The given figure consists of a semicircle and a rectangle. Given $GF = DC = 4$ cm and $AB = 22$ cm, we are asked to find: (i) The radius of the semicircle. (ii) The arc length of the semicircle. (iii) The length of $BC$, given that the perimeter of the whole figure is 76 cm. (iv) The area of the whole figure. (v) The number of bulbs needed to fix along the part $CDEFG$ with a gap of 2 cm.

1. Problem Description

The given figure consists of a semicircle and a rectangle. Given $GF = DC = 4$ cm and $AB = 22$ cm, we are asked to find:

(i) The radius of the semicircle.

(ii) The arc length of the semicircle.

(iii) The length of $BC$, given that the perimeter of the whole figure is 76 cm.

(iv) The area of the whole figure.

(v) The number of bulbs needed to fix along the part $CDEFG$ with a gap of 2 cm.

2. Solution Steps

(i) The radius of the sector (semicircle) is half of the diameter $GF$ or $DC$.

Since $GF = DC = 4$ cm, the radius $r$ is:

$r = \frac{4}{2} = 2$ cm

(ii) The arc length of the sector (semicircle) is half the circumference of a circle with radius $r$.

The formula for the circumference of a circle is $C = 2\pi r$.

The arc length $L$ of the semicircle is:

$L = \frac{1}{2} (2\pi r) = \pi r$

Since $r = 2$ cm, the arc length is:

$L = \pi (2) = 2\pi$ cm

Using $\pi \approx 3.14$, $L \approx 2(3.14) = 6.28$ cm

(iii) The perimeter of the whole figure is given as 76 cm.

The perimeter consists of the lengths $AB$, $BC$, $GA$, and the arc length $CDE$.

$Perimeter = AB + BC + GA + Arc \; length \; CDE$

$AB = 22$ cm, $GA = BC$, and the arc length $CDE = 2\pi$ cm $\approx 6.28$ cm.

We also know that $GF = 4$ cm, so $GD = 22 - 4 -4 = 14$.

So $CD+DE+EF+FG = CD+2\pi+FG = 76$.

$22+BC+BC+6.28 = 76$.

$44 + 2BC = 76 - 6.28$.

$2BC = 69.72 - 22 = 47.72$.

$BC = \frac{47.72}{2} = 23.86$.

The perimeter can also be written as:

$Perimeter = AB + BC + AG + \text{Arc length } CDE$

$76 = 22 + BC + BC + 2\pi$

$76 = 22 + 2BC + 2(3.14)$

$76 = 22 + 2BC + 6.28$

$76 - 22 - 6.28 = 2BC$

$47.72 = 2BC$

$BC = \frac{47.72}{2} = 23.86$ cm

(iv) The area of the whole figure is the sum of the area of the rectangle and the area of the semicircle.

Area of rectangle $= AB \times BC = 22 \times 23.86 = 524.92 \text{ cm}^2$

Area of semicircle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2^2) = \frac{1}{2} \pi (4) = 2\pi \approx 2(3.14) = 6.28 \text{ cm}^2$

Total area $= 524.92 + 6.28 = 531.2 \text{ cm}^2$

(v) Length of $CDEFG = CD + DE + EF + FG = DC + Arc \;Length \;E + GF$

$CD = GF = 4$ cm, $DE = EF$ is not relevant for length, Arc length $DE = \pi r = 2\pi$. Total $= CD + ArcLength + FG = 4 + 6.28 + 4 = 14.28 \text{ cm}$.

Gap between bulbs is 2 cm. Let $n$ be the number of bulbs. Then, $(n-1) \times 2$ is less than or equal to 14.

2

8. Then, $CDEFG \approx 14.28$. The number of gaps is $n-1$, so $2(n-1) = 14.28$. This is incorrect. Since they are placed every 2cm, we can divide the total length $CDEFG$ by the gap distance to find the approximate number of gaps:

$\frac{14.28}{2} \approx 7.14$.

Then $7.14 + 1 = 8.14$. It will require 8 bulbs.

However, we must consider $14.28 /2 = 7.14$. So, we place the first bulb at 0cm.

The second bulb is at 2cm. The third bulb is at 4cm. The fourth bulb is at 6cm. The fifth bulb is at 8cm. The sixth bulb is at 10cm. The seventh bulb is at 12cm. And the eighth bulb is at 14cm.

3. Final Answer

(i) The radius of the sector is 2 cm.

(ii) The arc length of the sector is $2\pi \approx 6.28$ cm.

(iii) The length of $BC$ is 23.86 cm.

(iv) The area of the whole figure is approximately 531.2 $\text{ cm}^2$.

(v) The total number of bulbs that needs is

8.