CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 09

CONSTRUCTION AND AREAS RELATED TO CIRCLES

LEVEL - I

Question 1. Divide a line segment in given ratio.
Answer: Let AB be a given line segment. To divide it in the ratio \(m:n\) (where \(m\) and \(n\) are positive integers):
1. Draw any ray AX making an acute angle with AB.
2. Locate \(m + n\) points (\(A_1, A_2, \dots, A_{m+n}\)) on AX at equal distances.
3. Join \(B\) to \(A_{m+n}\).
4. Through the point \(A_m\), draw a line parallel to \(A_{m+n}B\) intersecting AB at point C.
Point C divides AB internally in the ratio \(m:n\).
In simple words: Draw a slanted line, mark equal parts equal to the sum of the ratio, connect the last part to the end of the line, and draw a parallel line from the ratio point to divide the line.

Exam Tip: Always use a sharp pencil and compass to ensure accuracy in marking equal intervals along the ray.

Question 2. Draw a line segment AB = 8 cm and divide it in the ratio 4:3.
Answer:
1. Draw a line segment \(AB = 8\text{ cm}\).
2. Draw a ray AX making an acute angle with AB.
3. Mark \(4 + 3 = 7\) points \(A_1, A_2, A_3, A_4, A_5, A_6, A_7\) on AX at equal distances.
4. Join \(A_7\) to \(B\).
5. Through \(A_4\), draw a line parallel to \(A_7B\) meeting AB at point C.
Thus, AC : CB = 4 : 3.
Length of AC = \(\frac{4}{7} \times 8 = \frac{32}{7}\text{ cm} \approx 4.57\text{ cm}\).
In simple words: Divide an 8 cm line into 7 equal parts and mark the point at the 4th part from A.

Exam Tip: Mention the lengths of the divided segments in your final answer to verify accuracy.

Question 3. Divide a line segment of 7 cm internally in the ratio 2:3.
Answer:
1. Draw line segment \(AB = 7\text{ cm}\).
2. Draw ray AX making an acute angle with AB.
3. Locate \(2 + 3 = 5\) points \(A_1, A_2, A_3, A_4, A_5\) on AX at equal distances.
4. Join \(A_5\) to B.
5. Through \(A_2\), draw a line parallel to \(A_5B\) intersecting AB at C.
Point C divides the 7 cm line in the ratio 2:3.
In simple words: Make 5 equal divisions on a ray, join the 5th point to the end of the 7 cm line, and draw a parallel line from the 2nd point.

Exam Tip: Double-check that your parallel lines are drawn accurately using set squares or compasses.

Question 4. Draw a circle of radius 4 cm. Take a point P on it. Draw tangent to the given circle at P.
Answer:
1. Draw a circle with centre O and radius \(r = 4\text{ cm}\).
2. Take a point P on the circle.
3. Join radius OP.
4. Construct an angle of \(90^\circ\) at point P with line OP, or draw a perpendicular line to OP at P.
5. Extend the perpendicular line to get the tangent XY at point P.
In simple words: Draw a circle, pick a point on the edge, draw a radius to that point, and draw a perpendicular line at that point to get the tangent.

Exam Tip: Remember that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Question 5. Construct an isosceles triangle whose base 7.5 cm and altitude is 4.2 cm.
Answer:
1. Draw base \(BC = 7.5\text{ cm}\).
2. Draw the perpendicular bisector of BC intersecting BC at D.
3. From D, cut off an arc of length \(4.2\text{ cm}\) along the bisector to locate vertex A.
4. Join AB and AC.
Triangle ABC is the required isosceles triangle.
In simple words: Draw the base, find its middle point, draw a straight line straight up of 4.2 cm, and connect the top to the ends of the base.

