Mix Examples - Areas Related to Circles Questions in English

(A) Based on the definitions of circle parts:
$1.$ $\overline{OA} \cup \overline{OB} \cup \widehat{APB}$ represents the region bounded by two radii and the minor arc,which is the $Minor \ sector$ $(c)$.
$2.$ $\overline{AB} \cup \widehat{AQB}$ represents the region bounded by a chord and the major arc,which is the $Major \ segment$ $(d)$.
$3.$ $\overline{AB} \cup \widehat{APB}$ represents the region bounded by a chord and the minor arc,which is the $Minor \ segment$ $(b)$.
$4.$ $\overline{OA} \cup \overline{OB} \cup \widehat{AQB}$ represents the region bounded by two radii and the major arc,which is the $Major \ sector$ $(a)$.
Therefore,the correct matching is $(1-c), (2-d), (3-b), (4-a)$.

(A) The area of the circle $\odot(O, r)$ is given as $\pi r^2 = 240 \, cm^2$.
The formula for the area of a sector with central angle $\theta$ is $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Given $\theta = 45^\circ$ and $\pi r^2 = 240 \, cm^2$,we substitute these values into the formula:
$\text{Area of minor sector } OACB = \frac{45}{360} \times 240$
$\text{Area of minor sector } OACB = \frac{1}{8} \times 240 = 30 \, cm^2$.
Therefore,the area of the minor sector is $30 \, cm^2$.

(D) Given, the radius of the circle $r = 12 \text{ cm}$ and the central angle subtended by the minor arc $\widehat{ACB}$ is $\theta = 30^{\circ}$.
The circumference of the circle is $C = 2\pi r = 2 \times \pi \times 12 = 24\pi \text{ cm}$.
The length of the minor arc $\widehat{ACB}$ is given by the formula $L_{\text{minor}} = \frac{\theta}{360^{\circ}} \times 2\pi r$.
$L_{\text{minor}} = \frac{30^{\circ}}{360^{\circ}} \times 24\pi = \frac{1}{12} \times 24\pi = 2\pi \text{ cm}$.
The length of the major arc $\widehat{ADB}$ is the total circumference minus the length of the minor arc.
$L_{\text{major}} = 24\pi - 2\pi = 22\pi \text{ cm}$.

(D) Given: Radius $r = 6.3\, cm$ and central angle $\theta = 40^{\circ}$.
The formula for the area of a minor sector is $\text{Area} = \frac{\pi r^{2} \theta}{360^{\circ}}$.
Substituting the values,we get:
$\text{Area} = \frac{22}{7} \times (6.3)^{2} \times \frac{40^{\circ}}{360^{\circ}}$
$\text{Area} = \frac{22}{7} \times 6.3 \times 6.3 \times \frac{1}{9}$
$\text{Area} = 22 \times 0.9 \times 0.7$
$\text{Area} = 13.86\, cm^{2}$.

(A) The area of a sector of a circle is given by the formula $A = \frac{1}{2} r l$,where $r$ is the radius and $l$ is the length of the arc.
Given: $r = 10 \, cm$ and $A = 75 \, cm^2$.
Substituting the values into the formula:
$75 = \frac{1}{2} \times 10 \times l$
$75 = 5 \times l$
$l = \frac{75}{5}$
$l = 15 \, cm$.
Therefore,the length of the arc is $15 \, cm$.

(B) The formula for the length of an arc is $l = \frac{\pi r \theta}{180^{\circ}}$,where $l$ is the arc length,$r$ is the radius,and $\theta$ is the central angle in degrees.
Given: $l = 110 \,cm$ and $\theta = 150^{\circ}$.
Substituting the values into the formula:
$110 = \frac{22}{7} \times \frac{r \times 150}{180}$
$110 = \frac{22}{7} \times \frac{r \times 5}{6}$
$110 = \frac{110 \times r}{42}$
$r = \frac{110 \times 42}{110}$
$r = 42 \,cm$
Therefore,the radius of the circle is $42 \,cm$.

(A) The length of an arc is given by the formula $L = \frac{\theta}{360^\circ} \times 2\pi r$.
Here, the radius $r = 8 \, cm$ and the central angle $\theta = 45^\circ$.
Substituting these values into the formula:
$L = \frac{45^\circ}{360^\circ} \times 2 \times \pi \times 8$
$L = \frac{1}{8} \times 16\pi$
$L = 2\pi \, cm$.
Therefore, the length of the minor arc $\widehat{ACB}$ is $2\pi \, cm$.

