In fig the shaded area is radius 10cm
WebAB = AC + BC ( Pythagoras Theorem) AB = 6 2 + 8 2 AB = 36 + 64 AB = 100 = 10 cm AB is the diameter of the circle. π π Area of the circle = πr 2 = 22 7 × 5 2 ( radius = diameter 2) = 78. 57 cm 2 Step 3: Area of the shaded region. Area of shaded region = Area of the circle - Area of the triangle = 78. 57 - 24 = 54. 57 cm 2 WebWe have to find the area of the shaded region. From the figure, EF, FD and ED are the sectors made at the vertices A, B and C. Area of shaded region = 3 (area of sector) Here, radius, r = 10/2 = 5 cm Since ABC is an equilateral triangle, the corresponding angle θ is 60° Area of sector = πr²θ/360° = (3.14) (5)² (60°/360°) = (3.14) (25) (1/6)
In fig the shaded area is radius 10cm
Did you know?
Web(b) Find the area of the sector OAB. (2) The line AC shown in the diagram above is perpendicular to OA, and OBC is a straight line. (c) Find the length of AC, giving your answer to 2 decimal places. (2) The region H is bounded by the arc AB and the lines AC and CB. (d) Find the area of H, giving your answer to 2 decimal places. (3) (Total 9 marks) WebMar 29, 2024 · Ex 12.3, 2 Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°. Area of …
WebMar 22, 2024 · Transcript. Ex 12.3, 16 Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each. Area of designed region = Area of 1st quadrant + Area of 2nd quadrant Area of square Area of 1st quadrant = /360 2 = 90/360 22/7 82 = 1/4 22/7 8 8 = 22/7 2 8 = 352/7 cm2 For 2nd quadrant, As radius and ... WebApr 2, 2024 · of the shaded portion. Here, radius = 10cm. Area of major sector is given by formula, = 360 − θ 360 × π r 2 = 360 − 90 360 × 3.14 × 10 2 = 270 360 × 314 = 3 4 × 314 = 235.5 c m 2 . ∴ Area of minor sector =28.5 c m 2 . Area of major sector = 235.5 c m 2 . So, Area of major sector = 235.5 c m 2 .
Web3- Determine the moment of inertia of shaded area shown in Fig. below with respect to x-axis. 10cm 10 cm X Question Transcribed Image Text: 3- Determine the moment of inertia … WebThis means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is . Solution 2 Let the radius of the two smaller circles be . It follows that the area of one of the smaller circles is .
WebCalculate the area of the shaded region in the given figure, common between the two quadrants of the circles of radius 10 cm each. (use π = 3.14) A 60 cm2 B 57 cm2 C 58 …
WebJun 22, 2024 · Explanation: the area (A) of a circle is calculated using the formula. ∙ xA = πr2 ← r is the radius. here r = 10 thus. A = π× 102 = 100π ≈ 314.16 units2. Answer link. … django rest apiview postWebArea of Shaded region = Area of equilateral triangle ABO + Area of Major sector Area of Equilateral Triangle ABO = 4 3×a 2 = 4 3×12×12cm 2 =62.352cm 2 Area of major sector = … django rest framework javatpointWebIn this figure, AOB is a quarter circle of radius 10 and PQRO is a rectangle of perimeter 26. The perimeter of the shaded region is - A 13 + 5 π B 17 + 5 π C 7 + 10 π D 7 + 5 π Medium Solution Verified by Toppr Correct option is D) According to figure, r 2=l 2+b 2 Given r=10(l+b)=13 Perimeter of shaded region django rest gisWebIf OP = PQ = 10 cm, show that area of shaded region is 25 (√3- π /6)cm² Solution: OP = OQ = 10 cm PQ = 10 cm So, ΔOPQ is an equilateral triangle ∠POQ = 60° Area of segment PAQM … django rest project githubWebAnnulus Calculations: The following are the sets of formulas we used in our calculations: Calculate C1, C2, A1, A2, A0 Given r1, r2. Calculate outer circumference, inner circumference, area enclosed by the outer circle, area enclosed by the inner circle, area of the shaded region Given outer radius, inner radius. django rest apisWebMar 29, 2024 · Ex 12.3, 13 In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π= 3.14) Area of shaded region = Area of quadrant OPBQ – Area of square OABC Area of square Side of square = OA = 20 cm Area of square = (side)2 = (20)2 = 20×20 = 400 cm2 Area of quadrant, We need to find radius ... django rest projectWebFind the perimeter of the shaded region. Medium Solution Verified by Toppr We have, PS= Diameter of a circle of radius 6 m=12 cm ∴ PQ=QR=RS= 312=4 cm QS=QR+RS=(4+4)cm=8cm Hence required perimeter = Area of semicircle of radius 6 cm+ Arc of semicircle of radius 4 cm+ ARC of semicircle of radius 2 cm … django rest jwt logout