From each corner of a square of side $4\, cm$,a quadrant of a circle of radius $1\, cm$ is cut and also a circle of diameter $2\, cm$ is cut as shown in the figure. Find the area of the remaining portion of the square. [Use $\pi = \frac{22}{7}$]
(N/A) $1$. Area of the square $= (\text{side})^2 = (4\, cm)^2 = 16\, cm^2$.
$2$. Area of each quadrant of a circle with radius $r = 1\, cm$ is $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (1)^2 = \frac{22}{28} = \frac{11}{14}\, cm^2$.
$3$. Total area of $4$ quadrants $= 4 \times \frac{11}{14} = \frac{22}{7}\, cm^2$.
$4$. Area of the central circle with diameter $2\, cm$ (radius $r = 1\, cm$) $= \pi r^2 = \frac{22}{7} \times (1)^2 = \frac{22}{7}\, cm^2$.
$5$. Area of the remaining portion $= \text{Area of square} - (\text{Total area of } 4 \text{ quadrants} + \text{Area of central circle})$.
$6$. Area $= 16 - (\frac{22}{7} + \frac{22}{7}) = 16 - \frac{44}{7} = \frac{112 - 44}{7} = \frac{68}{7}\, cm^2$.
$2$. Area of each quadrant of a circle with radius $r = 1\, cm$ is $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (1)^2 = \frac{22}{28} = \frac{11}{14}\, cm^2$.
$3$. Total area of $4$ quadrants $= 4 \times \frac{11}{14} = \frac{22}{7}\, cm^2$.
$4$. Area of the central circle with diameter $2\, cm$ (radius $r = 1\, cm$) $= \pi r^2 = \frac{22}{7} \times (1)^2 = \frac{22}{7}\, cm^2$.
$5$. Area of the remaining portion $= \text{Area of square} - (\text{Total area of } 4 \text{ quadrants} + \text{Area of central circle})$.
$6$. Area $= 16 - (\frac{22}{7} + \frac{22}{7}) = 16 - \frac{44}{7} = \frac{112 - 44}{7} = \frac{68}{7}\, cm^2$.
Explore More
Similar Questions
$A$ horse is tied to a peg at one corner of a square-shaped grass field of side $15 \, m$ by means of a $5 \, m$ long rope (see figure). Find:
$(i)$ 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 $\pi = 3.14$)
$(i)$ 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 $\pi = 3.14$)
Difficult
View SolutionThe radii of two circles are $8 \, cm$ and $6 \, cm$ respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. [use $\pi = \frac{22}{7}$] (in $cm$)
Medium
View SolutionIn the given figure,$OACB$ is a quadrant of a circle with centre $O$ and radius $3.5 \, cm$. If $OD = 2 \, cm$,find the area of the:
$(i)$ quadrant $OACB$,
$(ii)$ shaded region. $\left[ \text{Use } \pi = \frac{22}{7} \right]$
$(i)$ quadrant $OACB$,
$(ii)$ shaded region. $\left[ \text{Use } \pi = \frac{22}{7} \right]$
Medium
View SolutionIn the figure,two circular flower beds are shown on two sides of a square lawn $ABCD$ of side $56 \, m$. If the centre of each circular flower bed is the point of intersection $O$ of the diagonals of the square lawn,find the sum of the areas of the lawn and the flower beds (in $m^2$).
Difficult
View SolutionIn the figure,$ABC$ is a quadrant of a circle of radius $14 \, cm$ and a semicircle is drawn with $BC$ as diameter. Find the area of the shaded region. (in $cm^2$) [Use $\pi = \frac{22}{7}$]
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo