NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.3#

$$\text{Area} = \dfrac{60}{360} \times \dfrac{22}{7} \times 7^2$$

$$\text{Area} = \dfrac{1}{6} \times 154$$

$$\text{Area} = \dfrac{77}{3}$$

∴ The area of the sector is $\dfrac{77}{3}$ cm², that is, about 25.67 cm².

First we find the radius. The circumference is 44 cm, so:

$$2 \times \dfrac{22}{7} \times r = 44$$

$$\Rightarrow r = 7 \text{ cm}$$

A quadrant is a quarter of the circle:

$$\text{Area} = \dfrac{1}{4} \times \dfrac{22}{7} \times 7^2$$

$$\text{Area} = \dfrac{1}{4} \times 154$$

$$\text{Area} = 38.5$$

∴ The area of the quadrant is 38.5 cm².

The minute hand turns through 360° in 60 minutes, so in 10 minutes it turns through:

$$\theta = \dfrac{10}{60} \times 360°$$

$$\theta = 60°$$

It sweeps a sector of radius 7 cm and angle 60°:

$$\text{Area} = \dfrac{60}{360} \times \dfrac{22}{7} \times 7^2$$

$$\text{Area} = \dfrac{77}{3}$$

∴ The area swept is $\dfrac{77}{3}$ cm², that is, about 25.67 cm².

$$\text{Area} = \dfrac{90}{360} \times 3.14 \times 10^2$$

$$\text{Area} = \dfrac{1}{4} \times 314$$

$$\text{Area} = 78.5$$

∴ The minor sector has area 78.5 cm².

$$\text{Area} = \dfrac{270}{360} \times 3.14 \times 10^2$$

$$\text{Area} = \dfrac{3}{4} \times 314$$

$$\text{Area} = 235.5$$

∴ The major sector has area 235.5 cm². (Check: $78.5 + 235.5 = 314$, the whole circle.)

Let the chord be $AB$ and the centre $O$. $OA = OB = 15$ cm and $\angle AOB = 60°$, so $\triangle OAB$ is equilateral with side 15 cm.

Area of the minor sector:

$$\text{Sector} = \dfrac{60}{360} \times 3.14 \times 15^2$$

$$\text{Sector} = \dfrac{1}{6} \times 706.5$$

$$\text{Sector} = 117.75 \text{ cm}^2$$

Area of the equilateral triangle, using $\dfrac{\sqrt{3}}{4}a^2$:

$$\text{Triangle} = \dfrac{1.73}{4} \times 15^2$$

$$\text{Triangle} = \dfrac{1.73 \times 225}{4}$$

$$\text{Triangle} = 97.3125 \text{ cm}^2$$

Minor segment:

$$\text{Minor segment} = 117.75 - 97.3125$$

$$\text{Minor segment} = 20.4375 \text{ cm}^2$$

Major segment = whole circle − minor segment:

$$\text{Major segment} = 3.14 \times 15^2 - 20.4375$$

$$\text{Major segment} = 706.5 - 20.4375$$

$$\text{Major segment} = 686.0625 \text{ cm}^2$$

∴ The minor segment is about 20.44 cm² and the major segment is about 686.06 cm².

Each blade sweeps a sector of radius 28 cm and angle 120°:

$$\text{One sector} = \dfrac{120}{360} \times \dfrac{22}{7} \times 28^2$$

$$\text{One sector} = \dfrac{1}{3} \times 2464$$

$$\text{One sector} = \dfrac{2464}{3} \text{ cm}^2$$

The wipers do not overlap, so the total is twice this:

$$\text{Total} = 2 \times \dfrac{2464}{3}$$

$$\text{Total} = \dfrac{4928}{3}$$

∴ The total area cleaned is $\dfrac{4928}{3}$ cm², that is, about 1642.67 cm².

