius. Also, find the area of circle.

Solution

Given that the side of the square = 22 cm Length of the wire = perimeter of the square = 4 x side = 4 x 22 = 88 cm Since the same wire is used to make the circle, both will have the same perimeter. So, perimeter of the circle = 88 Thus, radius of circle = 14 cm Now, area of circle =

Q7. The perimeter of a regular polygon of n side is given by:

• 1) n + (Side length)
• 2) 2n x (Side length)
• 3) n x (Side length)
• 4) 3n x (Side length)

Solution

Perimeter of a regular polygon of side n is n x (Side length).

Q8. In triangle ABC, AC = 10 cm, BC = 4 cm and AD = 6 cm. Find the length of BL.

Solution

Given that, in ABC, BC = base = 4 cm; AD = height = 6 cm Area of triangle ABC = x b x h = x 4 x 6 = 12 cm² Also, in ABC, AC = base = 10 cm; BL = height = h (say) Area = 12 cm² Area of triangle = x b x h 12 = x 10 x h 5h = 12 h = = 2.4 cm Thus, the height (BL) of the triangle = 2.4 cm

Q9. The length of a side of a hexagon is 2 inches. What is the perimeter?

Solution

A Hexagon is a polygon which has 6 equal sides. Here, the length of the side = 2 inches We find the perimeter by adding the lengths of the edges. Perimeter of the hexagon = 2 + 2 + 2 + 2 + 2 + 2 = 6 x 2 = 12 inches Thus, the perimeter of the hexagon is 12 inches.

Q10. The side of a square is 15 cm. Find the area of another square whose side is twice the first square.

• 1) 450 sq.cm
• 2) 900 sq.cm
• 3) None of the above
• 4) 225 sq.cm

Solution

Side of another square = 2 × 15 = 30 cm Area of square is side2 = 30² = 900 sq.cm

Q11. The length of a side of a square whose area is 441 sq cm is

• 1) 21 cm
• 2) 11 cm
• 3) 29 cm
• 4) 31 cm

Solution

Area of square = 441 sq cm Thus, (side)² = 441 = (21)² Then, side = 21 cm

Q12.

• 1) 7 m
• 2) 9 m
• 3) 16 m
• 4) 18 m

Solution

Q13. The base of a parallelogram is 5 inches and its height is twice its base. Find the area of parallelogram.

• 1) 50 square inches
• 2) 40 square inches
• 3) 30 square inches
• 4) 60 square inches

Solution

Base = 5 inches Thus, height = 2 x 5 = 10 inches Hence, area of parallelogram = base x height = 5 x 10 = 50 square inches

Q14. The area of a parallelogram of dimension 18 cm x 12 cm  is 144 sq cm. The height of this parallelogram along the longer side is

• 1) 12 cm
• 2) 18 cm
• 3) 9 cm
• 4) 8 cm

Solution

Q15. Find the value of s in the figure given below if perimeter is 24 km:

Solution

We know that perimeter is obtained by adding the lengths of the edges of the given polygon. Here, perimeter is given as 24 km. Perimeter of the polygon = 6 + s + 8 + s 24 = 14 + 2s    [swap both sides] 14 + 2s = 24 2s = 24 - 14 2s = 10 s = = 5 Thus, the value of s is 5 km.

Q16. The altitude of a right triangle is twice the base. If the area is 36 m², what are the dimensions of this triangle?

Solution

Here, let us assume the base of the triangle be b. Given that , Altitude (h) = 2 x base (b) = 2b Also, area of the triangle = 36 m² Substituting the known values in the formula for area: Area =  x b x h 36 = x b x 2b b² = 36 = 6² or b = 6 Thus, the base of the triangle = 6 m And, the height of the triangle = 2 x 6 = 12 m

Q17. Two sides of the parallelogram PQRS (below) are 15 cm and 12 cm. The height corresponding to the base RQ is 7 cm. Find (i) area of the parallelogram (ii) height corresponding to base PQ

Solution

Area of the parallelogram is given by: Area = base (b) x height (h) base (b) = 15 cm ; height (h) = 7 cm (i) Area of the parallelogram = 15 x 7 = 105 cm² (ii) base (b) = 12 cm ; height = a (say), Area = 105 cm² Area of the parallelogram = b x a cm² 105 = 12 x a Thus, a = 8.75 cm So, the height corresponding to base PQ is 8.75 cm.

Q18. The area of a circular cardboard sheet is 154 cm². Find the radius of sheet.