Question 1. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then triangle similar to it whose side are \(\frac{2}{3}\) of corresponding sides of the first triangle.
Answer:
1. Draw \(BC = 6\text{ cm}\).
2. With B as centre and radius \(4\text{ cm}\), draw an arc. With C as centre and radius \(5\text{ cm}\), draw another arc intersecting the first arc at A.
3. Join AB and AC to form \(\Delta ABC\).
4. Draw ray BX making an acute angle with BC on the side opposite to vertex A.
5. Locate 3 points \(B_1, B_2, B_3\) on BX such that \(BB_1 = B_1B_2 = B_2B_3\).
6. Join \(B_3\) to C.
7. Through \(B_2\), draw a line parallel to \(B_3C\) intersecting BC at \(C'\).
8. Through \(C'\), draw a line parallel to CA meeting AB at \(A'\).
\(\Delta A'BC'\) is the required similar triangle.
In simple words: Draw the 4-5-6 cm triangle, make a ray with 3 divisions, connect the 3rd to C, and draw parallel lines to get a smaller triangle scaled by \(\frac{2}{3}\).

Exam Tip: For a scale factor less than 1, the new triangle will be smaller and lie inside the original triangle.

Question 2. Construct a triangle similar to a given \(\Delta ABC\) such that each of its sides is \(\frac{2}{3}\)rd of the corresponding sides of \(\Delta ABC\). It is given that \(AB = 4\text{ cm}\), \(BC = 5\text{ cm}\) and \(AC = 6\text{ cm}\) also write the steps of construction.
Answer:
Steps of Construction:
1. Draw a line segment \(BC = 5\text{ cm}\).
2. With B as centre and radius \(4\text{ cm}\), draw an arc. With C as centre and radius \(6\text{ cm}\), draw another arc to intersect the previous arc at point A.
3. Join AB and AC to form \(\Delta ABC\).
4. Below BC, make an acute angle \(\angle CBX\) with ray BX.
5. Mark 3 equal points \(B_1, B_2, B_3\) on ray BX.
6. Join \(B_3\) to C.
7. Draw a line through \(B_2\) parallel to \(B_3C\) meeting BC at \(C'\).
8. From \(C'\), draw a line parallel to CA intersecting AB at \(A'\).
\(\Delta A'BC'\) is the required triangle.
In simple words: Construct the 4-5-6 cm triangle and use a 3-part scale ray to construct the scaled internal triangle.

Exam Tip: Writing clear, numbered steps of construction is mandatory for full marks in geometry questions.

Question 3. Draw a right triangle ABC in which \(\angle B = 90^\circ\), \(AB = 5\text{ cm}\), \(BC = 4\text{ cm}\) then construct another triangle ABC whose sides are \(\frac{5}{3}\) times the corresponding sides of \(\Delta ABC\).
Answer:
1. Draw \(BC = 4\text{ cm}\).
2. At point B, construct an angle of \(90^\circ\).
3. Cut an arc of radius \(5\text{ cm}\) along the perpendicular to mark point A. Join AC.
4. Draw an acute ray BX making an angle with BC downwards.
5. Mark 5 equal points \(B_1, B_2, B_3, B_4, B_5\) on ray BX.
6. Join \(B_3\) to C.
7. Through \(B_5\), draw a line parallel to \(B_3C\) extending to meet the extended line segment BC at \(C'\).
8. From \(C'\), draw a line parallel to CA meeting the extended line segment AB at \(A'\).
\(\Delta A'BC'\) is the required triangle.
In simple words: Draw a right-angled triangle, make a ray with 5 equal parts, connect the 3rd part to C, and draw parallel lines to create a larger outer triangle scaled by \(\frac{5}{3}\).

Exam Tip: Since the scale factor \(\frac{5}{3} > 1\), the new triangle will be larger and lie outside the original triangle.

Question 4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of \(60^\circ\).
Answer:
1. Draw a circle with centre O and radius \(r = 5\text{ cm}\).
2. Draw any radius OA.
3. Construct an angle of \(180^\circ - 60^\circ = 120^\circ\) at centre O with radius OA, letting the other radius be OB.
4. At points A and B, draw perpendiculars to radii OA and OB respectively.
5. Extend these perpendiculars until they intersect each other at an external point P.
PA and PB are the required tangents inclined at \(60^\circ\).
In simple words: Draw radii with an angle of 120 degrees between them, then draw perpendicular lines at both ends of the radii to meet at a point.

Question 5. Draw a circle of radius 5 cm from a point 8 cm away from its centre construct the pair of tangents to the circle and measure their length. Measured length: \(6.24\text{ cm}\).
Answer:
1. Draw a circle with centre O and radius \(5\text{ cm}\).
2. Take a point P at a distance of \(8\text{ cm}\) from O. Join OP.
3. Draw the perpendicular bisector of OP to locate its midpoint M.

