CBSE Class 8 Maths Chapter 11: Mensuration NCERT Solutions

We are given that a square field and a rectangular field have the same perimeter. Let's find the dimensions and areas of both.

For the Square Field:

Side of the square = $60\; \backslash text\{\; m\}$

Perimeter of the square = $4\; \backslash times\; \backslash text\{side\}\; =\; 4\; \backslash times\; 60\; =\; 240\; \backslash text\{\; m\}$

Area of the square = $\backslash text\{side\}^2\; =\; (60)^2\; =\; 3600\; \backslash text\{\; m\}^2$

For the Rectangular Field:

Length of the rectangle = $80\; \backslash text\{\; m\}$

Perimeter of the rectangle = $2(l+b)$

According to the problem, the perimeter of the rectangular field is equal to the perimeter of the square field.

$$2(l+b)\; =\; 240$$

Substitute the given length ($l=80$):

$$2(80+b)\; =\; 240$$

$$160\; +\; 2b\; =\; 240$$

Subtract 160 from both sides:

$$2b\; =\; 240\; -\; 160$$

$$2b\; =\; 80$$

Divide by 2 to find the breadth:

$$b\; =\; \backslash frac\{80\}\{2\}\; =\; 40\; \backslash text\{\; m\}$$

Now, calculate the area of the rectangular field:

Area of the rectangle = $l\; \backslash times\; b\; =\; 80\; \backslash times\; 40\; =\; 3200\; \backslash text\{\; m\}^2$

Comparison:

Area of square field = $3600\; \backslash text\{\; m\}^2$

Area of rectangular field = $3200\; \backslash text\{\; m\}^2$

Since $3600\; >\; 3200$, the square field has a larger area.

Answer: The square field has a larger area.

First, we need to find the area of the square plot and the area of the house.

Area of the Square Plot:

Side of the square plot = $25\; \backslash text\{\; m\}$

Area of the square plot = $\backslash text\{side\}^2\; =\; (25)^2\; =\; 625\; \backslash text\{\; m\}^2$

Area of the House:

Length of the house = $20\; \backslash text\{\; m\}$

Breadth of the house = $15\; \backslash text\{\; m\}$

Area of the house = $length\; \backslash times\; breadth\; =\; 20\; \backslash times\; 15\; =\; 300\; \backslash text\{\; m\}^2$

Area of the Garden:

The garden is developed around the house within the plot. So, the area of the garden is the difference between the area of the plot and the area of the house.

Area of garden = Area of square plot – Area of house

$$=\; 625\; \backslash text\{\; m\}^2\; -\; 300\; \backslash text\{\; m\}^2\; =\; 325\; \backslash text\{\; m\}^2$$

Total Cost of Developing the Garden:

The cost of developing the garden is given at the rate of ₹ 55 per m².

Total cost = Area of garden × Rate per m²

$$=\; 325\; \backslash text\{\; m\}^2\; \backslash times\; ₹55/\backslash text\{m\}^2$$

$$=\; ₹17,875$$

Answer: The total cost of developing the garden around the house is ₹ 17,875.

The garden consists of a rectangular part in the middle and two semi-circular parts at the ends. The diameter of the semi-circular ends is given as 7 m.

Dimensions:

Diameter of the semi-circles = $7\; \backslash text\{\; m\}$

Radius of the semi-circles = $\backslash frac\{diameter\}\{2\}\; =\; \backslash frac\{7\}\{2\}\; =\; 3.5\; \backslash text\{\; m\}$

The length of the rectangular part is the total length of the garden minus the radii of the two semi-circles at the ends.

Length of the rectangle = Total length - (radius of left semi-circle + radius of right semi-circle)

$$=\; 20\; \backslash text\{\; m\}\; -\; (3.5\; \backslash text\{\; m\}\; +\; 3.5\; \backslash text\{\; m\})$$

$$=\; 20\; \backslash text\{\; m\}\; -\; 7\; \backslash text\{\; m\}\; =\; 13\; \backslash text\{\; m\}$$

The breadth of the rectangle is equal to the diameter of the semi-circles, which is 7 m.

Breadth of the rectangle = $7\; \backslash text\{\; m\}$

Area of the Garden:

Area of the garden = Area of the rectangular part + Area of the two semi-circular parts.

Area of the rectangle = $length\; \backslash times\; breadth\; =\; 13\; \backslash times\; 7\; =\; 91\; \backslash text\{\; m\}^2$

The two semi-circular parts together form a full circle with radius $3.5\; \backslash text\{\; m\}$.

Area of the circle (two semi-circles) = $\backslash pi\; r^2\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (3.5)^2\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 12.25\; \backslash text\{\; m\}^2$

$$=\; 22\; \backslash times\; 1.75\; =\; 38.5\; \backslash text\{\; m\}^2$$

Total Area of the garden = Area of rectangle + Area of circle

$$=\; 91\; \backslash text\{\; m\}^2\; +\; 38.5\; \backslash text\{\; m\}^2\; =\; 129.5\; \backslash text\{\; m\}^2$$

Perimeter of the Garden:

The perimeter of the garden consists of the two straight sides of the rectangle and the curved boundaries of the two semi-circles.

Perimeter = Length of rectangle + Length of rectangle + Circumference of the two semi-circles

Perimeter = $13\; \backslash text\{\; m\}\; +\; 13\; \backslash text\{\; m\}\; +\; (\backslash text\{Circumference\; of\; a\; circle\; with\; radius\; \}\; 3.5\; \backslash text\{\; m\})$

Circumference of the circle = $2\; \backslash pi\; r\; =\; 2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; 3.5\; =\; 2\; \backslash times\; 22\; \backslash times\; 0.5\; =\; 22\; \backslash text\{\; m\}$

Total Perimeter = $13\; \backslash text\{\; m\}\; +\; 13\; \backslash text\{\; m\}\; +\; 22\; \backslash text\{\; m\}\; =\; 48\; \backslash text\{\; m\}$

Answer: The area of the garden is $129.5\; \backslash text\{\; m\}^2$ and the perimeter is $48\; \backslash text\{\; m\}$.