Free Areas Related to Circles 02 Practice Test - 10th Grade 

The circumference of a circle with radius r is given by
$2\pi r.$

Hence,
$2\pi r=528$

$r=\frac{528}{2\pi }=84cm$

$(\pi =\frac{22}{7})$

Here we need to find the area of the shaded part of the square. For finding the area of the shaded region we can find the area of the bigger square and the smaller square and then subtract the area of the smaller square from area of the larger square.

Area of the larger square =$x^2$
                                     = $8^2$
                                     = 64$c{m}^2$

Area of the smaller square = $b^2$
                                       = $4^2$
                                       = 16 $c{m}^2$

The area of the shaded region will be = Area of the bigger square – Area of the smaller square
                                 = 64 – 16
                                 = 48 $c{m}^2$
Therefore the area of the shaded region will be 48$c{m}^2$.

Area of a circle of radius $r$
$=\pi r^2$

Area
$=\pi$ x $\frac{7}{\sqrt{\pi }}$ x $\frac{7}{\sqrt{\pi }}=49{cm}^2$

The area of the sector = $\frac{120}{360}\times \pi \times {12}^2$ = 150.8 $c{m}^2$

Perpendicular from chord is drawn to the centre bisects the chord. The angle subtended by each triangle at the centre is ${60}^{\circ }$.

Height of perpendicular =  r x cos 60${}^{\circ }$ = 12 x 0.5 = 6 cm.

Length of chord = 2 × r × sin 60${}^{\circ }$ = 24 $\times \frac{\sqrt{3}}{2}$= 20.6 cm

The area of triangle = 0.5 × 20.6 × 6 = 61.8 $c{m}^2$

The area of segment = 150.8 - 61.8 = 89 $c{m}^2$

Circumference of the circle
$=2\pi r=528cm$ 
(where r is the radius of the circle)

$\Rightarrow r=\frac{528}{2\pi }=84cm$

Now, area of the circle = $\pi r^2$

 = $\frac{22}{7}$ x ${84}^2$ = 22176 $c{m}^2$

Area of a circle
$=\pi r^2$
 
From Figure, the diameter of circle is 14 cm. Two semi-circles make one full circle.

$\therefore$ The area of one full circle is
$=\frac{22}{7}\times 7^2=154c{m}^2$

The total area of square
$={14}^2=196c{m}^2$

The area of shaded portion = [Area of square- Area of full circle]
= 196 - 154 = $42c{m}^2$. 
Hence, area of shaded region 
$=42c{m}^2$

Area of shaded region = area of $\Delta$ABC - 3 (Area of sector BPR)

Let 'a' be the side of the equilateral $\Delta$ABC.

Using area of an equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$,
$\frac{\sqrt{3}}{4}{a}^2$ = 17320.5
Solving, $a^2=\frac{17320.5\times 4}{\sqrt{3}}$
$⟹a^2=\frac{17320.5\times 4}{1.73205}$
$⟹a^2=\frac{17320.5\times 4}{17320.5\times {10}^{-4}}$
$⟹a=2\times {10}^2$
$⟹$ a = 200 cm.

Radius of the circles = $\frac{1}{2}\times 200$ = 100 cm

Now, using area of a sector when the degree measure of the angle at the centre is $\theta$ = $\frac{\theta }{360}\pi r^2$
$\therefore$ Required area =17320.5 - 3[$\frac{60}{360}\times 3.14\times {100}^2$ ]

$\therefore$ Required area = 1620.5$c{m}^2$

Area of a square plate
$={\text{side}}^2$ 

Given length of the side of the square plate = 40 cm
Area of square plate
$={40}^2$
$=1600c{m}^2$

Area of a circle  
$=\pi r^2$ 
There are 336 holes of radius 1 cm each.

Total area of circles
$=336\times \frac{22}{7}\times 1^2$
 $=1056c{m}^2$

Remaining area =  [Area of square plate- Total area of circles]
$=1600-1056$
$=544c{m}^2$

$\therefore$ Area of remaining square plate
 $=544c{m}^2$

Area of the sector of angle  $\theta$ = $\frac{\theta }{360}$ × $\pi r^2$

A = $\frac{60}{360}$ × 3.14 $\times \left(6^2\right)$ = 18.85 $c{m}^2$

Perimeter of sector = 2r + Length of the arc of sector

From the given data,

Length of arc of sector = 27.2 - 2 × 5.7

Length of arc of sector = 27.2 - 11.4 = 15.8 m.