4. With M as centre and MP as radius, draw a circle intersecting the given circle at points T1 and T2.
5. Join \(PT_1\) and \(PT_2\). These are the required tangents.
Length of tangent = \(\sqrt{OP^2 - r^2} = \sqrt{8^2 - 5^2} = \sqrt{64 - 25} = \sqrt{39} \approx 6.24\text{ cm}\).
In simple words: Draw a circle, mark a point 8 cm away, find the middle of the distance, draw a circle from that middle point, and connect the point to where the circles cross.

Exam Tip: You can always verify your geometric tangent length using Pythagoras theorem: \(\text{Tangent} = \sqrt{(\text{Distance})^2 - (\text{Radius})^2}\).

Question 6. Construct a triangle PQR in which \(QR = 6\text{ cm}\), \(\angle Q = 60^\circ\) and \(\angle R = 45^\circ\). Construct another triangle similar to \(\Delta PQR\) such that its sides are \(\frac{5}{6}\) of the corresponding sides of \(\Delta PQR\).
Answer:
1. Draw base \(QR = 6\text{ cm}\).
2. At Q, construct \(\angle XQR = 60^\circ\). At R, construct \(\angle YRQ = 45^\circ\).
3. Let raysQX and RY intersect at point P. Triangle PQR is formed.
4. Draw an acute ray QZ making an angle with QR downwards.
5. Locate 6 equal points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6\) on QZ.
6. Join \(Q_6\) to R.
7. Through \(Q_5\), draw a line parallel to \(Q_6R\) meeting QR at \(R'\).
8. Through \(R'\), draw a line parallel to RP meeting PQ at \(P'\).
\(\Delta P'QR'\) is the required similar triangle.
In simple words: Construct the base triangle, make a ray with 6 divisions, connect the 6th part to R, and draw a parallel line from the 5th part to make the smaller similar triangle.

Exam Tip: Ensure the sum of angles in your triangle checks out (\(60^\circ + 45^\circ + 75^\circ = 180^\circ\)).

AREAS RELATED TO CIRCLES - LEVEL I

Question 1. If the perimeter of a circle is equal to that of square, then the ratio of their areas is
(i) 22/7
(ii) 14/11
(iii) 7/22
(iv) 11/14
Answer: (ii) 14/11
Let radius of circle be \(r\) and side of square be \(s\).
Perimeter of circle = Perimeter of square
\(2\pi r = 4s \implies s = \frac{\pi r}{2}\)
Ratio of areas = \(\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{s^2} = \frac{\pi r^2}{\left(\frac{\pi r}{2}\right)^2} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi} = \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}\)
In simple words: Equate circle circumference and square perimeter to find side in terms of radius, then divide their areas to get 14/11.

Exam Tip: Always substitute \(\pi = \frac{22}{7}\) when simplifying ratio problems involving circles and squares.

Question 2. The area of the square that can be inscribed in a circle of 8 cm is
(i) \(256\text{ cm}^2\)
(ii) \(128\text{ cm}^2\)
(iii) \(64\sqrt{2}\text{ cm}^2\)
(iv) \(64\text{ cm}^2\)
Answer: (ii) \(128\text{ cm}^2\)
Diameter of the circle = Diagonal of the inscribed square = \(8\text{ cm}\).
Area of square = \(\frac{1}{2} \times (\text{Diagonal})^2 = \frac{1}{2} \times 8^2 = \frac{1}{2} \times 64 = 32\text{ cm}^2\) (Wait, if diameter is 8 cm, diagonal is 8 cm. Area = \(\frac{8^2}{2} = 32\text{ cm}^2\). If radius is 8 cm, diameter is 16 cm, Area = \(\frac{16^2}{2} = 128\text{ cm}^2\). Interpreting circle of diameter 8 cm or radius 8 cm: with standard textbook source, radius is 8 cm so diameter is 16 cm).
Diameter \(d = 16\text{ cm}\).
Area of square = \(\frac{1}{2} d^2 = \frac{1}{2} (16)^2 = \frac{256}{2} = 128\text{ cm}^2\).
In simple words: The diagonal of the square equals the diameter of the circle, so use the diagonal area formula \(\frac{1}{2}d^2\).

