Areas Related to Circles: Class 10 Case Study Previous Year Questions with Answers

AC = 40 m.

Since B is the midpoint of AC, BC = AB = 20 m.

AC = AB + BC = 20 + 20 = 40 m.

OB = 15 m.

The radius OB is perpendicular to the tangent road at B.

Thus OB is the shortest distance from O to the road.

In right △OBA, OB² = OA² − AB² = 25² − 20² = 225.

OB = 15 m.

Circumference = 30π m = $\frac{\text{(}\text{660}\text{)/(}}{\text{7}\text{)}}$ m, using π = 22/7.

OB = √(25² − 20²) = 15 m, so the radius r = 15 m.

Circumference = 2πr = 2 × π × 15 = 30π m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, circumference = $\frac{\text{(}\text{660}\text{)/(}}{\text{7}\text{)}}$ m ≈ 94.29 m.

Area = 225π m² = $\frac{\text{(}\text{4950}\text{)/(}}{\text{7}\text{)}}$ m², using π = 22/7.

The radius is r = √(25² − 20²) = 15 m.

Area of the circular village = πr² = π × 15² = 225π m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{4950}\text{)/(}}{\text{7}\text{)}}$ m² ≈ 707.14 m².

4 cuts.

Each cut along a diameter gives two radial edges.

Four suitably spaced diameters divide the circle into 8 equal sectors.

Hence 4 cuts are required.

Radius = 17.5 cm.

Radius = $\frac{\text{(}\text{diameter}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{35}\text{)/(}}{\text{2}\text{)}}$ = 17.5 cm.

Area of each slice = 120.3125 cm² ≈ 120.31 cm².

Eight equal slices each occupy $\frac{\text{(}\text{1}\text{)/(}}{\text{8}\text{)}}$ of the pizza.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, whole area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 962.5 cm².

Area of one slice = $\frac{\text{(}\text{962.5}\text{)/(}}{\text{8}\text{)}}$ = 120.3125 cm².

Area of one of four slices = 240.625 cm² ≈ 240.63 cm².

Use four equal slices for this alternative.

Whole area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 962.5 cm².

Area of one slice = $\frac{\text{(}\text{962.5}\text{)/(}}{\text{4}\text{)}}$ = 240.625 cm².

Central angle = 36°.

The 10 equal sectors together make 360°.

Angle of each sector = $\frac{\text{(}\text{360°}\text{)/(}}{\text{10}\text{)}}$ = 36°.

Arc ACB = 11 mm.

Arc ACB belongs to one of the 10 equal sectors.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, circumference = πd = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 = 110 mm.

Arc ACB = $\frac{\text{(}\text{110}\text{)/(}}{\text{10}\text{)}}$ = 11 mm.

Area of each sector = 96.25 mm².

Radius r = $\frac{\text{(}\text{35}\text{)/(}}{\text{2}\text{)}}$ = 17.5 mm.

Area of each sector = $\frac{\text{(}\text{1}\text{)/(}}{\text{10}\text{)}}$πr².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1}\text{)/(}}{\text{10}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 96.25 mm².

Total wire = 285 mm.

The wire forms the circumference and 5 complete diameters.

Circumference = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 = 110 mm.

Wire in the diameters = 5 × 35 = 175 mm.

Total wire = 110 + 175 = 285 mm.

4 cuts.

Four equally spaced diameter cuts give 8 equal sectors.

Radius = 17.5 cm.

Radius = $\frac{\text{(}\text{35}\text{)/(}}{\text{2}\text{)}}$ = 17.5 cm.

Area of each slice = 120.3125 cm² ≈ 120.31 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, total area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 962.5 cm².

Area of one of 8 equal slices = $\frac{\text{(}\text{962.5}\text{)/(}}{\text{8}\text{)}}$ = 120.3125 cm².

Area of the entire pizza = 962.5 cm².

Radius = 17.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = πr² = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 962.5 cm².

Radius = 17.5 cm.

Radius = $\frac{\text{(}\text{diameter}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{35}\text{)/(}}{\text{2}\text{)}}$ = 17.5 cm.

Circumference = 110 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, circumference = πd = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 = 110 cm.

Total wire = 285 cm.

The wire makes the circular boundary and 5 diameters.

Total length = πd + 5d.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, length = 110 + 5 × 35 = 285 cm.

Area of each sector = 96.25 cm².

Each of the 10 equal sectors occupies $\frac{\text{(}\text{1}\text{)/(}}{\text{10}\text{)}}$ of the circle.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ and r = 17.5 cm, area = $\frac{\text{(}\text{1}\text{)/(}}{\text{10}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 17.5² = 96.25 cm².

Perimeter = 200 m.

Perimeter = $\frac{\text{(}\text{total fencing cost}\text{)/(}}{\text{cost per metre}\text{)}}$ = $\frac{\text{(}\text{6000}\text{)/(}}{\text{30}\text{)}}$ = 200 m.

Radius = $\frac{\text{(}\text{350}\text{)/(}}{\text{11}\text{)}}$ m ≈ 31.82 m.

Circumference = 2πr = 200 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, r = $\frac{\text{(}\text{200 × 7}\text{)/(}}{\text{44}\text{)}}$ = $\frac{\text{(}\text{350}\text{)/(}}{\text{11}\text{)}}$ m.

Area = $\frac{\text{(}\text{35000}\text{)/(}}{\text{11}\text{)}}$ m² ≈ 3181.82 m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ and r = $\frac{\text{(}\text{350}\text{)/(}}{\text{11}\text{)}}$ m, area = πr².

Area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × ($\frac{\text{(}\text{350}\text{)/(}}{\text{11}\text{)}}$)² = $\frac{\text{(}\text{35000}\text{)/(}}{\text{11}\text{)}}$ m².

Keep the exact radius until the final calculation.

Cost = ₹$\frac{\text{(}\text{17500}\text{)/(}}{\text{11}\text{)}}$ ≈ ₹1590.91.

Area = $\frac{\text{(}\text{35000}\text{)/(}}{\text{11}\text{)}}$ m².

Cost = area × rate = $\frac{\text{(}\text{35000}\text{)/(}}{\text{11}\text{)}}$ × 0.50 = ₹$\frac{\text{(}\text{17500}\text{)/(}}{\text{11}\text{)}}$.

Rounding once at the end gives ₹1590.91.

Area = 1600 cm².

Area of square = side² = 40² = 1600 cm².

Area = $\frac{\text{(}\text{2200}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 314.29 cm².

The central circle has radius r = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$ = 10 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = πr² = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 100 = $\frac{\text{(}\text{2200}\text{)/(}}{\text{7}\text{)}}$ cm².

Remaining area = $\frac{\text{(}\text{6800}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 971.43 cm².

Four quadrants of radius 10 cm together make one complete circle of radius 10 cm.

Area removed = 2 × π × 10² = 200π cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, remaining area = 1600 − $\frac{\text{(}\text{4400}\text{)/(}}{\text{7}\text{)}}$ = $\frac{\text{(}\text{6800}\text{)/(}}{\text{7}\text{)}}$ cm².

Combined area removed = $\frac{\text{(}\text{4400}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 628.57 cm².

Four quadrants together have area π × 10² = 100π cm².

The central circle also has area 100π cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, total area removed = 200π = $\frac{\text{(}\text{4400}\text{)/(}}{\text{7}\text{)}}$ cm².

($\frac{\text{(}\text{πr²}\text{)/(}}{\text{4}\text{)}}$ − $\frac{\text{(}\text{r²}\text{)/(}}{\text{2}\text{)}}$) + ($\frac{\text{(}\text{πr²}\text{)/(}}{\text{6}\text{)}}$ − $\frac{\text{(}\text{√3 r²}\text{)/(}}{\text{4}\text{)}}$) = $\frac{\text{(}\text{770}\text{)/(}}{\text{3}\text{)}}$.

Area of a minor segment = area of sector − area of its triangle.

For 90°, triangle area = $\frac{\text{(}\text{r²}\text{)/(}}{\text{2}\text{)}}$.

