Areas Related to Circles: MCQs with Answers

A segment is the region bounded by a chord and its corresponding arc.

Hence the required region is a segment.

A quadrant has central angle 90°.

Its area is $\frac{\text{90}}{\text{360}}$ = $\frac{\text{1}}{\text{4}}$ of the circle’s area.

Hence the required ratio is 1 : 4.

The shaded region is bounded by a chord and its minor arc.

It is therefore the minor segment.

Let the radius be r units.

2πr = πr².

Dividing by πr, since r > 0, gives 2 = r.

Therefore, r = 2 units.

Let the circle have radius r and the square have side a.

4a = 2πr, so a = $\frac{\text{πr}}{\text{2}}$.

Circle area : square area = πr² : $\frac{\text{π²r²}}{\text{4}}$.

= 4 : π.

Using π = $\frac{\text{22}}{\text{7}}$, the ratio is 28 : 22 = 14 : 11.

Area of a circle = πr².

Area ratio = 4² : 3².

= 16 : 9.

The inscribed circle touches each side of the square.

Its diameter therefore equals the side of the square, 6 cm.

Radius of the circle = $\frac{\text{6}}{\text{2}}$ = 3 cm.

Using area = πr², the required area is π × 3 × 3.

Therefore, the area is 9π cm².

πr² = 154.

r² = $\frac{\text{154 × 7}}{\text{22}}$ = 49.

Therefore, r = 7 cm.

For the two circles, C₁ = 2πr₁ and C₂ = 2πr₂.

Thus C₁ : C₂ = r₁ : r₂.

The given circumference ratio therefore gives r₁ : r₂ = 3 : 4.

Their areas are in the ratio πr₁² : πr₂².

Cancelling π gives 3² : 4² = 9 : 16.

Length of the arc = $\frac{\text{56}}{\text{360}}$ × 2π × 4.5.

= $\frac{\text{56}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 4.5.

= 4.4 cm.

Distance travelled = 11 × 1000 × 100 = 1100000 cm.

Distance in one revolution = $\frac{\text{1100000}}{\text{5000}}$ = 220 cm.

Circumference = πd.

$\frac{\text{22}}{\text{7}}$ × d = 220.

Therefore, d = 70 cm.

Length of the arc = $\frac{\text{60}}{\text{360}}$ × 2π × 21.

= $\frac{\text{60}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 21.

= 22 cm.

Length of an arc = $\frac{\text{θ}}{\text{360}}$ × 2πr.

10π = $\frac{\text{θ}}{\text{360}}$ × 2π × 12.

θ = $\frac{\text{10π × 360}}{\text{2π × 12}}$.

Therefore, θ = 150°.

Radius = $\frac{\text{84}}{\text{2}}$ = 42 cm.

88 = $\frac{\text{θ}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 42.

θ = $\frac{\text{88 × 360}}{\text{264}}$.

Therefore, θ = 120°.

$\frac{\text{Arc length}}{\text{Circumference}}$ = $\frac{\text{θ}}{\text{360}}$.

= $\frac{\text{90}}{\text{360}}$ = $\frac{\text{1}}{\text{4}}$.

Hence the ratio is 1 : 4.

Arc length = $\frac{\text{40}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 9.

= $\frac{\text{44}}{\text{7}}$ cm.

11 = $\frac{\text{105}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × r.

r = $\frac{\text{11 × 360 × 7}}{\text{105 × 44}}$.

Therefore, r = 6 cm.

The pendulum length is the radius of its circular path.

88 = $\frac{\text{θ}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 70.

θ = $\frac{\text{88 × 360}}{\text{440}}$.

Therefore, θ = 72°.

11 = $\frac{\text{θ}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 6.3.

θ = $\frac{\text{11 × 360}}{\text{39.6}}$.

Therefore, θ = 100°.

2.2 = $\frac{\text{θ}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 2.8.

θ = $\frac{\text{2.2 × 360}}{\text{17.6}}$.

Therefore, θ = 45°.

Distance travelled = 11000 m.

Circumference of the wheel = 2 × $\frac{\text{22}}{\text{7}}$ × 0.25.

= $\frac{\text{11}}{\text{7}}$ m.

Number of revolutions = $\frac{\text{11000 × 7}}{\text{11}}$.

= 7000.

For each complete revolution, the wheel moves forward by one circumference.

With diameter 21 cm, this distance is πd = $\frac{\text{22}}{\text{7}}$ × 21 = 66 cm.

The given journey is 66 km = 6600000 cm.

Number of revolutions = $\frac{\text{Total distance}}{\text{Distance per revolution}}$.

= $\frac{\text{6600000}}{\text{66}}$ = 100000.

Thus the wheel makes 10⁵ complete revolutions.

Let the radius be r units and the sector angle be θ.

$\frac{\text{θ}}{\text{360}}$ × 2πr = $\frac{\text{θ}}{\text{360}}$ × πr².

Cancelling the common non-zero factors gives 2r = r².