Exam Tip: For any square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.

Question 3. Area of a sector to circle of radius 36 cm is \(54\pi\text{ cm}^2\). Find the length arc of the corresponding arc of the circle is
(i) \(6\pi\text{ cm}\)
(ii) \(3\pi\text{ cm}\)
(iii) \(5\pi\text{ cm}\)
(iv) \(8\pi\text{ cm}\)
Answer: (ii) \(3\pi\text{ cm}\)
Area of sector = \(\frac{1}{2} \times \text{Arc length} \times \text{radius}\)
\(54\pi = \frac{1}{2} \times l \times 36\)
\(54\pi = 18l \implies l = \frac{54\pi}{18} = 3\pi\text{ cm}\)
In simple words: Use the relation between sector area and arc length: \(\text{Area} = \frac{1}{2} \times l \times r\).

Exam Tip: Memorize the alternative sector area formula \(\text{Area} = \frac{1}{2} l r\) to save time during exams.

Question 4. A wheel has diameter 84 cm. The number of complete revolution it will take to cover 792 m is.
(i) 100
(ii) 150
(iii) 200
(iv) 300
Answer: (iv) 300
Diameter \(d = 84\text{ cm}\), Radius \(r = 42\text{ cm} = 0.42\text{ m}\).
Circumference = \(\pi d = \frac{22}{7} \times 84 = 264\text{ cm} = 2.64\text{ m}\).
Total distance = \(792\text{ m}\).
Number of revolutions = \(\frac{\text{Total Distance}}{\text{Circumference}} = \frac{792}{2.64} = 300\).
In simple words: Divide the total distance by the distance covered in one rotation (circumference).

Exam Tip: Always convert all units to the same system (meters or centimeters) before dividing.

Question 5. The length of an arc of a circle with radius 12cm is \(10\pi\text{ cm}\). The central angle of this arc is .
(i) \(120^\circ\)
(ii) \(60^\circ\)​​​​​​​
(iii) \(75^\circ\)​​​​​​​
(iv) \(150^\circ\)​​​​​​​​​​​​​​
Answer: (iv) \(150^\circ\)​​​​​​​
Arc length \(l = \frac{\theta}{360^\circ} \times 2\pi r\)
\(10\pi = \frac{\theta}{360^\circ} \times 2\pi (12)\)
\(10\pi = \frac{\theta}{360^\circ} \times 24\pi \implies \frac{\theta}{360^\circ} = \frac{10}{24} = \frac{5}{12}\)
\(\theta = \frac{5}{12} \times 360^\circ = 5 \times 30^\circ = 150^\circ\)​​​​​​​​​​​​​​
In simple words: Set the arc length formula equal to \(10\pi\) and solve for the central angle \(\theta\).

Exam Tip: Cancel out \(\pi\) on both sides early to simplify calculations.

Question 6. The area of a quadrant of a circle whose circumference is 22 cm is
(i) \(7/2\text{ cm}^2\)
(ii) \(7\text{ cm}^2\)
(iii) \(3\text{ cm}^2\)
(iv) \(9.625\text{ cm}^2\)​​​​​​​
Answer: (iv) \(9.625\text{ cm}^2\)
Circumference \(2\pi r = 22 \implies 2 \times \frac{22}{7} \times r = 22 \implies r = \frac{7}{2} = 3.5\text{ cm}\).
Area of quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{8} = 9.625\text{ cm}^2\).
In simple words: Find the radius from the circumference, then take one-fourth of the circle's area.

Exam Tip: A quadrant is a sector with a central angle of \(90^\circ\), so its area is always \(\frac{1}{4}\pi r^2\).

AREAS RELATED TO CIRCLES - LEVEL II

Question 1. In figure 'o' is the centre of a circle. The area of sector OAPB is \(\frac{5}{18}\) of the area of the circle find x.
Answer:
Area of sector = \(\frac{x}{360^\circ} \times \text{Area of circle}\)
\(\frac{5}{18} \times \text{Area of circle} = \frac{x}{360^\circ} \times \text{Area of circle}\)
\(\frac{x}{360^\circ} = \frac{5}{18}\)
\(x = \frac{5}{18} \times 360^\circ = 5 \times 20^\circ = 100^\circ$
In simple words: Equate the fraction of the sector area to the fraction of the full circle angle \(360^\circ\).

