CBSE Class 10 Mathematics Area Relates to Circles Assignment Set 04

Question 1. In the adjoining figure \(\Delta ABC\) right angled triangle right angled at A. Semi circles are drawn on the sides of the triangle \(\Delta ABC\). Prove that area of the Shaded region is equal to area of \(\Delta ABC\)
Answer:
Let the sides of the right-angled triangle be \(a, b,\) and \(c\), where \(c\) is the hypotenuse opposite to angle A.
By Pythagoras theorem, \(b^2 + c^2 = a^2\).
The area of the right-angled triangle \(\Delta ABC\) is given by \(\frac{1}{2}bc\).
Semi-circles are drawn on sides \(b, c,\) and \(a\) as diameters. Their diameters are \(b, c,\) and \(a\), so their radii are \(\frac{b}{2}, \frac{c}{2},\) and \(\frac{a}{2}\).
Area of semi-circle on side \(b\) = \(\frac{1}{2}\pi \left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{8}\)
Area of semi-circle on side \(c\) = \(\frac{1}{2}\pi \left(\frac{c}{2}\right)^2 = \frac{\pi c^2}{8}\)
Area of semi-circle on hypotenuse \(a\) = \(\frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8}\)
Sum of areas of semi-circles on the two perpendicular sides = \(\frac{\pi b^2}{8} + \frac{\pi c^2}{8} = \frac{\pi (b^2 + c^2)}{8}\)
Since \(b^2 + c^2 = a^2\), the sum becomes \(\frac{\pi a^2}{8}\), which is exactly the area of the semi-circle on the hypotenuse.
The shaded region area = (Sum of areas of semi-circles on the two sides + Area of \(\Delta ABC\)) - (Area of semi-circle on the hypotenuse)
Area of shaded region = \(\left(\frac{\pi a^2}{8} + \text{Area of } \Delta ABC\right) - \frac{\pi a^2}{8} = \text{Area of } \Delta ABC\).
In simple words: The combined area of the semi-circles on the two smaller sides equals the semi-circle on the hypotenuse, leaving the shaded region equal to the triangle's area.

Exam Tip: Use Pythagoras theorem and express the areas of the semi-circles in terms of the triangle sides to easily prove such geometric identities.

Question 2. The sum of the diameters of two circles is 2.8 m and their difference of circumferences is 0.88m. Find the radii of the two circles
Answer:
Let the radii of the two circles be \(r_1\) and \(r_2\) where \(r_1 > r_2\).
Sum of diameters = \(2r_1 + 2r_2 = 2.8\text{ m} = 280\text{ cm}\)
\(r_1 + r_2 = 140\text{ cm}\) --- (1)
Difference of circumferences = \(2\pi r_1 - 2\pi r_2 = 0.88\text{ m} = 88\text{ cm}\)
\(2\pi (r_1 - r_2) = 88\)
\(r_1 - r_2 = \frac{88}{2\pi} = \frac{88 \times 7}{2 \times 22} = 14\text{ cm}\) --- (2)
Adding equation (1) and equation (2):
\(2r_1 = 140 + 14 = 154\)
\(r_1 = 77\text{ cm}\)
Substituting \(r_1\) in equation (1):
\(77 + r_2 = 140\)
\(r_2 = 140 - 77 = 63\text{ cm}\)
In simple words: Set up simultaneous linear equations using the given sum of diameters and difference of circumferences to find the two radii.

Exam Tip: Always convert all units to the same system (such as centimeters) before performing calculations to avoid calculation errors.

Question 3. Find the circumference of a circle whose area is 16 times the area of the circle with diameter 7cm
Answer:
Let the radius of the required circle be \(R\) and the radius of the given circle be \(r\).
Given diameter of the second circle = 7 cm, so \(r = \frac{7}{2}\text{ cm}\).
Area of the first circle = \(16 \times\) Area of the second circle
\(\pi R^2 = 16 \pi r^2\)
\(R^2 = 16 r^2\)
\(R^2 = 16 \times \left(\frac{7}{2}\right) \times \left(\frac{7}{2}\right)\)
\(R^2 = 16 \times \frac{49}{4} = 4 \times 49 = 196\)
\(R = \sqrt{196} = 14\text{ cm}\)
Circumference = \(2\pi R = 2 \times \frac{22}{7} \times 14 = 2 \times 22 \times 2 = 88\text{ cm}\)
In simple words: Use the area ratio to find the new radius, then apply the circumference formula.

Exam Tip: When given a ratio of areas, square roots are often needed to find the ratio of radii or the new radius directly.

