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Q 17.      Avneet buys 9 square paving slabs, each with a side of $\frac{1}{2}m$. He lays them in the form of a square.

(a) What is the perimeter of his arrangement ?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?

(c) Which has greater perimeter?

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this?

Given that, Length of each slab =  So, the length of the square =  (a) The first figure is a square So, the perimeter of the square = 4  side                                                    =  (b) The perimeter of the 2nd figure = Sum of all the sides =  = 10 m (c) Figure (ii) has a greater perimeter than fig (i) (d) Arrange all the slab in horizontally, it forms a rectangle whose...

Q 16.     What is the perimeter of each of the following figures? What do you infer from the answers?

(a) The perimeter of the square       = Sum of all the four side = 4   side       = 4   25       = 100 cm (b)The perimeter of the rectangle = SUm of all the sides = 2  [length + breadth ) = 2  (20 + 30) = 2  50 = 100 (c)The perimeter of the rectangle = Sum of all the sides = 2  [length + breadth ] = 2  (40 + 10) = 2  50 = 100 cm (d)The perimeter of the triangle = Sum of all three...

Q 15.     Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?

Given that, Length of the square park = 75m Length and breadth of the rectangular park is 60 m and 45m respectively Therefore, The perimeter of the square park  = 4  75  =300 m Also, the Perimeter of the rectangular park = 2 ( 60 + 45) =2 105m =210m Hence Bulbul covers the less distance.

Q 14.     Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of rupees 12 per metre.

Given that, The length of the rectangular park is 15 m and the breadth is 125m. Cost of fencing rate = Rs 12 per meter Therefore, Perimeter of the park =2 [L + B] = 2 [175 +125] =2  300m =600m So, the total cost of fencing = 600   12  = Rs. 7,200

Q 13.       Find the cost of fencing a square park of side 250 m at the rate of rupees 20 per metre.

Given that, Length of the square park  = 250 m Cost of fencing = rs. 20 per meter Therefore, the Perimeter of the square park = = 4  250 =1000 m = 1km So, total cost = 20   1000 = 20,000 Rs

Q 12.     Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Given that, The two sides of the triangle are 12 cm and 14 cm. let the third side be A. And the perimeter of the triangle is 36cm. Therefore, the Perimeter of the triangle   = sum of all sides of the triangle   = 12cm +14cm + A = 36cm   = A + 26cm =36cm   = A = 10cm

Q 11.     A piece of string is 30 cm long. What will be the length of each side if the string is used to form :

(a) a square?             (b) an equilateral triangle?             (c) a regular hexagon?

Given that, Length of the string is 30 cm (a)For square, Number of sides = 4 Therefore, Length of one side = 30 /4 = 7.5 cm (b)For equilateral triangle, Number of sides = 3 Therefore, length of sides = 30/3 = 10cm (c) For a regular hexagon, Number of sides = 6 Therefore, Length of one side = 30/6 = 5cm

Q 10.     The perimeter of a regular pentagon is 100 cm. How long is its each side?

Given, The perimeter of a regular pentagon is 100cm and, no. of sides in regular pentagon is five (5) therefore, the length of each side

Q 9.     Find the side of the square whose perimeter is 20 m.

Given that, The perimeter of the square is 20m According to question, The perimeter of square = 4  side

Q 8.     Find the perimeter of a regular hexagon with each side measuring 8 m.

It is known that, A regular hexagon has six equal sides. So, the perimeter of the regular hexagon   = 6  side   = 6  8m = 48m

Q 7.     Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

We all know that,  Perimeter of a triangle = sum of all the sides of the triangle                              =10cm + 14cm +15cm                              = 39 cm

Q 6.     Find the perimeter of each of the following shapes :

(a) A triangle of sides 3 cm, 4 cm and 5 cm.

(b) An equilateral triangle of side 9 cm.

(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

(a) The perimeter of a triangle = Sum of all the three sides                                                     = 3cm + 4cm + 5cm                                                      = 12 cm  (b) The perimeter of a eqilateral triangle       = Sum of all the sides of  the triangle        = (9cm +9cm +9cm)        = 27cm (c) Perimeter of a given isosceles triangle       = Sum of all...

Q 5.     A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

We have,   A rectangular piece of land measures 0.7 km by 0.5 km and side is to be fenced with 4 rows of wires. As per the question, Required length of wires = 4 2 [ 0.7 km + 0.5 km] = 8  1.2 km = 9.6 km

Q 4.     What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

We have, Length and breadth of a photograph is 32 cm and 21 cm respectively. So, According to question, Length of wooden strip required to frame the photograph = 2 [L + B]                                                                                             = 2 [32 + 21] cm                                                                                             = 2   53 cm  = 106...

Q 3.     A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

We have table-top measures 2 m 25 cm by 1 m 50 cm. Therefore, the Perimeter of the table-top = sum of all sides                                                                   = 2(L+B)                                                                  = 2 ((2m + 0.25m) + (1m + 0.5m))                                                                  = 7.5 m

Q 2.     The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

We have, Length = 40 cm and, Breadth = 10 cm So, the length of tape required = perimeter of the rectangular box                                                   = 2 (L + B)                                                   = 2 50 cm                                                   = 100 cm Hence 100 cm tape is required.

Q 1.     Find the perimeter of each of the following figures :

The perimeter of the following figures- Perimeter = sum of all sides  Therefore,  (a) Perimeter = 4 cm + 2cm +1 cm +5cm = 12 cm (b) Perimeter = 23 cm + 35 cm + 40cm + 35cm = 133 cm (c) Required perimeter = 15cm + 15 cm + 15 cm +15 cm = 60 cm (d) Required perimeter= (4 + 4 + 4 + 4 + 4) cm = 20 cm (e) Perimeter = 4cm + 0.5 cm +2.5cm +2.5cm + 0.5cm +4cm +1cm = 15cm (f) Perimeter = Sum of all...
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