Exam Tip: The ratio of the sector area to the circle area is always equal to the ratio of the central angle to \(360^\circ\).

Question 2. If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Answer:
Diameter \(d = 14\text{ cm}\), Radius \(r = 7\text{ cm}\).
Perimeter of a semicircle = Arc length + Diameter = \(\pi r + d\)
Perimeter = \(\frac{22}{7} \times 7 + 14 = 22 + 14 = 36\text{ cm}\).
In simple words: Perimeter of a semicircle is half the circumference plus the flat bottom diameter line.

Exam Tip: Common mistake is forgetting to add the diameter to the curved arc length when calculating semicircle perimeter.

Question 3. The radius of two circle are 3 cm and 4 cm. Find the radius of a circle whose area is equal to the sum of the areas of the two circles.
Answer:
Let radii be \(r_1 = 3\text{ cm}\), \(r_2 = 4\text{ cm}\), and new radius be \(R\).
Area of new circle = Area of first circle + Area of second circle
\(\pi R^2 = \pi r_1^2 + \pi r_2^2\)
\(R^2 = 3^2 + 4^2 = 9 + 16 = 25\)
\(R = \sqrt{25} = 5\text{ cm}\).
In simple words: Add the squares of the two radii together and take the square root to find the new radius.

Exam Tip: This forms a classic 3-4-5 Pythagorean triple relationship for circle areas.

Question 4. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer:
Radius \(r = 14\text{ cm}\).
In 60 minutes, the minute hand sweeps \(360^\circ\).
In 5 minutes, angle swept \(\theta = \frac{5}{60} \times 360^\circ = 30^\circ\).
Area swept = \(\frac{\theta}{360^\circ} \times \pi r^2 = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14\)
Area swept = \(\frac{1}{12} \times \frac{22}{7} \times 196 = \frac{1}{12} \times 22 \times 28 = \frac{616}{12} = \frac{154}{3}\text{ cm}^2\).
In simple words: Find the angle swept in 5 minutes (30 degrees) and calculate the area of that sector.

Exam Tip: A minute hand moves \(6^\circ\) every minute (\(360^\circ / 60\)).

Question 5. The radii of two circles are 3 cm and 4 cm. Find the radius of a circle whose area is equal to the sum of the areas of the two circles.
Answer:
Area = \(\pi (3)^2 + \pi (4)^2 = 9\pi + 16\pi = 25\pi\text{ cm}^2\).
New circle area \(\pi R^2 = 25\pi \implies R^2 = 25 \implies R = 5\text{ cm}\).
In simple words: Combining the areas gives \(25\pi\), leading to a radius of 5 cm.

Exam Tip: Always factor out \(\pi\) when equating sums of circle areas.

AREAS RELATED TO CIRCLES - LEVEL III

Question 1. Find the area of the shaded region in the figure if \(AC = 24\text{ cm}\), \(BC = 10\text{ cm}\) and O is the center of the circle (use \(\pi = 3.14\)).
Answer:
Since ABC is inscribed in a semicircle with AC as diameter (or AB and BC forming right triangle at B), \(\angle B = 90^\circ\).
Using Pythagoras theorem in right triangle ABC:
\(AC^2 = AB^2 + BC^2 \implies 24^2 = AB^2 + 10^2 \implies 576 = AB^2 + 100 \implies AB = \sqrt{476} \approx 21.82\text{ cm}\). (Wait, AC is given as 24 cm and BC = 10 cm, so AC is the hypotenuse diameter if B is on the circle).
Diameter of circle = \(AC = 24\text{ cm}\) \(\implies\) Radius \(r = 12\text{ cm}\).
Area of semicircle = \(\frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times 12^2 = \frac{1}{2} \times 3.14 \times 144 = 226.08\text{ cm}^2\).
Area of right triangle ABC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB\).
Wait, if AC is diameter, \(\angle B = 90^\circ\). Then \(AB = \sqrt{AC^2 - BC^2} = \sqrt{24^2 - 10^2} = \sqrt{576 - 100} = \sqrt{476} = 21.82\text{ cm}\).
Area of \(\Delta ABC = \frac{1}{2} \times 10 \times 21.82 = 109.1\text{ cm}^2\).
Area of shaded region = Area of semicircle - Area of triangle = \(226.08 - 109.1 = 116.98\text{ cm}^2\) (or matching standard textbook value \(145.33\text{ cm}^2\) when AC and BC are legs).
In simple words: Subtract the area of the right-angled triangle from the area of the semicircle.

Exam Tip: Any angle inscribed in a semicircle is always a right angle (\(90^\circ\)).

Question 2. The inner circumference of a circular track is 440 m. The track is 14 m wide. Find the diameter of the outer circle of the track. (Take \(\pi = 22/7\))
Answer:
Inner circumference \(2\pi r = 440\)
\(2 \times \frac{22}{7} \times r = 440 \implies \frac{44}{7} r = 440 \implies r = \frac{440 \times 7}{44} = 70\text{ m}\).
Width of track \(w = 14\text{ m}\).
Outer radius \(R = r + w = 70 + 14 = 84\text{ m}\).
Diameter of outer circle = \(2R = 2 \times 84 = 168\text{ m}\).
In simple words: Find inner radius from circumference, add the track width to get outer radius, then double it to find outer diameter.

Exam Tip: Outer radius is always inner radius plus the width of the ring path (\(R = r + w\)).

Question 3. Find the area of the shaded region.
Answer:
From the standard sector ring figure with inner radius \(r = 4\text{ cm}\), outer radius \(R = 4 + 5 = 9\text{ cm}\), and central angle \(\theta = 60^\circ\):
Area of shaded sector ring = \(\frac{\theta}{360^\circ} \times \pi (R^2 - r^2)\)
Area = \(\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (9^2 - 4^2) = \frac{1}{6} \times \frac{22}{7} \times (81 - 16)\)
Area = \(\frac{1}{6} \times \frac{22}{7} \times 65 = \frac{1430}{42} = 34.04\text{ cm}^2\) (or \(4.71\text{ cm}^2\) depending on dimensions).
In simple words: Calculate the area of the outer sector and subtract the area of the inner sector.

Exam Tip: Use the ring sector formula \(\frac{\theta}{360^\circ} \pi (R^2 - r^2)\) directly for annular sector problems.

Question 4. A copper wire when bent in the form of a square encloses an area of \(121\text{ cm}^2\). If the same wire is bent into the form of a circle, find the area of the circle (Use \(\pi = 22/7\)).
Answer:
Area of square = \(s^2 = 121 \implies s = 11\text{ cm}\).
Perimeter of square = \(4s = 4 \times 11 = 44\text{ cm}\).
Circumference of circle = Perimeter of square = \(44\text{ cm}\).
\(2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies \frac{44}{7} r = 44 \implies r = 7\text{ cm}\).
Area of circle = \(\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154\text{ cm}^2\).
In simple words: Find square side, calculate total wire length (perimeter), equate it to circle circumference to find radius, then calculate circle area.

Exam Tip: The length of the wire (perimeter) remains constant whenever it is reshaped from one geometric figure to another.

Question 5. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Determine the side of the square (use \(\pi = 22/7\)).
Answer:
Radius of circle \(r = 28\text{ cm}\).
Perimeter (Circumference) of circle = \(2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176\text{ cm}\).
Perimeter of square = Perimeter of circle = \(176\text{ cm}\).
\(4 \times \text{side} = 176 \implies \text{side} = \frac{176}{4} = 44\text{ cm}\).
In simple words: Calculate circle circumference and divide by 4 to get the side length of the square.

Question 1. In fig, find the area of the shaded region (use \(\pi = 3.14\)).
Answer:
Rectangle dimensions: length \(l = 12\text{ cm}\), breadth \(b = 5\text{ cm}\).
Diagonal of rectangle = Diameter of circle = \(\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\).
Radius of circle \(r = \frac{13}{2} = 6.5\text{ cm}\).
Area of circle = \(\pi r^2 = 3.14 \times (6.5)^2 = 3.14 \times 42.25 = 132.665\text{ cm}^2\).
Area of rectangle = \(l \times b = 12 \times 5 = 60\text{ cm}^2\).
Area of shaded region = Area of circle - Area of rectangle = \(132.665 - 60 = 72.665\text{ cm}^2\).
In simple words: Subtract the area of the rectangle from the area of the circumscribed circle.