Question 4. Find the area enclosed between two concentric circles of radii 3.5cm, 7cm. A third concentric circle is drawn outside the circle so that the area enclosed between it and the 7cm circle is same as that between two inner circles. Find the radius of the third circle
Answer:
Radius of inner circle \(r_1 = 3.5\text{ cm}\)
Radius of middle circle \(r_2 = 7\text{ cm}\)
Let the radius of the third outer circle be \(R\).
Area between first two circles = \(\pi r_2^2 - \pi r_1^2 = \pi (7)^2 - \pi (3.5)^2 = 49\pi - 12.25\pi = 36.75\pi\)
Area between third circle and 7 cm circle = \(\pi R^2 - \pi (7)^2 = \pi R^2 - 49\pi\)
According to the question, these two areas are equal:
\(\pi R^2 - 49\pi = 49\pi - 12.25\pi\)
\(\pi R^2 = 98\pi - 12.25\pi\)
\(\pi R^2 = 85.75\pi\)
\(R^2 = 85.75 = \frac{343}{4}\)
\(R = \sqrt{\frac{343}{4}} = \frac{\sqrt{343}}{2}\text{ cm}\)
In simple words: Equate the area of the outer ring to the area of the inner ring to find the outer radius.

Exam Tip: Keep \(\pi\) in symbolic form during intermediate steps to simplify calculations before substituting its numerical value at the end.

Question 5. Two circles touch externally. The sum of their areas is \(58\pi\text{ cm}^2\) and the distance between their centres is 10 cm. Find the radii of the two circles.
Answer:
Let the radii of the two circles be \(r\) and \(10 - r\).
Sum of areas = \(\pi r^2 + \pi (10 - r)^2 = 58\pi\)
\(\pi [r^2 + (100 - 20r + r^2)] = 58\pi\)
\(2r^2 - 20r + 100 = 58\)
\(2r^2 - 20r + 42 = 0\)
\(r^2 - 10r + 21 = 0\)
\((r - 7)(r - 3) = 0\)
Therefore, \(r = 7\text{ cm}\) or \(r = 3\text{ cm}\).
The radii of the two circles are 7 cm and 3 cm.
In simple words: Use a quadratic equation formed by the sum of circle areas and the distance between their centers to find the radii.

Exam Tip: Factoring quadratic equations quickly saves time in exams; look for factors that multiply to the constant term and add to the middle coefficient.

Question 6. From a sheet of cardboard in the shape of a square of side 14 cm, a piece in the shape of letter B is cut off. The curved side of the letter consists of two equal semicircles & the breadth of the rectangular piece is 1 cm. Find the area of the remaining part of cardboard.
Answer:
Side of the square = 14 cm
Area of the square = \(14 \times 14 = 196\text{ cm}^2\)
Breadth of the rectangular strip = 1 cm
Total height of the square = 14 cm, which means the diameter of each semicircle is \(\frac{14}{2} = 7\text{ cm}\), so radius \(r = 3.5\text{ cm}\).
Area of two semi-circles = Area of one full circle = \(\pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\text{ cm}^2\)
Area of the rectangular attachment = \(\text{length} \times \text{breadth} = 14 \times 1 = 14\text{ cm}^2\)
Area of remaining portion = Area of square - (Area of two semi-circles + Area of rectangle)
Area of remaining portion = \(196 - 38.5 - 14 = 143.5\text{ cm}^2\)
In simple words: Subtract the area of the cut-out shapes (two semicircles and a rectangle) from the total area of the square.

Exam Tip: Two identical semicircles sharing the same diameter combine to form one complete circle, simplifying area calculations.

Question 7. A piece of cardboard in the shape of a trapezium ABCD & AB \(\parallel\) DE, \(\angle BCD = 90^\circ\), quarter circle BFEC is removed. Given AB = BC = 3.5 cm, DE = 2 cm. Calculate the area of remaining piece of cardboard.
Answer:
Area of trapezium ABCD = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Parallel sides are \(AB\) and \(CD\). Here \(AB = 3.5\text{ cm}\), \(CD = CE + ED = 3.5 + 2 = 5.5\text{ cm}\).
Height \(BC = 3.5\text{ cm}\).
Area of trapezium = \(\frac{1}{2} \times (3.5 + 5.5) \times 3.5 = \frac{1}{2} \times 9 \times 3.5 = 15.75 \times 1.75 = 15.75 - \frac{19.25}{2} = 15.75 - 9.625 = 6.125\text{ cm}^2\)
Area of the removed quarter circle = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625\text{ cm}^2\) (Wait, let us compute precisely: \(\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} = 9.625\)).
Area of remaining portion = Area of trapezium - Area of quarter circle
Area = \(15.75 - 9.625 = 6.125\text{ cm}^2\)
In simple words: Subtract the area of the quarter circle from the area of the trapezium.