(B) The circumference of a circle is given by $C = 2\pi r$,where $r$ is the radius.
The length of an arc $L$ subtending an angle $\theta$ at the centre is given by $L = \frac{\theta}{360^\circ} \times 2\pi r$.
According to the problem,the length of the arc is one-eighth of the circumference:
$L = \frac{1}{8} \times 2\pi r$.
Equating the two expressions for $L$:
$\frac{\theta}{360^\circ} \times 2\pi r = \frac{1}{8} \times 2\pi r$.
Dividing both sides by $2\pi r$:
$\frac{\theta}{360^\circ} = \frac{1}{8}$.
Solving for $\theta$:
$\theta = \frac{360^\circ}{8} = 45^\circ$.
Thus,the measure of the angle subtended at the centre is $45^\circ$.

(C) Given that the radius of the circle $r = OA = OB = 4 \, cm$.
The length of the chord $AB = 4 \, cm$.
In $\triangle AOB$,we have $OA = OB = AB = 4 \, cm$.
Since all three sides of the triangle are equal,$\triangle AOB$ is an equilateral triangle.
In an equilateral triangle,all internal angles are equal to $60^\circ$.
Therefore,$m \angle AOB = 60^\circ$.

(C) Let the original radius of the circle be $r$.
The original area of the circle is $A_1 = \pi r^2$.
If the radius is increased by $10 \%,$ the new radius $r'$ becomes:
$r' = r + 0.10r = 1.1r$.
The new area of the circle $A_2$ is given by:
$A_2 = \pi (r')^2 = \pi (1.1r)^2 = \pi (1.21r^2) = 1.21 \pi r^2$.
Therefore,the area of the new circle is $1.21 \pi r^2$.

(C) Let the radius of the semicircle be $r = 10 \, cm$.
For a triangle inscribed in a semicircle,the base of the triangle is the diameter of the semicircle,which is $d = 2r = 2 \times 10 = 20 \, cm$.
The height of the triangle is the perpendicular distance from the diameter to the circumference,which is maximum when the vertex lies at the midpoint of the arc,making the height equal to the radius $r = 10 \, cm$.
The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Substituting the values: $\text{Area} = \frac{1}{2} \times 20 \times 10 = 100 \, cm^2$.
Thus,the maximum area is $100 \, cm^2$.

(A) The length of the minute hand is the radius of the circle,$r = 14 \, cm$.
When the minute hand moves from $1$ to $10$,it covers $9$ divisions out of $12$ on the clock face.
The angle subtended by the minute hand at the center is $\theta = \frac{9}{12} \times 360^\circ = \frac{3}{4} \times 360^\circ = 270^\circ$.
The area covered by the minute hand is the area of the sector with angle $\theta = 270^\circ$.
Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$.
Area $= \frac{270}{360} \times \frac{22}{7} \times 14 \times 14$.
Area $= \frac{3}{4} \times 22 \times 2 \times 14$.
Area $= 3 \times 11 \times 14 = 462 \, cm^2$.

(C) The length of an arc $L$ is given by the formula $L = \frac{\theta}{360^{\circ}} \times 2\pi r$,where $\theta$ is the angle subtended at the centre and $2\pi r$ is the circumference of the circle.
Given that the length of the minor arc $\widehat{AB}$ is $\frac{1}{4}$ of the circumference,we have:
$L = \frac{1}{4} \times (2\pi r)$
Equating the two expressions for $L$:
$\frac{\theta}{360^{\circ}} \times 2\pi r = \frac{1}{4} \times 2\pi r$
Dividing both sides by $2\pi r$:
$\frac{\theta}{360^{\circ}} = \frac{1}{4}$
$\theta = \frac{360^{\circ}}{4} = 90^{\circ}$
Therefore,the measure of the angle subtended by the minor arc $\widehat{AB}$ at the centre is $90^{\circ}$.

(A) Given,the area of the circle $A = 38.5\, m^2$.
The formula for the area of a circle is $A = \pi r^2$.
Substituting the value of $\pi = \frac{22}{7}$,we get:
$38.5 = \frac{22}{7} \times r^2$
$r^2 = \frac{38.5 \times 7}{22}$
$r^2 = \frac{269.5}{22} = 12.25$
$r = \sqrt{12.25} = 3.5\, m$.
The circumference of the circle is given by $C = 2\pi r$.
$C = 2 \times \frac{22}{7} \times 3.5$
$C = 2 \times 22 \times 0.5 = 22\, m$.
Therefore,the circumference is $22\, m$.

(A) The correct matches are as follows:
$1.$ The length of a minor arc $(l)$ is given by the formula $l = \frac{\theta}{360} \times 2\pi r = \frac{\pi r \theta}{180}$. Thus,$1$ matches with $c$.
$2.$ The area of a minor sector $(A)$ is given by the formula $A = \frac{\theta}{360} \times \pi r^{2}$. Thus,$2$ matches with $d$.
$3.$ The area of a circle $(A)$ is given by the formula $A = \pi r^{2}$. Thus,$3$ matches with $b$.
$4.$ The circumference of a circle $(C)$ is given by the formula $C = 2\pi r$. Thus,$4$ matches with $a$.
Therefore,the correct matching is $(1-c), (2-d), (3-b), (4-a)$.