Solution

Given that, Area of the circular cardboard sheet = 154 cm² Hence, the radius is 7 cm.

Q19. In the figure given below, the area of the parallelogram is 24 cm² and the base is 4 cm. Find the height.

Solution

Area of the parallelogram is given as 24 cm² base (b) = 4 cm ; Let the height be h cm Area = base (b) x height (h) 24 = 4 x h (swap both the sides)4 x h = 24 h = 6 cm Thus, the height of the parallelogram is 6 cm.

Q20. A rectangle whose area is 24 m² has a length that is 2 m longer than the width. What are the dimensions of the rectangle?

Solution

Given that area of the rectangle = 24 m² Let the width of the rectangle be 'b' m Thus, length (l) of the rectangle = (b + 2) m Area = length x width 24 = (b + 2) x b 24 = b² + 2b (using distributive law) b² + 2b = 24 (swap the left side with the right side) b² + 2b - 24 = 0 (b + 6) x (b - 4 ) = 0 b = -6 and b = 4 Since, the width of the rectangle cannot be negative, Width = 4 m and hence, length = 4 + 2 = 6 m

Q21. The circumference of a circle of diameter 10 cm is _____.

• 1) 31.4 cm
• 2) 314 cm
• 3) 3.14 cm
• 4) 0.314 cm

Solution

Circumference of circle = d = 3.14 x 10 = 31.4 cm

Q22. In the figure given below, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle CEF to that of the rectangle ABCD?

Solution

We know that, Rectangle is a quadrilateral with all four angles equal to 90^o. Now, since triangle CEF is inscribed inside the rectangle, we have Base = EF; height = BC Thus, Area of triangle CEF = x EF x BC And, area of rectangle ABCD = AB x BC But, given that, AE = EF = FB So, AB = 3 x EF = 3EF  ....(1) Thus, area of rectangle ABCD = 3EF x BC [from (1)] So, required ratio = = 1 : 6

Q23. Find the area of an equilateral triangle with side equal to 10 cm.

Solution

Let A, B and C be the vertices of the equilateral triangle and M the midpoint of segment BC. Since the triangle is equilateral, AMC is a right triangle.   Let us find h, the height of the triangle using Pythagorean theorem. h² + 5² = 10² h² = 100 - 25 h² = 75 h = cm We now find the area using the formula : Area = x b x h Thus, the area of the equilateral triangle is cm²

Q24. Find AB, if the area of the triangle ABC is 48 m² and the height CD is 12 m.    

Solution

Area of the triangle ABC = 48 m²   Height,CD = 12 m   Area =  x base x height   48= x AB x 12   6 x AB = 48   AB = 8 m   Thus, base AB = 8 m    

Q25. Find the area and circumference of a circular park whose radius is 20 cm.

Solution

Q26. Find the area of the given parallelogram.

Solution

Q27. A rectangular park is 38 m long and 15 m wide. A path 3.5 m wide is constructed outside the park. Find the outer perimeter of  the path.

Solution

The above data can be shown in a figure as follows:Let PQRS represent the rectangular park and the shaded region represent the path 3.5 m wide. Thus, to find the length AB and breadth BC, we have to add 3.5 m to both sides of rectangular park whose dimensions are 38 x 15. So, the length and breadth of the path are: Length, AB = (38 + 3.5 + 3.5) m = 45 m Breadth, BC = (15 + 3.5 + 3.5) m = 22 m So, outer perimeter of the path = 2 x (l + b) = 2 x (45 + 22) = 2 x 67 = 134 m Thus, outer perimeter of the path is 134 m.

Q28. Raju wants to fence the garden in his house which is in the shape of a rectangle having dimensions 25 m x 18 m. Find the cost of fencing at the rate of Rs 200 per metre.

Solution

The length of the fence required is the perimeter of the rectangular garden. Given that: Length (l) = 25 m; Breadth (b) = 18 m i.e.  Perimeter = 2 x (l + b)                     = 2 x (25 + 18)                      = 2 x 43                     = 86 m So, the perimeter of the garden or the length of the fence required is 86 m. Cost of fencing 1 m = Rs 200 Thus, Cost of fencing 86 m = Rs 200 x 86 = Rs 17,200

Q29. What is the area of the figure given below?

Solution

The above figure is in the form of a triangle with base = 4.2 cm and height = 3 cm. Use these numbers in the formula.   Area = x b x h   = x 4.2 x 3   = 2.1 x 3   = 6.3 cm² Thus, the area of the given figure is 6.3 cm²  

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