WebIn 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 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 of shaded area shown in Fig. below with respect to x-axis. 10cm 10 cm Зст X
C2 Trigonometry: Arc Length & Sector Area …
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 ... WebFrom Fig. A area of the shaded region is A A ( 4 − π) Similarly with Fig B the area is A B = ( 4 − π) Now when you subtract it from the whole you get the four shaded areas in the given question. The shaded area is A = 4 − ( 4 − π) − ( 4 − π) = 2 π − 4 Share Cite Follow edited Oct 29, 2014 at 14:35 Null 1,292 3 17 21 answered Oct 29, 2014 at 14:16 eric newton tiger properties
35. In the Fig if AB=16 cm and BC=12 cm Calculate the area of the shaded
WebMay 10, 2016 · Area of shaded ring = π (b^2 – a^2) In this particular problem we are given that the area of the shaded ring is 3 times the area of the smaller circular region. We know … 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 WebNow, area of the shaded region = Area of ∆ABC – Area of the inscribed circle = [ √3 4 ×(12)2 – π(2√3)2]cm2 = [36√3−12π]cm2 = [36 ×1.73 – 12 × 3.14] cm2 = [62.28 – 37.68] cm2 = 24.6 cm2 Therefore, the area of the shaded region is 24.6 cm2. Suggest Corrections 3 … eric newton reading