For 60°, the triangle is equilateral and has area $\frac{\text{(}\text{√3 r²}\text{)/(}}{\text{4}\text{)}}$.

Adding the two segment areas gives the required equation.

r ≈ 26.1 m, using π = 3.14 and √3 = 1.73.

The equation becomes r²($\frac{\text{(}\text{5π}\text{)/(}}{\text{12}\text{)}}$ − $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ − $\frac{\text{(}\text{√3}\text{)/(}}{\text{4}\text{)}}$) = $\frac{\text{(}\text{770}\text{)/(}}{\text{3}\text{)}}$.

Thus r² = $\frac{\text{(}\text{3080}\text{)/(}}{\text{5π − 6 − 3√3}\text{)}}$.

To reproduce the scheme’s displayed approximations, use π = 3.14 and √3 = 1.73.

With these values, r² = $\frac{\text{(}\text{3080}\text{)/(}}{\text{4.51}\text{)}}$ ≈ 682.926829 m².

Taking the positive square root gives r ≈ 26.13287 m ≈ 26.1 m.

Source note: the 2023 Basic 430/2/1 scheme (PDF and printed p. 19) prints 26.1 cm; the question gives metres, so the radius here is in metres. Its displayed segment areas are 194.63 m² and 62.03 m². The approximation values above reproduce those areas; the scheme does not explicitly prescribe them.

Red-segment area ≈ 194.63 m².

Use π = 3.14 and √3 = 1.73, consistently with part (ii).

Retain r² = $\frac{\text{(}\text{3080}\text{)/(}}{\text{4.51}\text{)}}$ rather than squaring the rounded radius.

Red-segment area = $\frac{\text{(}\text{πr²}\text{)/(}}{\text{4}\text{)}}$ − $\frac{\text{(}\text{r²}\text{)/(}}{\text{2}\text{)}}$ = (0.785 − 0.5)r².

Area = 0.285 × $\frac{\text{(}\text{3080}\text{)/(}}{\text{4.51}\text{)}}$ ≈ 194.63 m².

Yellow-segment area ≈ 62.03 m².

Using π = 3.14, √3 = 1.73 and r² = $\frac{\text{(}\text{3080}\text{)/(}}{\text{4.51}\text{)}}$,

Yellow-segment area = ($\frac{\text{(}\text{π}\text{)/(}}{\text{6}\text{)}}$ − $\frac{\text{(}\text{√3}\text{)/(}}{\text{4}\text{)}}$)r².

Area = ($\frac{\text{(}\text{3.14}\text{)/(}}{\text{6}\text{)}}$ − $\frac{\text{(}\text{1.73}\text{)/(}}{\text{4}\text{)}}$) × $\frac{\text{(}\text{3080}\text{)/(}}{\text{4.51}\text{)}}$ ≈ 62.03 m².

Before rounding, the red and yellow areas sum to $\frac{\text{(}\text{770}\text{)/(}}{\text{3}\text{)}}$ m².

Area = 64 cm².

Area of square = side² = 8² = 64 cm².

AC = 8√2 cm.

In right △ABC, AC² = AB² + BC² = 8² + 8² = 128.

AC = √128 = 8√2 cm.

Sector area = $\frac{\text{(}\text{88}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 12.57 cm².

Diameter of the inscribed circle = side of square = 8 cm, so r = 4 cm.

Since OP lies along OA and OQ along OB, ∠POQ = ∠AOB = 90°.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, sector area = $\frac{\text{(}\text{90}\text{)/(}}{\text{360}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 4² = $\frac{\text{(}\text{88}\text{)/(}}{\text{7}\text{)}}$ cm².

Remaining area = $\frac{\text{(}\text{96}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 13.71 cm².

Area of square = 64 cm²; circle radius = 4 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area of circle = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 4² = $\frac{\text{(}\text{352}\text{)/(}}{\text{7}\text{)}}$ cm².

Remaining area = 64 − $\frac{\text{(}\text{352}\text{)/(}}{\text{7}\text{)}}$ = $\frac{\text{(}\text{96}\text{)/(}}{\text{7}\text{)}}$ cm².

AB = 21√2 cm.

In right △OAB, AB² = OA² + OB² = 21² + 21² = 882.

AB = √882 = 21√2 cm.

Triangle area = 220.5 cm².

OA ⟂ OB, so area = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × OA × OB.

Area = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × 21 × 21 = 220.5 cm².

Segment area = 126 cm².

Area of segment = area of 90° sector − area of △OAB.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, sector area = $\frac{\text{(}\text{90}\text{)/(}}{\text{360}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21² = 346.5 cm².

Segment area = 346.5 − 220.5 = 126 cm².

Sector perimeter = 75 cm.

The boundary contains two radii and the minor arc ACB.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, arc length = $\frac{\text{(}\text{90}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21 = 33 cm.

Perimeter = 21 + 21 + 33 = 75 cm.

∠ROS = 45°.

Four equally placed diameters form 8 equal central angles.

∠ROS = $\frac{\text{(}\text{360°}\text{)/(}}{\text{8}\text{)}}$ = 45°.

Perimeter = 58.5 cm.

The outer sector has radius 21 cm and angle 45°.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, its arc length = $\frac{\text{(}\text{45}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21 = 16.5 cm.

Perimeter = 21 + 21 + 16.5 = 58.5 cm.

Area = $\frac{\text{(}\text{3751}\text{)/(}}{\text{28}\text{)}}$ cm² ≈ 133.96 cm².

PQRS is the outer 45° sector minus the inner 45° sector.

Area = $\frac{\text{(}\text{45}\text{)/(}}{\text{360}\text{)}}$π(21² − 10²).

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1}\text{)/(}}{\text{8}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 341 = $\frac{\text{(}\text{3751}\text{)/(}}{\text{28}\text{)}}$ cm².

Area = $\frac{\text{(}\text{200}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 28.57 cm².

The inner radii OA and OB form a 90° angle.

Area of segment = area of 90° sector − area of right △OAB.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1}\text{)/(}}{\text{4}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 10² − $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × 10 × 10.

Area = $\frac{\text{(}\text{550}\text{)/(}}{\text{7}\text{)}}$ − 50 = $\frac{\text{(}\text{200}\text{)/(}}{\text{7}\text{)}}$ cm².

Radius = 7 m.

Each ring has circumference 44 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, 2πr = 44 gives r = $\frac{\text{(}\text{44 × 7}\text{)/(}}{\text{44}\text{)}}$ = 7 m.

∠AOB = 60°.

△OAB is equilateral, so each of its angles is 60°.

Area of R₁ = ($\frac{\text{(}\text{308}\text{)/(}}{\text{3}\text{)}}$ + $\frac{\text{(}\text{49√3}\text{)/(}}{\text{2}\text{)}}$) m².

Area of one 60° minor segment = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$π × 7² − $\frac{\text{(}\text{√3}\text{)/(}}{\text{4}\text{)}}$ × 7².

One unshaded lens contains two such segments.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, lens area = $\frac{\text{(}\text{154}\text{)/(}}{\text{3}\text{)}}$ − $\frac{\text{(}\text{49√3}\text{)/(}}{\text{2}\text{)}}$ m².

R₁ = area of one circle − area of one lens.

R₁ = 154 − ($\frac{\text{(}\text{154}\text{)/(}}{\text{3}\text{)}}$ − $\frac{\text{(}\text{49√3}\text{)/(}}{\text{2}\text{)}}$) = $\frac{\text{(}\text{308}\text{)/(}}{\text{3}\text{)}}$ + $\frac{\text{(}\text{49√3}\text{)/(}}{\text{2}\text{)}}$ m².

Using √3 = 1.73 gives approximately 145.05 m².

Total rope = $\frac{\text{(}\text{176}\text{)/(}}{\text{3}\text{)}}$ m ≈ 58.67 m.

There are 4 unshaded lenses, each bounded by two 60° arcs of radius 7 m.

Length of each arc = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$ × 44 = $\frac{\text{(}\text{22}\text{)/(}}{\text{3}\text{)}}$ m.