Exam Tip: Ensure all dimensions of parallel sides and height are correctly identified from the given figure components before applying formulas.

Question 8. In the figure, ABCD is a square inside a circle with centre O. The Centre of the square coincides with O & the diagonal AC is horizontal of AP, DO are vertical & AP = 45 cm, DQ = 25 cm. Find a) the radius of the circle b) c) area of shaded region (use \(\pi = 3.14, \sqrt{2} = 1.41\))
Answer:
a) Radius of the circle = 53 cm
b) Side length / dimensions = 39.48 cm
c) Area of shaded region = \(7252.26\text{ cm}^2\)
In simple words: Find circle and square dimensions using vertical and horizontal coordinates to determine the shaded area.

Exam Tip: Break complex figures involving inscribed squares and circles into simpler geometric segments.

Question 9. The area enclosed between two concentric circles is \(770\text{ cm}^2\). If the radius of the outer circle is 21cm, find the radius of the inner circle.
Answer:
Let the radius of the inner circle be \(r\).
Area between concentric circles = \(\pi R^2 - \pi r^2 = 770\)
\(\pi (21^2 - r^2) = 770\)
\(\frac{22}{7} (441 - r^2) = 770\)
\(441 - r^2 = \frac{770 \times 7}{22} = 35 \times 7 = 245\)
\(r^2 = 441 - 245 = 196\)
\(r = \sqrt{196} = 14\text{ cm}\)
In simple words: Subtract the inner circle area from the outer circle area to find the inner radius using the ring area formula.

Exam Tip: Cancel out common factors on both sides of the equation early to simplify arithmetic.

Question 10. A circular disc of 6 cm radius is divided into three sectors with central angles \(120^\circ, 150^\circ, 90^\circ\). What part of the circle is the sector with central angles \(120^\circ\). Also give the ratio of the areas of three sectors.
Answer:
Fraction of circle for \(120^\circ\) sector = \(\frac{120^\circ}{360^\circ} = \frac{1}{3}\) (i.e., \(\frac{1}{3}\) of the area of the circle).
Ratio of the areas of the three sectors = \(\frac{120}{360} \pi r^2 : \frac{150}{360} \pi r^2 : \frac{90}{360} \pi r^2\)
= \(120 : 150 : 90\)
= \(4 : 5 : 3\)
In simple words: Divide the central angle by \(360^\circ\) to find the fraction, and simplify the ratio of angles to find the area ratio.

Exam Tip: The ratio of sector areas is always equal to the ratio of their central angles when the radius is constant.

Question 11. If the minute hand of a big clock is 1.05 m long, find the rate at which its tip is moving in cm per minute.
Answer:
Radius \(r = 1.05\text{ m} = 105\text{ cm}\)
Distance covered by the tip in one complete revolution (60 minutes) = Circumference of the circle = \(2\pi r = 2 \times \frac{22}{7} \times 105 = 2 \times 22 \times 15 = 660\text{ cm}\)
Rate of movement per minute = \(\frac{660\text{ cm}}{60\text{ minutes}} = 11\text{ cm/min}\)
In simple words: Calculate the total circumference covered in one hour and divide by 60 to get the speed per minute.

Exam Tip: Pay close attention to unit conversions (meters to centimeters) as requested in the final answer format.

Question 12. ABC is a right angled triangle in which \(\angle A = 90^\circ\). Find the area of the shaded region if AB = 6 cm, BC = 10cm & I is the centre of the Incircle of \(\Delta ABC\).
Answer:
In right-angled triangle ABC, \(\angle A = 90^\circ\).
\(BC = 10\text{ cm}\), \(AB = 6\text{ cm}\).
By Pythagoras theorem, \(AC = \sqrt{BC^2 - AB^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}\).
Area of \(\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24\text{ cm}^2\)
Let the inradius of the incircle be \(r\).
The area of the triangle can also be written as the sum of areas of three triangles formed by the incentre: \(\frac{1}{2} \times AB \times r + \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC \times r = 24\)
\(\frac{1}{2} r (6 + 10 + 8) = 24\)
\(\frac{1}{2} r (24) = 24\)
\(12r = 24 \implies r = 2\text{ cm}\)
Area of the incircle = \(\pi r^2 = \frac{22}{7} \times (2)^2 = \frac{88}{7}\text{ cm}^2\)
Area of shaded region = Area of \(\Delta ABC - \text{Area of incircle} = 24 - \frac{88}{7} = \frac{168 - 88}{7} = \frac{80}{7}\text{ cm}^2\)
In simple words: Find the inradius by equating the triangle area to the sum of three sub-triangle areas, then subtract the circle's area from the triangle's area.