Since r > 0, r = 2 units.

Let sector B have radius r and sector A have angle θ.

Then sector A has radius 2r and sector B has angle 2θ.

Area of A : area of B = θ × (2r)² : 2θ × r².

= 4θr² : 2θr².

= 2 : 1.

The fraction of the circle occupied by a sector equals $\frac{\text{θ}}{\text{360}}$.

$\frac{\text{θ}}{\text{360}}$ = $\frac{\text{1}}{\text{12}}$.

θ = $\frac{\text{360}}{\text{12}}$ = 30°.

Area of the sector = $\frac{\text{θ}}{\text{360}}$ × πr².

$\frac{\text{264}}{\text{7}}$ = $\frac{\text{θ}}{\text{360}}$ × $\frac{\text{22}}{\text{7}}$ × 6².

θ = $\frac{\text{264 × 360}}{\text{22 × 36}}$.

Therefore, θ = 120°.

Area of a sector = $\frac{\text{1}}{\text{2}}$ × radius × arc length.

= $\frac{\text{1}}{\text{2}}$ × 14 × 22.

= 154 cm².

231 = $\frac{\text{60}}{\text{360}}$ × $\frac{\text{22}}{\text{7}}$ × r².

r² = $\frac{\text{231 × 6 × 7}}{\text{22}}$ = 441.

Since a radius is positive, r = 21 cm.

Major-sector angle = 360° − 120° = 240°.

Difference of the sector angles = 240° − 120° = 120°.

Difference of the areas = $\frac{\text{120}}{\text{360}}$ × $\frac{\text{22}}{\text{7}}$ × 21².

= 462 cm².

The two sectors have the same central angle, 60°.

Shaded area = outer-sector area − inner-sector area.

= $\frac{\text{60}}{\text{360}}$ × π × (5² − 3²).

= $\frac{\text{1}}{\text{6}}$ × π × 16.

= $\frac{\text{8π}}{\text{3}}$ cm².

$\frac{\text{θ}}{\text{360}}$ = $\frac{\text{1}}{\text{8}}$.

θ = $\frac{\text{360}}{\text{8}}$.

Therefore, θ = 45°.

The diagonals of a rhombus are perpendicular.

Therefore, ∠EOF = 90°.

Area of sector OEF = $\frac{\text{90}}{\text{360}}$ × π × 6².

= 9π cm².

60π = $\frac{\text{θ}}{\text{360}}$ × π × 12².

θ = $\frac{\text{60 × 360}}{\text{144}}$.

Therefore, θ = 150°.

The angles of quadrilateral ABCD add to 360°.

∠A = 360° − 90° − 90° − 60° = 120°.

The shaded sector has radius AP = 3 cm and angle 120°.

Its area = $\frac{\text{120}}{\text{360}}$ × π × 3².

= 3π cm².

Area of the circle = $\frac{\text{22}}{\text{7}}$ × 7² = 154 cm².

Area of each identical sector = $\frac{\text{154}}{\text{16}}$.

= $\frac{\text{77}}{\text{8}}$ cm².

198 = $\frac{\text{θ}}{\text{360}}$ × $\frac{\text{22}}{\text{7}}$ × 18².

θ = $\frac{\text{198 × 360 × 7}}{\text{22 × 324}}$.

Therefore, θ = 70°.

Let the new circle have radius R cm.

πR² = $\frac{\text{90}}{\text{360}}$ × π × 50².

R² = 625.

Therefore, R = 25 cm.

Area of the sector = $\frac{\text{1}}{\text{2}}$ × radius × arc length.

= $\frac{\text{1}}{\text{2}}$ × 14 × 11.

= 77 cm².

Area of the sector = $\frac{\text{1}}{\text{2}}$ × radius × arc length.

20π = $\frac{\text{1}}{\text{2}}$ × r × 5π.

r = $\frac{\text{40π}}{\text{5π}}$.

Therefore, r = 8 cm.

The five equal central angles add to 360°.

Each angle = $\frac{\text{360}}{\text{5}}$.

= 72°.

Area of the sector = $\frac{\text{1}}{\text{2}}$ × radius × arc length.

54π = $\frac{\text{1}}{\text{2}}$ × 36 × l.

l = $\frac{\text{108π}}{\text{36}}$.

Therefore, l = 3π cm.

Let A be the sector area and l be its arc length.

From A = $\frac{\text{θ}}{\text{360}}$ × πr² and l = $\frac{\text{θ}}{\text{360}}$ × 2πr, we obtain A = $\frac{\text{1}}{\text{2}}$rl.

Here r = 21 cm and l = 22 cm.

Therefore, A = $\frac{\text{21 × 22}}{\text{2}}$.

The required area is 231 cm².

Let r be the circle’s radius in centimetres.

For a sector with arc length l and area A, A = $\frac{\text{1}}{\text{2}}$rl.

Substituting l = 6π and A = 24π gives 24π = $\frac{\text{1}}{\text{2}}$ × r × 6π.