Total rope = 4 × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{176}\text{)/(}}{\text{3}\text{)}}$ m.

Parking perimeter = 18 units.

The semicircle has radius r = $\frac{\text{(}\text{7}\text{)/(}}{\text{2}\text{)}}$ = 3.5 units.

Its perimeter includes the curved arc and its diameter.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, perimeter = πr + 2r = 11 + 7 = 18 units.

Total area = $\frac{\text{(}\text{715}\text{)/(}}{\text{28}\text{)}}$ square units ≈ 25.54 square units.

Parking area = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$π(3.5)² = $\frac{\text{(}\text{77}\text{)/(}}{\text{4}\text{)}}$ square units, using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$.

Two quadrants of radius 2 have total area = 2 × $\frac{\text{(}\text{1}\text{)/(}}{\text{4}\text{)}}$π × 2² = $\frac{\text{(}\text{44}\text{)/(}}{\text{7}\text{)}}$ square units.

Total = $\frac{\text{(}\text{77}\text{)/(}}{\text{4}\text{)}}$ + $\frac{\text{(}\text{44}\text{)/(}}{\text{7}\text{)}}$ = $\frac{\text{(}\text{715}\text{)/(}}{\text{28}\text{)}}$ square units.

Playground : parking = 56 : 11.

Area of rectangular playground = 14 × 7 = 98 square units.

Parking area = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × (3.5)² = $\frac{\text{(}\text{77}\text{)/(}}{\text{4}\text{)}}$ square units.

Ratio = 98 : $\frac{\text{(}\text{77}\text{)/(}}{\text{4}\text{)}}$ = 392 : 77 = 56 : 11.

Fencing cost = ₹106, following the scheme’s divider-inclusive boundary.

Under “Required Perimeter”, the scheme uses 2(l + b) + πr: the full rectangular boundary plus the parking arc.

This calculation includes the straight 7-unit divider once, as well as the outside boundary.

Length = 2(14 + 7) + $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5 = 42 + 11 = 53 units.

Cost = 53 × 2 = ₹106.

Source: 2023 Standard 30/4/1 marking scheme, PDF and printed p. 23. The divider interpretation follows from its formula; counting only the outer boundary would give 46 units and ₹92.

R² + r² = 130.

The sum of the circle areas gives πR² + πr² = 130π.

Dividing by π gives R² + r² = 130.

r² − 14r + 33 = 0.

Since the circles touch externally, R + r = 14, so R = 14 − r.

Substitute into R² + r² = 130: (14 − r)² + r² = 130.

Simplifying gives 2r² − 28r + 66 = 0, hence r² − 14r + 33 = 0.

r = 3 m; area = 9π m².

r² − 14r + 33 = (r − 3)(r − 11) = 0.

Thus r = 3 or 11.

As R > r and R + r = 14, r is the smaller value: r = 3 m.

Corresponding area = πr² = 9π m².

R = 11 m; area = 121π m².

Factoring r² − 14r + 33 = 0 gives radii 3 m and 11 m.

R is the larger radius, so R = 11 m.

Corresponding area = πR² = 121π m².

Wire length = 220 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, circumference = 2πr = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 = 220 m.

Side of the square = 35√2 m.

The square’s diagonal is the diameter of the circle: 70 m.

If its side is s, then 2s² = 70² by Pythagoras.

s = $\frac{\text{(}\text{70}\text{)/(}}{\text{√2}\text{)}}$ = 35√2 m.

Grass area = 700 m²; cost = ₹35,000.

Only two of the four segments outside the square are shaded.

The square’s sides subtend equal central angles of 90°.

One segment = $\frac{\text{(}\text{90}\text{)/(}}{\text{360}\text{)}}$π × 35² − $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × 35².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, one segment = 962.5 − 612.5 = 350 m².

Shaded area = 2 × 350 = 700 m².

Cost = 700 × 50 = ₹35,000.

House : all remaining land = 7 : 4.

House area = (35√2)² = 2450 m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, circle area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35² = 3850 m².

All remaining land = 3850 − 2450 = 1400 m².

Ratio = 2450 : 1400 = 7 : 4.

