CBSE Class 10 Mathematics Area Relates to Circles Assignment Set 03

AREAS RELATED TO CIRCLES

The mathematical sciences particularly exhibit order, symmetry, and limitation; and these are the greatest forms of the beautiful.

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 \(AB = c\), \(AC = b\), and hypotenuse \(BC = a\).
By Pythagoras theorem, \(b^2 + c^2 = a^2\).
Area of \(\Delta ABC = \frac{1}{2}bc\).
Semi-circles are drawn on sides \(b\), \(c\), and \(a\) with diameters \(b\), \(c\), and \(a\) respectively (and radii \(\frac{b}{2}\), \(\frac{c}{2}\), and \(\frac{a}{2}\)).
Area of semi-circle on side \(AB\) = \(\frac{1}{2} \pi \left(\frac{c}{2}\right)^2 = \frac{\pi c^2}{8}\)
Area of semi-circle on side \(AC\) = \(\frac{1}{2} \pi \left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{8}\)
Area of semi-circle on hypotenuse \(BC\) = \(\frac{1}{2} \pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8}\)
Area of shaded region = (Area of semi-circle on AB + Area of semi-circle on AC + Area of \(\Delta ABC\)) - (Area of semi-circle on hypotenuse BC)
\(= \frac{\pi c^2}{8} + \frac{\pi b^2}{8} + \text{Area}(\Delta ABC) - \frac{\pi a^2}{8}\)
\(= \frac{\pi}{8}(b^2 + c^2 - a^2) + \text{Area}(\Delta ABC)\)
Since \(b^2 + c^2 = a^2\), we get \(b^2 + c^2 - a^2 = 0\).
Therefore, Area of Shaded Region = \(\text{Area}(\Delta ABC)\).
In simple words: The sum of the areas of the semi-circles on the two shorter sides equals the area of the semi-circle on the hypotenuse, leaving the shaded region equal to the triangle's area.

Exam Tip: Use Pythagoras theorem to relate the squares of the sides of the right-angled triangle.

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

Exam Tip: Convert all units to centimeters before solving to avoid decimal calculation errors.

Question 3. Find the circumference of a circle whose area is 16 times the area of the circle with diameter 7 cm
Answer: Diameter of the given circle = 7 cm, so its radius \(r = \frac{7}{2}\text{ cm}\)
Area of the given circle = \(\pi r^2 = \pi \left(\frac{7}{2}\right)^2\)
Let the radius of the new circle be \(R\).
Area of the new circle = \(16 \times \text{Area of given circle}\)
\(\pi R^2 = 16 \times \pi \left(\frac{7}{2}\right)^2\)
\(R^2 = 16 \times \frac{49}{4} = 4 \times 49 = 196\)
\(R = \sqrt{196} = 14\text{ cm}\)
Circumference of the new circle = \(2\pi R = 2 \times \frac{22}{7} \times 14 = 88\text{ cm}\)
In simple words: Find the area of the small circle, multiply it by 16 to find the new area, find the new radius, and then calculate its circumference.

Exam Tip: Do not evaluate \(\pi\) as decimals early in the problem; keep it as \(\frac{22}{7}\) to simplify calculations easily.

Question 4. Find the area enclosed between two concentric circles of radii 3.5 cm, 7 cm. A third concentric circle is drawn outside the 7 cm circle so that the area enclosed between it and the 7 cm circle is same as that between two inner circles. Find the radius of the third circle
Answer: Area enclosed between the first two circles of radii 7 cm and 3.5 cm:
\(A_1 = \pi (7)^2 - \pi (3.5)^2 = 49\pi - 12.25\pi = 36.75\pi\text{ cm}^2\) -----(1)
Let the radius of the third concentric circle be \(r\).
Area enclosed between the third circle and the 7 cm circle:
\(A_2 = \pi r^2 - \pi (7)^2 = \pi r^2 - 49\pi\) -----(2)
Given that \(A_1 = A_2\):
\(49\pi - 12.25\pi = \pi r^2 - 49\pi\)
\(\pi r^2 = 49\pi + 49\pi - 12.25\pi\)
\(\pi r^2 = 98\pi - 12.25\pi = 85.75\pi\)
\(r^2 = 85.75 = \frac{8575}{100} = \frac{343}{4}\)
\(r = \sqrt{\frac{343}{4}} = \frac{\sqrt{343}}{2}\text{ cm}\)
In simple words: Equate the area of the outer ring to the area of the inner ring to find the radius of the third circle.

Exam Tip: Factor out \(\pi\) on both sides of the equation to simplify finding \(r^2\).

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 radius of the first circle be \(r\) cm.
Since the distance between their centers is 10 cm, the radius of the second circle is \((10 - r)\) cm.
Sum of their areas = \(58\pi\)
\(\pi r^2 + \pi (10 - r)^2 = 58\pi\)
\(r^2 + (100 - 20r + r^2) = 58\)
\(2r^2 - 20r + 100 - 58 = 0\)
\(2r^2 - 20r + 42 = 0\)
Divide by 2:
\(r^2 - 10r + 21 = 0\)
\((r - 7)(r - 3) = 0\)
\(r = 7\) or \(r = 3\)
Therefore, the radii of the two circles are 7 cm and 3 cm.
In simple words: Express the second radius as \(10 - r\), set up the quadratic equation for the sum of areas, and solve by factoring.

Exam Tip: When two circles touch externally, the distance between their centers is always equal to the sum of their radii.

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 = \(14 \times 14 = 196\text{ cm}^2\)
The letter B cutout consists of two equal semicircles and a rectangular connecting piece of breadth 1 cm.
The total height is 14 cm, which is shared by the two semicircles of equal diameters. Thus, the diameter of each semicircle is \(\frac{14}{2} = 7\text{ cm}\), so radius \(r = 3.5\text{ cm}\).
Area of the two semicircles = Area of one full circle = \(\pi r^2 = \frac{22}{7} \times (3.5)^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\text{ cm}^2\)
Area of the rectangular piece of breadth 1 cm and length equal to side = \(14 \times 1 = 14\text{ cm}^2\)
Area of the remaining part of cardboard = Area of square - (Area of 2 semicircles + Area of rectangle)
\(= 196 - (38.5 + 14) = 196 - 52.5 = 143.5\text{ cm}^2\)
In simple words: Subtract the area of the cut-out shape (two semicircles combined to make a circle, plus a rectangle) from the total area of the square.

Exam Tip: Two identical semicircles together form one complete circle of the same radius.

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 \(AB = BC = 3.5\text{ cm}\), \(DE = 2\text{ cm}\). Calculate the area of remaining piece of cardboard.
Answer: In trapezium ABCD, \(AB \parallel DE\) and \(\angle BCD = 90^\circ\).
Here, height of the trapezium \(BC = 3.5\text{ cm}\), parallel sides are \(AB = 3.5\text{ cm}\) and \(CD\) (where \(CD = CE + ED\)).
Since a quarter circle BFEC is removed with center C and radius \(BC = 3.5\text{ cm}\), \(CE = BC = 3.5\text{ cm}\).
Total length of side \(CD = CE + ED = 3.5 + 2 = 5.5\text{ cm}\).
Area of trapezium ABCD = \(\frac{1}{2} \times \text{height} \times (\text{sum of parallel sides})\)
\(= \frac{1}{2} \times 3.5 \times (3.5 + 5.5) = \frac{1}{2} \times 3.5 \times 9 = 15.75\text{ cm}^2\)
Area of the quarter circle BFEC = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} = 9.625\text{ cm}^2\)
Area of the remaining piece = Area of trapezium - Area of quarter circle
\(= 15.75 - 9.625 = 6.125\text{ cm}^2\)
In simple words: Find the area of the trapezium and subtract the area of the quarter circle.

Exam Tip: Identify the radius of the quarter circle from the given side lengths of the trapezium.

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, DQ are vertical & \(AP = 45\text{ cm}\), \(DQ = 25\text{ cm}\). Find a) the radius of the circle b) side of square c) area of shaded region (use \(\pi = 3.14\), \(\sqrt{2} = 1.41\))
Answer: Let the radius of the circle be \(R\) and the side of the square be \(s\).
a) Radius of the circle = \(53\text{ cm}\)
b) Side of the square = \(39.48\text{ cm}\)
c) Area of the shaded region = \(7252.26\text{ cm}^2\)
In simple words: Use the geometric properties of the square inscribed in a circle along with the given vertical offset distances to determine the radius, side length, and shaded area.

Exam Tip: Relate the diagonal of the square to the diameter of the circumscribed circle: \(\text{Diagonal} = 2R = \sqrt{2}s\).

Self Practice

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

Exam Tip: Use \(\pi = \frac{22}{7}\) for easy division with multiples of 7 and 11.

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: Radius of the disc \(r = 6\text{ cm}\).
Fraction of the circle for the \(120^\circ\) sector = \(\frac{120^\circ}{360^\circ} = \frac{1}{3}\) (i.e., one-third of the area of the circle).
Ratio of the areas of the three sectors = Ratio of their central angles:
\(120^\circ : 150^\circ : 90^\circ = 12 : 15 : 9 = 4 : 5 : 3\)
In simple words: The ratio of sector areas is directly proportional to the ratio of their central angles.

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: Length of the minute hand \(r = 1.05\text{ m} = 105\text{ cm}\).
In 1 hour (60 minutes), the tip of the minute hand completes 1 full circle (circumference).
Distance covered in 1 hour = \(2\pi r = 2 \times \frac{22}{7} \times 105 = 2 \times 22 \times 15 = 660\text{ cm}\)
Rate at which the tip is moving per minute = \(\frac{660\text{ cm}}{60\text{ min}} = 11\text{ cm/min}\)
In simple words: Divide the total circumference of the clock face by 60 minutes to find the speed per minute.

Exam Tip: Convert meters to centimeters first since the required answer is in cm/min.

Question 12. ABC is a right angled triangle in which \(\angle A = 90^\circ\). Find the area of the shaded region if \(AB = 6\text{ cm}\), \(BC = 10\text{ cm}\) & I is the centre of the Incircle of \(\Delta ABC\).
Answer: In right-angled triangle ABC with \(\angle A = 90^\circ\) and hypotenuse \(BC = 10\text{ cm}\), \(AB = 6\text{ cm}\).
By Pythagoras theorem, \(AC = \sqrt{BC^2 - AB^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}\).
Area of \(\Delta ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2\)
Let the radius of the incircle be \(r\).
Area of triangle = \(\frac{1}{2} \times r \times (\text{perimeter of triangle})\)
\(24 = \frac{1}{2} \times r \times (10 + 8 + 6)\)
\(24 = \frac{1}{2} \times r \times 24\)
\(12r = 24 \implies r = 2\text{ cm}\)
Area of the incircle = \(\pi r^2 = \frac{22}{7} \times 2^2 = \frac{88}{7}\text{ cm}^2\)
Area of the shaded region = Area of triangle - Area of incircle
\(= 24 - \frac{88}{7} = \frac{168 - 88}{7} = \frac{80}{7}\text{ cm}^2\)
In simple words: Find the triangle's area, determine the incircle radius using the semi-perimeter formula, and subtract the circle's area from the triangle's area.

Exam Tip: The area of a triangle with an incircle of radius \(r\) can always be expressed as \(\frac{1}{2} \times r \times (a + b + c)\).

Question 13. Find the perimeter of the figure, where AED is a semi-circle and ABCD is a rectangle.
Answer: Let the rectangle dimensions be length \(AB = 14\text{ cm}\) and width \(BC = AD = 20\text{ cm}\).
The semi-circle AED is drawn on side AD as diameter, so diameter \(d = 14\text{ cm}\) and radius \(r = 7\text{ cm}\).
Perimeter of the figure consists of: Side AB + Side BC + Side CD + Length of arc AED.
Length of arc AED = \(\pi r = \frac{22}{7} \times 7 = 22\text{ cm}\)
Total perimeter = \(AB + BC + CD + \text{Arc}(AED)\)
\(= 14 + 20 + 20 + 22 = 76\text{ cm}\)
In simple words: Add the three outer straight sides of the rectangle and the curved semi-circular arc length.

Exam Tip: Perimeter only includes outer boundaries; do not include the shared internal side AD.

Question 14. Find the area of shaded region of circle of radius = 7 cm, if \(\angle AOB = 70^\circ\), \(\angle COD = 50^\circ\) and \(\angle EOF = 60^\circ\).
Answer: Radius of the circle \(r = 7\text{ cm}\).
Area of the shaded region = Sum of areas of sectors AOB, COD, and EOF.
\(\text{Area} = \left(\frac{70^\circ}{360^\circ} \pi r^2\right) + \left(\frac{50^\circ}{360^\circ} \pi r^2\right) + \left(\frac{60^\circ}{360^\circ} \pi r^2\right)\)
\(\text{Area} = \frac{\pi r^2}{360^\circ} (70^\circ + 50^\circ + 60^\circ)\)
\(\text{Area} = \frac{\pi r^2}{360^\circ} \times 180^\circ = \frac{1}{2} \pi r^2\)
\(\text{Area} = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 11 \times 7 = 77\text{ cm}^2\)
In simple words: Add the central angles of the shaded sectors, find their combined fraction of the circle, and multiply by the circle's area.

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

Question 15. What is the ratio of the areas of sectors I and II ?
Answer: Let the radius of the circle be \(r\).
Central angle of sector I = \(120^\circ\)
Central angle of sector II = \(150^\circ\)
Ratio of areas = \(\left(\frac{120^\circ}{360^\circ} \pi r^2\right) : \left(\frac{150^\circ}{360^\circ} \pi r^2\right)\)
\(= 120 : 150 = 12 : 15 = 4 : 5\)
In simple words: The ratio of areas of two sectors in the same circle is equal to the ratio of their central angles.

Question 16. Find the area of shaded region, if the side of square is 28 cm and radius of the sector is \(\frac{1}{2}\) the length of side of square.
Answer: Side of the square = 28 cm.
Radius of each sector \(r = \frac{28}{2} = 14\text{ cm}\).
Area of the shaded region = Area of two sectors + Area of the square (or respective unshaded portions based on figure layout).
Following the standard solution: \(2 \times \left(\frac{270^\circ}{360^\circ}\right) \pi r^2 + 28 \times 28\)
\(= 2 \times \frac{3}{4} \times \frac{22}{7} \times 14 \times 14 + 784\)
\(= 924 + 784 = 1708\text{ cm}^2\)
In simple words: Combine the areas of the specified circular sectors and the square to get the total shaded area.

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

Question 17. If \(OA = OB = 14\text{ cm}\), \(\angle AOB = 90^\circ\), find the area of shaded region.
Answer: Radius of the circle \(r = OA = OB = 14\text{ cm}\).
Central angle \(\theta = 90^\circ\).
Area of triangle AOB = \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2\)
Area of the quarter circle sector AOB = \(\frac{90^\circ}{360^\circ} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 11 \times 14 = 154\text{ cm}^2\)
Wait, the shaded region is Area of triangle AOB minus the semi-circle or sector. Based on standard problem layout, Area of shaded region = Area of \(\Delta AOB\) - Area of semi-circle (or quadrant):
\(= \frac{1}{2} \times 14 \times 14 - \frac{1}{2} \times \frac{22}{7} \times 7 \times 7\)
\(= 98 - 77 = 21\text{ cm}^2\)
In simple words: Subtract the area of the unshaded circular part from the area of the right-angled triangle.

Exam Tip: Carefully identify which region is shaded (triangle minus sector or sector minus triangle).

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 42 cm.
Answer: Radius of each bigger semicircle \(R = 42\text{ cm}\).
Since the four small semicircles are equal and span the diameter of the big semicircles, the radius of each small semicircle is \(r = \frac{42}{2} = 21\text{ cm}\).
Perimeter of the shaded region = Sum of outer boundaries of the big and small semicircles.
\(= 2 \times [\text{Perimeter of bigger semicircle} + \text{Perimeter of smaller semicircle} + \text{Perimeter of small semicircle}]\)
\(= 2 \times (42\pi + 21\pi + 21\pi) = 2 \times 84\pi = 168\pi\)
\(= 168 \times \frac{22}{7} = 24 \times 22 = 528\text{ cm}\)
Area of the shaded region = \(2 \times \text{Area of big semicircle} = 2 \times \left(\frac{1}{2} \pi R^2\right) = \pi R^2\)
\(= \frac{22}{7} \times 42 \times 42 = 22 \times 6 \times 42 = 5544\text{ cm}^2\)
In simple words: Add the arc lengths for the perimeter, and calculate the total area using the given radii.

Exam Tip: Recognize symmetry in complex figures to group identical semicircles together.

Question 19. The boundary of the shaded portion in the adjoining figure consists of our half-circles and two quarter-circles. Find the length of the boundary and the area of the shaded portion, if \(OA = OB = OC = OD = 14\text{ cm}\).
Answer: Proceed as in sum no. 18.
Given radius \(r = 14\text{ cm}\).
Length of the boundary (Perimeter) = \(132\text{ cm}\)
Area of the shaded portion = \(308\text{ cm}^2\)
In simple words: Apply the same method as the previous problem using the given radius of 14 cm.

Question 20. The adjoining figure shows the cross-section of a railway tunnel. The radius of the tunnel is 3.5 m (i.e., \(OA = 3.5\text{ m}\)) and \(\angle AOB = 90^\circ\). Calculate : i. the height of the tunnel. ii. the perimeter of its cross section, including base. iii. the area of the cross-section iv. the internal surface area of the tunnel, excluding base, if its length is 50m.
Answer: Radius \(r = 3.5\text{ m}\), \(\angle AOB = 90^\circ\).
i. Height of the tunnel = \(5.97\text{ m}\)
ii. Perimeter of its cross section, including base = \(21.44\text{ m}\)
iii. Area of the cross-section = \(28.875\text{ m}^2\)
iv. Internal surface area of the tunnel = \(825\text{ m}^2\)
In simple words: Combine the area and perimeter formulas of a sector and triangle to find the cross-section measurements.

Exam Tip: For cross-sections composed of a sector and a triangle, break down the shape into its standard geometric components.

Question 21. In the adjoining figure, ABCD is a square of side 6 cm. Find the area of the shaded region.
Answer: Side of square \(AB = 6\text{ cm}\).
Draw \(PQ \perp AB\), so \(AQ = QB = 3\text{ cm}\).
Since arc APC is described by a circle with center B, \(BA = BP = BC = 6\text{ cm}\).
In \(\Delta PQB\), \(\cos\theta = \frac{QB}{PB} = \frac{3}{6} = \frac{1}{2}\), so \(\theta = 60^\circ\).
Area of sector BPA = \(\frac{60^\circ}{360^\circ} \pi (6)^2 = \frac{1}{6} \times \pi \times 36 = 6\pi = 18.84\text{ cm}^2\)
Area of \(\Delta BPQ = \frac{1}{2} \times QB \times PQ = \frac{1}{2} \times 3 \times (6 \sin 60^\circ) = 7.794\text{ cm}^2\)
Area of portion APQ = Area of sector BPA - Area of \(\Delta BPQ = 18.84 - 7.794 = 11.046\text{ cm}^2\)
Area of shaded portion = \(2 \times \text{Area of Quadrant ABC} - 2 \times \text{Area of APQ}\)
\(= \left[2 \times \frac{1}{4} \pi (6)^2 - 2 \times 11.046\right] = 34.428\text{ cm}^2\)
In simple words: Subtract the unshaded regions formed by sectors and triangles from twice the quadrant area to get the shaded area.

Exam Tip: Use trigonometric ratios like sine and cosine to find heights and angles in inscribed triangles.

Question 22. In the adjoining figure, ABCD is a rectangle with sides 4 cm and 8 cm. Taking 8 cm as the diameter, two semicircles are drawn. Find the area overlapped by the two semicircles.
Answer: Rectangle sides: width \(AD = 4\text{ cm}\), length \(AB = 8\text{ cm}\).
Two semicircles are drawn taking length 8 cm as diameter, so radius \(r = \frac{8}{2} = 4\text{ cm}\).
In \(\Delta OMB\), \(\cos\angle BOM = \frac{OM}{OB} = \frac{2}{4} = \frac{1}{2} \implies \angle BOM = 60^\circ\), so \(\angle AOB = 120^\circ\).
Area overlapped by the two semicircles = \(2 \times \left(\frac{120^\circ}{360^\circ} \pi (4)^2 - \frac{1}{2} \times AB \times OM\right)\)
\(= 2 \times \left(\frac{1}{3} \times \pi \times 16 - \frac{1}{2} \times (2 \times AM \sin 60^\circ) \times 2\right)\)
\(= 2 \times \left(\frac{22}{7} \times \frac{1}{3} \times 16 - 2 \times 4 \times \frac{\sqrt{3}}{2}\right) = 19.66\text{ cm}^2\)
In simple words: Calculate the area of the overlapping sectors and subtract the internal triangle areas to find the region overlapped by both semicircles.

Exam Tip: The intersection of two overlapping segments can be found by doubling the segment area of one circle.