Exam Tip: The area of a triangle with an inscribed circle is always equal to \(\frac{1}{2} \times \text{perimeter} \times \text{inradius}\).

Question 13. Find the perimeter of the figure, where AED is a semi-circle and ABCD is a rectangle.
Answer:
Dimensions: Rectangle sides = 20 cm and 14 cm.
Diameter of semi-circle = 14 cm, so radius \(r = 7\text{ cm}\).
Perimeter of the figure = \(AB + BC + CD + \text{Length of arc AED}\)
\(AB = 20\text{ cm}\), \(BC = 14\text{ cm}\), \(CD = 20\text{ cm}\)
Length of arc AED = \(\pi r = \frac{22}{7} \times 7 = 22\text{ cm}\)
Total Perimeter = \(20 + 14 + 20 + 22 = 76\text{ cm}\)
In simple words: Add the lengths of the three outer straight sides of the rectangle and the curved arc of the semicircle.

Exam Tip: Perimeter only includes the outer exposed boundaries; do not include the line where the semicircle meets the rectangle.

Question 14. Find the area of shaded region of circle of radius = 7cm, if \(\angle AOB = 70^\circ\), \(\angle COD = 50^\circ\) and \(\angle EOF = 60^\circ\).
Answer:
Area of shaded region = Area(Sector AOB + Sector COD + Sector EOF)
\(= \frac{70^\circ}{360^\circ} \pi r^2 + \frac{50^\circ}{360^\circ} \pi r^2 + \frac{60^\circ}{360^\circ} \pi r^2\)
\(= \pi (7)^2 \left( \frac{70 + 50 + 60}{360} \right)\)
\(= 49\pi \left( \frac{180}{360} \right) = 49\pi \times \frac{1}{2} = 49 \times \frac{22}{7} \times \frac{1}{2} = 7 \times 11 = 77\text{ cm}^2\)
In simple words: Combine the central angles of all shaded sectors and calculate the total area using a single fraction.

Exam Tip: When multiple sectors share the same radius, you can add their central angles together before calculating the area.

Question 15. What is the ratio of the areas of sectors I and II?
Answer:
Central angle of sector I = \(120^\circ\)
Central angle of sector II = \(150^\circ\)
Ratio of areas = \(\frac{120^\circ}{360^\circ} \pi r^2 : \frac{150^\circ}{360^\circ} \pi r^2\)
\(= 120 : 150 = 4 : 5\)
In simple words: The ratio of sector areas is directly proportional to the ratio of their central angles.

Exam Tip: Cancel common terms like \(\frac{\pi r^2}{360^\circ}\) immediately when finding ratios of sectors in the same circle.

Question 16. Find the area of shaded region, if the side of square is 28cm and radius of the sector is \(\frac{1}{2}\) the length of side of square.
Answer:
Side of square = 28 cm
Radius of sector = \(\frac{1}{2} \times 28 = 14\text{ cm}\)
Area of shaded region = Area of square + Area of sectors
Let us compute: \(2 \times \left(\frac{270^\circ}{360^\circ}\right) \pi (14)^2 + 28 \times 28\)
\(= 2 \times \frac{3}{4} \times \frac{22}{7} \times 14 \times 14 + 784\)
\(= 924 + 784 = 1708\text{ cm}^2\)
In simple words: Add the area of the square to the areas of the attached sectors to get the total shaded area.

Exam Tip: Check whether sectors are inside or outside the main shape to correctly add or subtract their areas.

Question 17. If OA = OB = 14cm, \(\angle AOB = 90^\circ\), find the area of shaded region.
Answer:
Radius \(r = 14\text{ cm}\)
Area of shaded region = Area of \(\Delta AOB\) - Area of quadrant
Area of \(\Delta AOB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2\)
Area of quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 77\text{ cm}^2\)
Area of shaded region = \(98 - 77 = 21\text{ cm}^2\)
In simple words: Subtract the area of the circular quadrant from the area of the right-angled triangle.

Exam Tip: A sector with a \(90^\circ\) central angle is exactly one-fourth of a circle (a quadrant).

Question 18. The given figure consists of four small semicircles and two big semicircles. If the smaller semicircles are equal in radii and the bigger semicircles are also equal in radii, find the perimeter and the area of the shaded portion of the figure. Given that radius of each bigger semicircle is 42cm.
Answer:
Radius of each bigger semicircle \(R = 42\text{ cm}\).
Radius of each smaller semicircle \(r = 21\text{ cm}\).
Perimeter of the shaded region = Perimeter of 2 bigger semicircles + Perimeter of 4 smaller semicircles
\(= 2(\pi R) + 4(\pi r) = 2 \left(\frac{22}{7} \times 42\right) + 4 \left(\frac{22}{7} \times 21\right) = 2(132) + 4(66) = 264 + 264 = 528\text{ cm}\)
Area of shaded region = Area of 2 bigger semicircles + Area of 4 smaller semicircles
\(= 2 \left(\frac{1}{2} \pi R^2\right) + 4 \left(\frac{1}{2} \pi r^2\right) = \pi (42)^2 + 2 \pi (21)^2 = \frac{22}{7} \times 1764 + 2 \times \frac{22}{7} \times 441 = 5544\text{ cm}^2\)
In simple words: Add the circumferences and areas of all the constituent semicircles to find the total perimeter and area.

Exam Tip: Two semicircles of the same radius combine to form one full circle, making perimeter and area calculations faster.

Question 19. The boundary of the shaded portion in the adjoining figure consists of four half-circles and two quarter-circles. Find the length of the boundary and the area of the shaded portion, if OA=OB=OC=OD=14cm.
Answer:
Radius \(r = 14\text{ cm}\)
Length of boundary (Perimeter) = \(132\text{ cm}\)
Area of shaded portion = \(308\text{ cm}^2\)
In simple words: Sum the arc lengths of all semicircles and quarter circles for the boundary, and add their areas for the total shaded region.

Answer:
Radius \(r = 3.5\text{ m}\)
i. Height of the tunnel = Radius + Height of triangle = \(3.5 + \sqrt{3.5^2 - 1.75^2} \approx 5.97\text{ m}\)
ii. Perimeter of cross section = \(21.44\text{ m}\)
iii. Area of cross-section = \(28.875\text{ m}^2\)
iv. Internal surface area = \(825\text{ m}^2\)
In simple words: Calculate the geometric properties of the circular sector and triangular base combined to find height, perimeter, area, and surface area.

Exam Tip: For tunnel surface area problems, multiply the perimeter of the cross-section (excluding the floor) by the length of the tunnel.

Question 21. In the adjoining figure, ABCD is a square of side 6cm. Find the area of the shaded region.
Answer:
Side of square = 6 cm
Area of square = \(6 \times 6 = 36\text{ cm}^2\)
From P, draw \(PQ \perp AB\). Then \(AQ = QB = 3\text{ cm}\).
In \(\Delta PQB\), \(\cos \theta = \frac{QB}{PB} = \frac{3}{6} = \frac{1}{2}\), so \(\theta = 60^\circ\).
Area of sector BPA = \(\frac{60^\circ}{360^\circ} \pi (6)^2 = \frac{1}{6} \times 36\pi = 6\pi = 18.84\text{ cm}^2\)
Area of \(\Delta BPQ = \frac{1}{2} \times QB \times PQ = \frac{1}{2} \times 3 \times (3\sqrt{3}) = 7.794\text{ cm}^2\)
Area of shaded portion = \(34.428\text{ cm}^2\)
In simple words: Subtract the unshaded sector and triangle areas from the area of the square to find the shaded region.

Exam Tip: Use trigonometric ratios like sine and cosine in right-angled triangles to find unknown heights and angles in composite figures.

Question 22. In the adjoining figure, ABCD is a rectangle with sides 4cm and 8cm. Taking 8cm as the diameter, two semicircles are drawn. Find the area overlapped by the two semicircles.
Answer:
Rectangle dimensions: length = 8 cm, breadth = 4 cm.
Diameter = 8 cm, so radius of each semicircle \(r = 4\text{ cm}\).
In \(\Delta OMB\), \(\cos \angle BOM = \frac{OM}{OB} = \frac{2}{4} = \frac{1}{2}\), so \(\angle BOM = 60^\circ\).
\(\angle AOB = 120^\circ\).
Area overlapped by semi circles = \(2 \left( \frac{120^\circ}{360^\circ} \pi (4)^2 - \frac{1}{2} \times AB \times OM \right)\)
\(= 2 \left( \frac{16\pi}{3} - \frac{1}{2} (8 \times 2\sqrt{3}) \right) = 19.66\text{ cm}^2\)
In simple words: Calculate the area of the intersecting sectors and subtract the triangle parts to find the overlapped region.

Exam Tip: Intersection area problems often require finding sector areas and subtracting triangle components.