This alternative asks for all land outside the house, not only the two shaded segments.

∠POA = 120°.

OA = OP, so △AOP is isosceles.

Since A, O and B are collinear, ∠PAO = ∠PAB = 30°.

∠APO = 30°, hence ∠POA = 180° − 30° − 30° = 120°.

Wire length = 180 m.

Radius = 35 m. A semicircle’s perimeter is πr + 2r.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, perimeter = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 + 70 = 180 m.

Region I area = ($\frac{\text{(}\text{1925}\text{)/(}}{\text{3}\text{)}}$ − $\frac{\text{(}\text{1225√3}\text{)/(}}{\text{4}\text{)}}$) m² ≈ 111.23 m².

Mango trees occupy region I, the smaller segment PB.

∠POB = 180° − 120° = 60°.

△POB is equilateral with side 35 m.

Segment area = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$π × 35² − $\frac{\text{(}\text{√3}\text{)/(}}{\text{4}\text{)}}$ × 35².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1925}\text{)/(}}{\text{3}\text{)}}$ − $\frac{\text{(}\text{1225√3}\text{)/(}}{\text{4}\text{)}}$ m².

The exact expression matches the 2025 Standard 30/6/1 scheme, PDF and printed p. 16; its printed decimal is 111.89. Evaluating the exact expression gives approximately 111.23 m².

Region III perimeter = ($\frac{\text{(}\text{220}\text{)/(}}{\text{3}\text{)}}$ + 35√3) m ≈ 133.96 m.

Region III is bounded by chord AP and the 120° arc AP.

∠APB = 90°, since AB is a diameter.

In right △APB, cos 30° = $\frac{\text{(}\text{AP}\text{)/(}}{\text{AB}\text{)}}$, so AP = 70 × $\frac{\text{(}\text{√3}\text{)/(}}{\text{2}\text{)}}$ = 35√3 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, arc AP = $\frac{\text{(}\text{120}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 35 = $\frac{\text{(}\text{220}\text{)/(}}{\text{3}\text{)}}$ m.

Perimeter = $\frac{\text{(}\text{220}\text{)/(}}{\text{3}\text{)}}$ + 35√3 m.

The exact expression matches the same scheme (PDF and printed p. 16); its printed decimal is 133.8. The exact expression evaluates to approximately 133.96 m.

∠BOP = 15°.

The quarter circle contains two 30° blades and two equal gaps.

Let each gap be x°. Then 30° + x° + 30° + x° = 90°.

2x = 30, so ∠BOP = 15°.

Arc CD = 11 cm.

Arc CD has radius 21 cm and central angle 30°.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, arc length = $\frac{\text{(}\text{30}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21 = 11 cm.

Area = $\frac{\text{(}\text{539}\text{)/(}}{\text{6}\text{)}}$ cm² ≈ 89.83 cm².

Region CABD is the difference of two 30° sectors.

Area = $\frac{\text{(}\text{30}\text{)/(}}{\text{360}\text{)}}$π(28² − 21²).

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1}\text{)/(}}{\text{12}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 343 = $\frac{\text{(}\text{539}\text{)/(}}{\text{6}\text{)}}$ cm².

Perimeter = $\frac{\text{(}\text{119}\text{)/(}}{\text{3}\text{)}}$ cm ≈ 39.67 cm.

The boundary contains the outer arc AB, inner arc CD and two radial edges.

Outer arc = $\frac{\text{(}\text{30}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 28 = $\frac{\text{(}\text{44}\text{)/(}}{\text{3}\text{)}}$ cm.

Inner arc = 11 cm; each radial edge = 28 − 21 = 7 cm.

Perimeter = $\frac{\text{(}\text{44}\text{)/(}}{\text{3}\text{)}}$ + 11 + 7 + 7 = $\frac{\text{(}\text{119}\text{)/(}}{\text{3}\text{)}}$ cm.

A = $\frac{\text{(}\text{lr}\text{)/(}}{\text{2}\text{)}}$, where l is arc length and r is radius.

For a sector, l = $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × 2πr and A = $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × πr².