The chord and the two radii form a triangle with two sides $r$ and an angle of 60° between them, so it is equilateral with side $r$.

$$\text{Sector} = \dfrac{60}{360} \times \pi r^2$$

$$\text{Sector} = \dfrac{1}{6}\pi r^2$$$$\text{Triangle} = \dfrac{\sqrt{3}}{4}r^2$$$$\text{Minor segment} = \dfrac{1}{6}\pi r^2 - \dfrac{\sqrt{3}}{4}r^2$$

$$\text{Minor segment} = r^2\left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)$$

This is the correct area. The printed answer $\pi r^2\left(\dfrac{1}{6} - \dfrac{\sqrt{3}}{4}\right)$ has a misprint: it multiplies $\dfrac{\sqrt{3}}{4}$ by $\pi$ as well, which would make the segment negative ($\dfrac{1}{6} < \dfrac{\sqrt{3}}{4}$). The intended result is

$$\text{Minor segment} = r^2\left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right) \approx 0.0906\,r^2$$

Join the centre $O$ to the three vertices. This splits the triangle into three equal triangles, each with two sides $r$ and an angle of $\dfrac{360°}{3} = 120°$ at $O$.

Let one small triangle be $OAB$ with $\angle AOB = 120°$, and let $M$ be the midpoint of $AB$. Then $OM \perp AB$, $\angle AOM = 60°$, so $\triangle OMA$ is half of an equilateral triangle of side $r$:

$$OM = \dfrac{r}{2}$$

$$AM = \dfrac{\sqrt{3}}{2}r$$

So the side of the big triangle is $AB = 2AM = \sqrt{3}\,r$, and its area is:

$$\text{Triangle} = \dfrac{\sqrt{3}}{4}(\sqrt{3}\,r)^2$$

$$\text{Triangle} = \dfrac{3\sqrt{3}}{4}r^2$$$$\dfrac{\text{Triangle}}{\text{Circle}} = \dfrac{\dfrac{3\sqrt{3}}{4}r^2}{\pi r^2}$$

$$\dfrac{\text{Triangle}}{\text{Circle}} = \dfrac{3\sqrt{3}}{4\pi} \approx 0.413$$

Hence shown.

The diagonals of the inscribed square are diameters of the circle, so each diagonal is $2r$. The area of a square (a rhombus) is half the product of its diagonals:

$$\text{Square} = \dfrac{1}{2} \times 2r \times 2r$$

$$\text{Square} = 2r^2$$$$\dfrac{\text{Square}}{\text{Circle}} = \dfrac{2r^2}{\pi r^2}$$

$$\dfrac{\text{Square}}{\text{Circle}} = \dfrac{2}{\pi} \approx 0.637$$

Hence shown.

A regular hexagon inscribed in a circle of radius $r$ is made of six equilateral triangles of side $r$ (Chapter 5):

$$\text{Hexagon} = 6 \times \dfrac{\sqrt{3}}{4}r^2$$

$$\text{Hexagon} = \dfrac{3\sqrt{3}}{2}r^2$$$$\dfrac{\text{Hexagon}}{\text{Circle}} = \dfrac{\dfrac{3\sqrt{3}}{2}r^2}{\pi r^2}$$

$$\dfrac{\text{Hexagon}}{\text{Circle}} = \dfrac{3\sqrt{3}}{2\pi} \approx 0.827$$

Why it is twice the answer to Question 8: label the hexagon’s vertices $1$ to $6$ and join vertices $1$, $3$, $5$. This gives the inscribed equilateral triangle of Question 8. The hexagon is this triangle plus three outer triangles ($1$-$2$-$3$, $3$-$4$-$5$, $5$-$6$-$1$). Each outer triangle is congruent to one of the three pieces $O$-$1$-$3$, $O$-$3$-$5$, $O$-$5$-$1$ that make up the big triangle (each is a rhombus $O$-$1$-$2$-$3$ cut along its diagonal $1$-$3$). So the outer triangles together have the same area as the big triangle, and

$$\text{Hexagon} = 2 \times \text{Triangle}$$