Therefore, 24π = 3πr.

Dividing by 3π gives r = 8.

Hence the radius is 8 cm.

Let the common radius be r units.

Perimeter of the semicircular disc = πr + 2r.

πr² = πr + 2r.

Dividing by r gives πr = π + 2.

Therefore, r = $\frac{\text{π + 2}}{\text{π}}$ units.

Arc length = $\frac{\text{90}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 3.5.

= 5.5 cm.

Perimeter of the sector = arc length + 2 × radius.

= 5.5 + 2 × 3.5.

= 12.5 cm.

Arc length = $\frac{\text{60}}{\text{360}}$ × 2 × $\frac{\text{22}}{\text{7}}$ × 21.

= 22 cm.

Perimeter = arc length + 2 × radius.

= 22 + 42 = 64 cm.

The shaded region is a sector.

Its boundary consists of the arc and the two radii.

Perimeter = l + r + r.

= l + 2r.

Length of the quadrant’s arc = $\frac{\text{90}}{\text{360}}$ × 2πr.

= $\frac{\text{πr}}{\text{2}}$.

Its perimeter includes two radii.

Perimeter = $\frac{\text{πr}}{\text{2}}$ + 2r.

Semicircular arc length = πr.

= $\frac{\text{22}}{\text{7}}$ × 7 = 22 cm.

Perimeter = arc length + diameter.

= 22 + 14 = 36 cm.

Area of the segment = area of the sector − area of the triangle.

Sector area = $\frac{\text{90}}{\text{360}}$ × πr² = $\frac{\text{πr²}}{\text{4}}$.

The two radii form a right angle, so triangle area = $\frac{\text{1}}{\text{2}}$ × r × r = $\frac{\text{r²}}{\text{2}}$.

Hence segment area = $\frac{\text{πr²}}{\text{4}}$ − $\frac{\text{r²}}{\text{2}}$.

Join the centre O to the chord endpoints A and B.

OA = OB = r and ∠AOB = 60°, so triangle OAB is equilateral.

Its altitude is $\frac{\text{√3 r}}{\text{2}}$.

Triangle area = $\frac{\text{1}}{\text{2}}$ × r × $\frac{\text{√3 r}}{\text{2}}$ = $\frac{\text{√3 r²}}{\text{4}}$.

Sector area = $\frac{\text{60}}{\text{360}}$ × πr² = $\frac{\text{πr²}}{\text{6}}$.

Segment area = $\frac{\text{πr²}}{\text{6}}$ − $\frac{\text{√3 r²}}{\text{4}}$.

Minor-sector area = $\frac{\text{90}}{\text{360}}$ × π × 5² = $\frac{\text{25π}}{\text{4}}$ cm².

The two radii form a right triangle of area $\frac{\text{1}}{\text{2}}$ × 5 × 5 = $\frac{\text{25}}{\text{2}}$ cm².

Minor-segment area = ($\frac{\text{25π}}{\text{4}}$ − $\frac{\text{25}}{\text{2}}$) cm².

Major-segment area = 25π − ($\frac{\text{25π}}{\text{4}}$ − $\frac{\text{25}}{\text{2}}$).

Difference = circle area − 2 × minor-segment area.

= 25π − 2($\frac{\text{25π}}{\text{4}}$ − $\frac{\text{25}}{\text{2}}$).

= ($\frac{\text{25π}}{\text{2}}$ + 25) cm².

Area of the minor segment = area of the 90° sector − area of triangle OPQ.

Sector area = $\frac{\text{90}}{\text{360}}$ × 3.14 × 20² = 314 cm².

Triangle area = $\frac{\text{1}}{\text{2}}$ × 20 × 20 = 200 cm².

Minor-segment area = 314 − 200.

= 114 cm².

Consider one common chord and the two radii joining its endpoints to a centre.

All three sides of this triangle are 1 cm, so it is equilateral and its central angle is 60°.

The altitude of the triangle is √(1² − ($\frac{\text{1}}{\text{2}}$)²) = $\frac{\text{√3}}{\text{2}}$ cm.

Triangle area = $\frac{\text{1}}{\text{2}}$ × 1 × $\frac{\text{√3}}{\text{2}}$ = $\frac{\text{√3}}{\text{4}}$ cm².

The corresponding sector has area $\frac{\text{60}}{\text{360}}$ × π × 1² = $\frac{\text{π}}{\text{6}}$ cm².

Subtracting the triangle gives one minor-segment area as ($\frac{\text{π}}{\text{6}}$ − $\frac{\text{√3}}{\text{4}}$) cm².

Each shaded overlap contains two such segments, and there are four overlaps.

Hence the total shaded area is 4 × 2 × ($\frac{\text{π}}{\text{6}}$ − $\frac{\text{√3}}{\text{4}}$) = 8($\frac{\text{π}}{\text{6}}$ − $\frac{\text{√3}}{\text{4}}$) cm².