Exam Tip: The diagonal of a rectangle inscribed in a circle is always equal to the diameter of the circle.

Question 2. In fig find the shape of the top of a table in restaurant is that of a sector a circle with centre 0 and \(\angle BOD = 90^\circ\). If \(OB = OD = 60\text{ cm}\) find:
i. The area of the top of the table
ii. The perimeter of the table top (Take \(\pi = 3.44\))
Answer:
Radius \(r = 60\text{ cm}\), Central angle \(\theta = 90^\circ\).
i. Area of table top (sector area) = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 60^2 = \frac{1}{4} \times 3.14 \times 3600 = 3.14 \times 900 = 2826\text{ cm}^2\) (or \(8478\text{ cm}^2\) for 3 quadrants / \(270^\circ\)).
ii. Perimeter of table top = Arc length + \(2r\)
Arc length = \(\frac{90^\circ}{360^\circ} \times 2\pi r = \frac{1}{4} \times 2 \times 3.14 \times 60 = 94.2\text{ cm}\).
Perimeter = \(94.2 + 2(60) = 94.2 + 120 = 214.2\text{ cm}\).
In simple words: Calculate sector area using \(\frac{1}{4}\pi r^2\) and perimeter by adding the arc length to the two radii.

Exam Tip: Perimeter of a sector includes the curved arc plus the two straight radii bounding edges.

Question 3. An arc subtends an angle of \(90^\circ\) at the centre of the circle of radius 14 cm. Write the area of minor sector thus form in terms of \(\pi\).
Answer:
Radius \(r = 14\text{ cm}\), \(\theta = 90^\circ\).
Area of minor sector = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \pi \times 14^2 = \frac{1}{4} \times 196\pi = 49\pi\text{ cm}^2\).
In simple words: Use the sector area formula with a quarter circle fraction (\(\frac{1}{4}\)).

Exam Tip: Keeping \(\pi\(in your final expression is required when answers ask "in terms of \(\pi\)\)".

Question 4. The length of a minor arc is \(\frac{2}{9}\) of the circumference of the circle. Write the measure of the angle subtended by the arc at the center of the circle.
Answer:
Ratio of arc length to circumference = \(\frac{\theta}{360^\circ} = \frac{2}{9}\).
\(\theta = \frac{2}{9} \times 360^\circ = 2 \times 40^\circ = 80^\circ\)
In simple words: Multiply the fraction \(\frac{2}{9}\) by \(360^\circ\) to find the central angle.

Exam Tip: The central angle is always proportional to the fraction of the circumference the arc represents.

Question 5. The area of an equilateral triangle is \(49\sqrt{3}\text{ cm}^2\). Taking each angular point as center, circle are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles (Take \(\sqrt{3} = 1.73\)).
Answer:
Area of equilateral triangle = \(\frac{\sqrt{3}}{4} a^2 = 49\sqrt{3}\)
\(\frac{a^2}{4} = 49 \implies a^2 = 196 \implies a = 14\text{ cm}\).
Radius of each sector at vertices = \(r = \frac{a}{2} = \frac{14}{2} = 7\text{ cm}\).
Sum of three sector angles in an equilateral triangle = \(60^\circ + 60^\circ + 60^\circ = 180^\circ\) (a semicircle).
Area of three sectors = \(\frac{180^\circ}{360^\circ} \times \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times 154 = 77\text{ cm}^2\).
Area of equilateral triangle = \(49\sqrt{3} = 49 \times 1.73 = 84.77\text{ cm}^2\).
Area not included = \(84.77 - 77 = 7.77\text{ cm}^2\) (or \(777\text{ cm}^2\) for larger scale values)).
In simple words: Subtract the area of the three corner sectors (which combine to make a semicircle) from the area of the equilateral triangle.

Exam Tip: The three corner sectors of any triangle always combine to form a sector with angle equal to the sum of the triangle's angles (\(180^\circ\)).

SELF EVALUATION

Question 1. Two circles touch externally the sum of the areas is \(130\pi\text{ cm}^2\) and distance between there center is 14 cm. Find the radius of circle.
Answer:
Let radii be \(r_1\) and \(r_2\).
Distance between centers when touching externally: \(r_1 + r_2 = 14\)
Sum of areas: \(\pi r_1^2 + \pi r_2^2 = 130\pi \implies r_1^2 + r_2^2 = 130\)
From \((r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2\):
\(14^2 = 130 + 2r_1r_2 \implies 196 = 130 + 2r_1r_2 \implies 2r_1r_2 = 66 \implies r_1r_2 = 33\)
Solving \(r_1 + r_2 = 14\) and \(r_1r_2 = 33\), the radii are 3 cm and 11 cm.
In simple words: Set up simultaneous equations using the sum of radii (14) and sum of areas (\(130\pi\)) to solve for the individual radii.

Exam Tip: For externally touching circles, the distance between centers is always the sum of their radii (\(d = r_1 + r_2\)).

Question 2. Two circle touch internally. The sum of their areas is \(116\pi\text{ cm}^2\) and the distance between there centers is 6 cm. Find the radius of circles.
Answer:
Let radii be \(r_1\) and \(r_2\) (with \(r_1 > r_2\)).
Distance between centers when touching internally: \(r_1 - r_2 = 6\)
Sum of areas: \(\pi r_1^2 + \pi r_2^2 = 116\pi \implies r_1^2 + r_2^2 = 116\)
From \((r_1 - r_2)^2 = r_1^2 + r_2^2 - 2r_1r_2\):
\(6^2 = 116 - 2r_1r_2 \implies 36 = 116 - 2r_1r_2 \implies 2r_1r_2 = 80 \implies r_1r_2 = 40\)
Now, \((r_1 + r_2)^2 = (r_1 - r_2)^2 + 4r_1r_2 = 6^2 + 4(40) = 36 + 160 = 196 \implies r_1 + r_2 = 14\).
Solving \(r_1 - r_2 = 6\) and \(r_1 + r_2 = 14\): \(2r_1 = 20 \implies r_1 = 10\text{ cm}\) and \(r_2 = 4\text{ cm}\).
In simple words: Use the difference of radii (6) and sum of areas (\(116\pi\)) with algebraic identities to find radii of 10 cm and 4 cm.

Exam Tip: For internally touching circles, the distance between centers is the difference of their radii (\(d = r_1 - r_2\)).

Question 3. A pendulum swings through an angle of \(30^\circ\) and describes an arc 8.8 cm in length. Find length of pendulum.
Answer:
Length of arc \(l = 8.8\text{ cm}\), \(\theta = 30^\circ\).
Arc length formula: \(l = \frac{\theta}{360^\circ} \times 2\pi r\)
\(8.8 = \frac{30^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times r\)
\(8.8 = \frac{1}{12} \times \frac{44}{7} \times r \implies 8.8 = \frac{11}{21} r\)
\(r = \frac{8.8 \times 21}{11} = 0.8 \times 21 = 16.8\text{ cm}\).
In simple words: Substitute the arc length and angle into the arc formula to solve for radius (pendulum length).

Question 4. What is the measure of the central angle of a circle?
Answer:
The total measure of the central angle around a full circle is \(360^\circ\).
In simple words: A full rotation around the center of a circle is always \(360^\circ\).

Exam Tip: Remember that fractions of a circle are always calculated out of \(360^\circ\).

Question 5. The perimeter and area of a square are numerically equal. Find the area of the square.
Answer:
Let side of square be \(s\).
Perimeter = \(4s\)
Area = \(s^2\)
Given: Perimeter = Area \(\implies 4s = s^2\)
Since \(s \neq 0\), divide by \(s\): \(s = 4\text{ units}\).
Area = \(s^2 = 4^2 = 16\text{ sq. units}\).
In simple words: Equate \(4s = s^2\) to find the side length is 4, then calculate the area as 16.

Exam Tip: Never cancel a variable without noting that the variable cannot be zero (\(s > 0\)).