# Area & Preimeter

## Exercise 11.4

**Question 1:** A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

**Solution:**Given; Length of the garden = 90m, Width of the garden = 75m, Width of the path around the garden = 5m

Area of rectangle = Length `xx` Width

`= 90 xx75= 6750` m^{2}

We know that, 10000 m^{2} = 1 hectare

Hence; 6750 m^{2} `= 6750 ÷ 10000 = 0.675` hectare

Length of the garden with path `= 90 m + 5m + 5m = 100 m`

Width of the garden with path `= 75 m + 5m + 5m = 95 m`

Area of the garden with path = Length `xx` Width

`= 100xx95= 9500` m^{}

Now, Area of the path = Area of the garden with path — Area of the garden

= 9500 m^{2} — 6750 m^{2} = 2750 m^{2}

Hence, Area of the path = 2750 m^{2}, and area of the garden = 0.675 hectare

**Question 2:** A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.

**Solution:** Given, Length of the park = 125m, Width of the park = 65m, Width of the path around the park = 3m

Therefore, Length of the park with path = 125m + 3m + 3m = 131m

Width of the park with park = 65m + 3m + 3m = 71m

Area of a rectangle = Length `xx` Width

Area of the park `= 125 xx 65 = 8125` m^{2}

Area of the park with path = Length `xx` Width

`= 131xx71 = 9301` m^{2}

Now, Area of the path = Area of the park with path — Area of the park

= 9301m^{2} — 8125m^{2} = 1176 m^{2}

**Question 3:** A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

**Solution:** Given, Length of the cardboard = 8cm, Width of the cardboard = 5cm, Margin left along each side of cardboard = 1.5cm

Hence, Length of the cardboard without margin = 8cm — 1.5cm — 1.5cm = 5cm

Width of the cardboard without margin = 5cm — 1.5cm — 1.5cm = 2cm

Area of the cardboard = Length `xx` Width

`= 8xx5= 40` cm^{2}

Area of the cardboard without margin = Length × Width

`= 5xx2= 10` cm^{2}

Therefore, Area of margin = Area of the cardboard — Area of the cardboard without margin

= 40cm^{2} — 10cm^{2}= 30cm^{2}

**Question 4:** A verandah of width 2.25m is constructed all along outside a room which is 5.5m long and 4m wide. Find

(i) The area of the verandah.

(ii) The cost of cementing the floor of the verandah at the rate of Rs 200 per m^{2}.

**Solution:** Given, Length of room = 5.5m, Width of room = 4m, Width of verandah which is constructed all along outside of room = 2.25m, Cost of cementing of verandha = Rs 200 per m^{2}

Therefore, Length of room with verandah `= 5.5m + 2.25m + 2.25m = 10m`

Width of room with verandah `= 4m + 2.25m + 2.25m = 8.5m`

Area of room = Length `xx` width

`= 5.5 xx 4= 22.0` m^{2}

Area of room with verandah = Length `xx` Width

`=10xx8.5 = 85` m^{2}

Area of verandah = Area of room with verandah — Area of room

= 85m^{2} —22m^{2} = 63 m^{2}

Cost of cementing of verandah = Area × Rate

`Rs. 200 xx 63 = Rs. 12600`