Question : Find the area of the shaded portion of an equilateral triangle with sides 6 units shown in the following figure. A circle of radius 1 unit is centred at the midpoint of a side of the triangle.
Option 1: $\frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Option 2: $\frac{1}{4}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit $^2$
Option 3: $\frac{1}{2}\left(6 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Option 4: $\frac{1}{2}\left(9 \sqrt{3}-\frac{22}{7}\right)$ unit$^2$
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $\frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ unit$^2$
Solution : Area of equilateral triangle = $\frac{\sqrt 3}{4}(side)^2$ $= \frac{\sqrt3}{4}(6)^2 = 9\sqrt{3}\ \mathrm{unit}^2$ Area of circle = $\pi r^2 = \pi\ \mathrm{unit^2}$ where $r$ = radius of circle = 1 unit. Area of shaded region = $\frac{1}{2}$ Area of triangle - $\frac{1}{4}$ Area of circle Area of shaded region = $\frac{9\sqrt{3}}{2} - \frac{\pi}{4}$ $= \frac{9\sqrt{3}}{2} - \frac{22}{7\times 4}$ $= \frac{9\sqrt{3}}{2} - \frac{11}{7\times 2}$ $= \frac{1}{2}\left(9 \sqrt{3}-\frac{11}{7}\right)$ Hence, the correct answer is $\frac{1}{2}(9 \sqrt{3}-\frac{11}{7})$ unit 2 .
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : The value of $\left(2 \frac{6}{7}\right.$ of $\left.4 \frac{1}{5} \div \frac{2}{3}\right) \times 5 \frac{1}{9} \div\left(\frac{3}{4} \times 2 \frac{2}{3}\right.$ of $\left.\frac{1}{2} \div \frac{1}{4}\right)$ is:
Question : The value of $\left(1 \frac{1}{3} \div 2 \frac{6}{7}\right.$ of $\left.5 \frac{3}{5}\right) \times\left(6 \frac{2}{5} \div 4 \frac{1}{2}\right.$ of $\left.5 \frac{1}{3}\right) \div\left(\frac{3}{4} \times 2 \frac{2}{3} \div \frac{5}{9}\right.$ of
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Question : If the side of a square is $\frac{1}{2}(x+1)$ units and its diagonal is $\frac{3-x}{\sqrt{2}}$ units, then the length of the side of the square would be:
Question : The value of $9 \div\left\{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6} \div\left(\frac{3}{4}-\frac{1}{3}\right)\right.$ of $\left.\frac{2}{9}\right\}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile