Solution: As we know, AB = BC = CD = AD = 8 cm The area of the triangle ABC = Area of the triangle ADC = (½)×8×8 = 32 cm2 The area of quadrant AECB = the area of quadrant AFCD = (¼)×22/7×82 = 352/7...

Solution: The radius of the circle's quadrant ABC = 14 cm AC = AB = 14 cm The diameter of a semicircle is BC. The triangle ABC is right angled triangle. Using the Pythagoras theorem in ΔABC, BC2 =...

Solution: 21 cm = Radius (R) of the larger circle,  7 cm = Radius (r) of the smaller circle, 30°= Both concentric circles' sectors form an angle. Formula of sector = $\frac{\theta }{{{360}^{\circ...

Solution: Squares’ side = AB = OA = 20 cm Quadrants’ radius = OB The OAB triangle is a right-angled triangle. Using the Pythagoras theorem in ΔOAB, OB2 = AB2+OA2 ⇒ OB2 = 202 +202 ⇒...

(i) quadrant OACB, (ii) shaded region. Solution: Quadrants’ radius = 3.5 cm or 7/2 cm (i) The area of the quadrant OACB = (πR2)/4 cm2 = (22/7)×(7/2)×(7/2)/4 cm2 = 77/8 cm2 (ii) The area of triangle...

Solution: 9 = Number of circular designs 7 cm = Radius of the circular design One side of a square handkerchief has three circles. The side of the square = 3×diameter of circle = 3×14 = 42 cm...

Solution: An equilateral triangle is ABC. As a result, ∠ A = ∠ B = ∠ C = 60° There are three sectors, each of which makes a 60°. The area of ΔABC = 17320.5 cm2 ⇒ √3/4 ×(side)2 = 17320.5 ⇒...

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find: (i) the distance around the track along its inner edge (ii) the area...

Solution: Square's side $A B C D=14 \mathrm{~cm}$ Square's area $=(\text { side })^{2}=14 \times 14=196 \mathrm{~cm}^{2}$ The circle's radius drawn at vertex $=\frac{1}{2}(\text { side of square...

Solution:

Solution: 4 cm = side of the square 1 cm = radius of the circle Four quadrants of a circle are cut from the corner, and one radius circle is cut from the middle. The area of square (side)2 = 42 = 16...

Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. Solution: It is assumed...

 Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles. Solution: 14 cm = Side of the square ABCD (as given) As a result, the area...

Solution: Provided to us , 40°= Angle made by sector, 7 cm = Radius the inner circle(r), 14 cm = Radius(R) of the outer circle We are aware that, The area of sector = (θ/360°)×πr2 As a result, the...

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Solution: P is in the semi-circle in this case, P = 90° As a result, QR is the circle's...

Area of a sector of angle p (in degrees) of a circle with radius R is (A) p/180 × 2πR (B) p/180 × π R2 (C) p/360 × 2πR(D) p/720 × 2πR2 Solution:  (θ/360°)×πr2 = The area of a sector Provided...

Solution: The total number of designs that are equal is 6. AOB= 360°/6 = 60° 28 cm = Radius of the cover ₹ 0.35 per cm2 = Cost of making design The two arms of the triangle are equal since they...

(Use π = 3.14) Solution: Let the position of Lighthouse be O. The distance over which light spreads will be the radius here. Radius (r) = 16.5 km (given) Sector angle (θ) = 80° (given) Now, area...

Solution: Provided, 25 cm = Radius (r) 115° = Sector angle (θ) Because there are 2 blades, The total area of the sector made by wiper = 2×(θ/360°)×π r2 = 2×(115/360)×(22/7)×252 = 2×158125/252 cm2 =...

Solution: When the umbrella is flat, its radius (r) is 45 cm. As a result, the circle's area (A) = πr2 = (22/7)×(45)2 =6364.29 cm2 In total the number of ribs (n) = 8 ∴ The area between...

(i) the total length of the silver wire required. (ii) the area of each sector of the brooch. Solution: 35 mm = Diameter (D) The total number of diameters that must be considered is 5. Now, the...

(i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14) Solution: Because the horse is...

Solution: Radius, r = 12 cm Draw a perpendicular OD on chord AB now, and it will bisect the chord AB. So, AD = DB Now, the area of the minor sector = (θ/360°)×πr2 = (120/360)×(22/7)×122 = 150.72 cm2...

Solution: Provided, Radius of the circle = 15 cm θ = 60° So, The area of sector OAPB = (60°/360°)×πr2 cm2 = 225/6 πcm2 Now, ΔAOB is equilateral because two sides are circle radii and hence...

(i) area of the segment formed by the corresponding chord Solution: Provided, Radius of the circle = 21 cm θ = 60° (i)  Area of sector OAPB – Area of ΔOAB = Area of segment APB OAB is an...

(i) minor segment (ii) major sector. (Use π = 3.14) Solution: Here, AB is the chord that forms a 90° angle at the centre O. The circle's radius (r) is given as 10 cm. (i) The area of...

Solution: Radius of the clock (circle) = length of minute hand Therefore, as given the radius (r) of the circle = 14 cm In 60 minutes, the minute hand swept the angle = 360° As a result, the minute...

Solution: C = 22 cm as the circumference of the circle (given) It's noted that a quadrant of a circle is a sector that forms a 90° angle. r be the radius of the circle As C = 2πr = 22, R =...

Solution: The sector's angle is assumed to be 60°. We are aware of the sector's area = (θ/360°)× πr2  Therefore, area of the sector having angle 60° = (60°/360°)×πr2 cm2 = (36/6)π cm2 =...

(A) 2 units (B) π units (C) 4 units (D) 7 units Solution: Because the circle's perimeter equals its area,   2πr = πr2 (or r = 2) As a result, option (A) is correct, and the circle's radius...

Solution: As D = 80 cm, the radius of a car's wheel is 80/2 = 40 cm. Thus, 2πr = 80π cm is the circumference of the wheels. The distance covered in a single revolution is now 80 cm...

Solution: r1 = 21/2 cm is the radius of the first circle (as diameter D is given as 21 cm) As a result, the area of the gold region = π r12 = π(10.5)2 = 346.5 cm2. Given that each of the...

Solution: 1st circle radius = 8 cm (given) ∴ The area of the first circle is π(8)2 = 64. The radius of the second circle is 6 cm (given) Area of the second circle = π(6)2 = 36π As a result, the sum...

Solution: The first circle's radius is 19 cm (given) ∴ 2π×19 = 38 cm is the circumference of the first circle. The second circle's radius is 9 cm (given) The circumference of the second circle is...