Brilliant Games - Quarter Quell 1

Area of quarter = pi r^2 / 4 = 0.785.................i Area of right triangle = 1/2 b*h = 0.5......................ii . Now subtracting ii from i we get 0.285.

(Pi)/4 - 1/2 = 0.285398

The area of the quarter circle = pir^2 (90/360) or pi / 4 = 3.14 / 4 = 0.785

The area of the triangle = 1/2bh = 1/2 (1)(1) = 1/2 or 0.5

Area of the blue region = Area of the quarter circle - Area of the triangle

= 0.785 - 0.5 = 0.285

0.285 is the area of the blue region

here the radius of quarter circle is 1 so the area of this quarter circle is pir^2/4=.7854.& The area of the triangle is 1/2 x1x1=.5. so the area of the blue shaded region remains .7854_.5=.285.

$\begin{aligned} \text {Area of the quarter circle} & = \dfrac {1}{4} \pi r^2\\ & = \dfrac{1 × 22 × {(1)}^2}{4 × 7}\\ & = \require {cancel}{\dfrac {1 × \cancel{22} 11 × 11}{\cancel{4} 2 × 7}}\\ & = \dfrac{11}{14} \rightarrow {eq}^{n} 1 \end{aligned}$

$\begin{aligned} \text {Area of red triangle} & = \dfrac {1}{2} × \text {Base} × \text {Height}\\ & = \dfrac {1 × 1 × 1}{2}\\ & = \dfrac {1}{2} \rightarrow {eq}^{n} 2 \end{aligned}$

$\begin{aligned} \text {Area of blue shaded region} & = \text {Area of quarter circle} - \text {Area of red triangle}\\ & = {eq}^n 1 - {eq}^n 2\\ & = \dfrac{11}{14} - \dfrac {1}{2}\\ & = \dfrac{11 - 7}{14}\\ & = \dfrac {4}{14}\\ & = \require {cancel}{\dfrac {\cancel{4}2}{\cancel{14}7}}\\ & = \dfrac{2}{7}\\ & \approx \boxed{0.285} \end{aligned}$

Its damn easy. Just take out the area of quarter circle i.e (pi r r)/4=(22/7 1 1)4 =22/28 which is equal to approx. 0.78571. Now subtract the area of red part i.e of triangle whose are is 1/2 h b=1/2 1 1 =1/2...Now subtract 0.78571-0.5 = approx. 0.28571, which is the answer :D. HAVE A GOOD DAY .

Area of a circle = (pi)(r)^2 -Since it is only a quarter circle the area is 1/4. Area of the quarter circle = [(pi)(r)^2]/4 -The area of a triangle is: [(b)(h)]/2 -Therefore the area of the shaded region is [(pi)(r)^2]/4 - [(b)(h)]/2