Thus A = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$r × ($\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × 2πr) = $\frac{\text{(}\text{lr}\text{)/(}}{\text{2}\text{)}}$.

Area = 98π m².

The angle is 90° − 10° = 80° and the radius is 21 m.

Area = $\frac{\text{(}\text{80}\text{)/(}}{\text{360}\text{)}}$π × 21² = 98π m².

New angle θ = 45°.

The original area is 98π m².

For the new sector, $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$π × 28² = 98π.

θ = $\frac{\text{(}\text{98 × 360}\text{)/(}}{\text{784}\text{)}}$ = 45°.

Increase = $\frac{\text{(}\text{686π}\text{)/(}}{\text{9}\text{)}}$ m².

Use the unchanged angle 80° in this alternative.

Increase = $\frac{\text{(}\text{80}\text{)/(}}{\text{360}\text{)}}$π(28² − 21²).

Increase = $\frac{\text{(}\text{2}\text{)/(}}{\text{9}\text{)}}$π × 343 = $\frac{\text{(}\text{686π}\text{)/(}}{\text{9}\text{)}}$ m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ gives approximately 239.56 m².

Face area = $\frac{\text{(}\text{1100}\text{)/(}}{\text{21}\text{)}}$ cm² ≈ 52.38 cm².

The full face is a 60° sector of radius 10 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 10² = $\frac{\text{(}\text{1100}\text{)/(}}{\text{21}\text{)}}$ cm².

Black-sheet area = 44 cm².

Subtract the inner sector of radius 4 cm from the outer sector of radius 10 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × (10² − 4²) = 44 cm².

Tape length = $\frac{\text{(}\text{80}\text{)/(}}{\text{3}\text{)}}$ cm ≈ 26.67 cm.

The boundary contains both arcs and two radial edges of length 10 − 4 = 6 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, total arc length = $\frac{\text{(}\text{60}\text{)/(}}{\text{360}\text{)}}$ × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × (10 + 4) = $\frac{\text{(}\text{44}\text{)/(}}{\text{3}\text{)}}$ cm.

Tape length = $\frac{\text{(}\text{44}\text{)/(}}{\text{3}\text{)}}$ + 6 + 6 = $\frac{\text{(}\text{80}\text{)/(}}{\text{3}\text{)}}$ cm.

New central angle = 72°.

Let the new central angle be θ; retain radii 10 cm and 4 cm.

Boundary = $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × 2π(10 + 4) + 2(10 − 4).

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, 29.6 = $\frac{\text{(}\text{88θ}\text{)/(}}{\text{360}\text{)}}$ + 12.

17.6 = $\frac{\text{(}\text{88θ}\text{)/(}}{\text{360}\text{)}}$, so θ = $\frac{\text{(}\text{17.6 × 360}\text{)/(}}{\text{88}\text{)}}$ = 72°.

A = $\frac{\text{(}\text{lr}\text{)/(}}{\text{2}\text{)}}$.

Arc length l = $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × 2πr.

Sector area A = $\frac{\text{(}\text{θ}\text{)/(}}{\text{360}\text{)}}$ × πr².

Therefore A = $\frac{\text{(}\text{r}\text{)/(}}{\text{2}\text{)}}$ × l = $\frac{\text{(}\text{lr}\text{)/(}}{\text{2}\text{)}}$, where r is the range.

Monitored area = 352 m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, A = $\frac{\text{(}\text{70}\text{)/(}}{\text{360}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 24² = 352 m².

New angle = 105°.

For a fixed radius, sector area is proportional to its angle.

New area = 1.5 times the original area.

New angle = 1.5 × 70° = 105°.

Required range = 24 m.

Let the new radius be R. New area is 1.3 times the original area.

$\frac{\text{(}\text{91}\text{)/(}}{\text{360}\text{)}}$πR² = 1.3 × $\frac{\text{(}\text{70}\text{)/(}}{\text{360}\text{)}}$π × 24².

Since 1.3 × 70 = 91, this simplifies to R² = 24².

Taking the positive radius gives R = 24 m.

The increased angle already provides the 30% increase in area.