Here, P is in the semi-circle, and so,
P = 90°
So, it can be concluded that QR is the hypotenuse of the circle and is equal to the diameter of the circle.
QR = D
Using the Pythagorean theorem,
$QR^2 = PR^2+PQ^2$
Or, $QR^2 = 7^2+24^2$
QR= 25 cm = Diameter
Hence, the radius of the circle = $\dfrac{25}{2}$ cm
Now, the area of the semicircle = $\dfrac{πR^2}{2}$
= $\dfrac{\dfrac{22}{7}×\dfrac{25}{2}×\dfrac{25}{2}}{2}$ $cm^2$
= $\dfrac{13750}{56}$ $cm^2$ = 245.54 $cm^2$
Also, the area of the ΔPQR = $\dfrac{1}{2}×PR×PQ$
=$\dfrac{1}{2}×7×24$ $cm^2$
= 84 $cm^2$
Hence, the area of the shaded region = 245.54 $cm^2$ - 84 $cm^2$
= 161.54 $cm^2$
Given,
Angle made by sector = 40°,
Radius the inner circle = r = 7 cm, and
Radius of the outer circle = R = 14 cm
We know,
Area of the sector = $\dfrac{θ}{360°}×πr^2$
So, Area of OAC = $\dfrac{40°}{360°}×πr^2 cm^2$
=$\dfrac{1}{9}×\dfrac{22}{7}×14^2$ = 68.44 $cm^2$
Area of the sector OBD = $\dfrac{40°}{360°}×πr^2$ $cm^2$
=$\dfrac{1}{9}×\dfrac{22}{7}×7^2 = 17.11$ $cm^2$
Now, the area of the shaded region ABDC = Area of OAC – Area of the OBD
= 68.44 $cm^2$ – 17.11 $cm^2$ = 51.33 $cm^2$
Side of the square ABCD (as given) = 14 cm
So, the Area of ABCD = $a^2$
= 14×14 $cm^2$ = 196 $cm^2$
We know that the side of the square = diameter of the circle = 14 cm
So, the side of the square = diameter of the semicircle = 14 cm
the radius of the semicircle = 7 cm
Now, the area of the semicircle = $\dfrac{πR^2}{2}$
= $\dfrac{(\dfrac{22}{7}×7×7)}{2}$ $cm^2$
= 77 $cm^2$
he area of two semicircles = 2×77 $cm^2$ = 154 $cm^2$
Hence, the area of the shaded region = Area of the Square – Area of two semicircles
= 196 $cm^2$ -154 $cm^2$
= 42 $cm^2$
It is given that OAB is an equilateral triangle having each angle as 60°
The area of the sector is common in both.
The radius of the circle = 6 cm
Side of the triangle = 12 cm
Area of the equilateral triangle = $\dfrac{\sqrt3}{4} (OA)^2= \dfrac{\sqrt3}{4}×12^2 = 36\sqrt3$ $cm^2$
Area of the circle = $πR^2 = \dfrac{22}{7}×6^2 = \dfrac{792}{7}$ $cm^2$
Area of the sector making angle 60° = $\dfrac{60°}{360°} ×πr^2$ $cm^2$
= $\dfrac{1}{6}×\dfrac{22}{7}× 6^2$ $cm^2$ = $\dfrac{132}{7}$ $cm^2$
Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
= $36\sqrt3$ $cm^2$ +$\dfrac{792}{7}$ $cm^2$- $\dfrac{132}{7}$ $cm^2$
= $36\sqrt3+\dfrac{660}{7} cm^2$
Side of the square = 4 cm
The radius of the circle = 1 cm
Four quadrants of a circle are cut from the corner, and one circle of radius are cut from the middle.
Area of the square = $(side)^2= 4^2 = 16$ $cm^2$
Area of the quadrant = $\dfrac{πR^2}{4}$ $cm^2$ = $\dfrac{22}{7}×\dfrac{1^2}{4} = \dfrac{11}{14}$ $cm^2$
Total area of the 4 quadrants =$4 × \dfrac{11}{14}$ $cm^2$ = $\dfrac{22}{7}$ $cm^2$
Area of the circle = $πR^2 cm^2 = (\dfrac{22}{7}×1^2) = \dfrac{22}{7} cm^2$
Area of the shaded region = Area of the square – (Area of the 4 quadrants + Area of the circle)
= $16 cm^2-\dfrac{22}{7} cm^2 – \dfrac{22}{7}$ $cm^2$
= $\dfrac{68}{7}$ $cm^2$
The radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
BD = $\dfrac{AB}{2}$
Since, AD is the median of the triangle
AO = Radius of the circle = =$\dfrac{2}{3}$ AD
$\dfrac{2}{3}$ AD = 32 cm
AD = 48 cm
In ΔADB,
By Pythagoras’ theorem,
$AB^2 = AD^2 +BD^2$
$AB^2 = 48^2+(\dfrac{AB}{2})^2$
$AB^2 = 2304+\dfrac{AB^2}{4}$
$\dfrac{3}{4} (AB^2)$= 2304
$AB^2 = 3072$
$AB= 32\sqrt{3}$ cm
Area of ΔADB = $\sqrt{3}/4 ×(32\sqrt{3})^2 cm^2 = 768\sqrt{3}$ $cm^2$
Area of the circle = $πR^2 = \dfrac{22}{7}×32×32 = \dfrac{22528}{7}$ $cm^2$
Area of the design = Area of the circle – Area of ΔADB
= $\dfrac{22528}{7} – 768\sqrt{3}$ $cm^2$
Side of square = 14 cm
Four quadrants are included in the four sides of the square.
radius of the circles = $\dfrac{14}{2}$ cm = 7 cm
Area of the square ABCD = $14^2$ = 196 $cm^2$
Area of the quadrant = $\dfrac{πR^2}{4}$ $cm^2$ = $\dfrac{22}{7} ×\dfrac{7^2}{4}$ $cm^2$
= $\dfrac{77}{2}$ $cm^2$
Total area of the quadrant = $\dfrac{4×77}{2}$ $cm^2$ = 154$cm^2$
Area of the shaded region = Area of the square ABCD – Area of the quadrant
= 196 $cm^2$ – 154 $cm^2$
= 42 $cm^2$
find :
(i) the distance around the track along its inner edge
(ii) the area of the track.
Width of the track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
The radius of the inner semicircle, r = OD = O’C
= $\dfrac{60}{2}$ m = 30 m
The radius of the outer semicircle, R = OA = O’B
= 30+10 m = 40 m
Also, AB = CD = EF = GH = 106 m
(i)the distance around the track along its inner edge
Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)
= 106+106+(2×πr) m = $212+\dfrac{2×22}{7×30}$ m
= $212+\dfrac{1320}{7} $m = $\dfrac{1484+1320}{7} $m = $\dfrac{2804}{7}$ m
Area of the track
Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)
= $(AB×CD)+(EF×GH)+2×\dfrac{πr^2}{2} -2×\dfrac{πr^2}{2}$ $m^2$
= $(106×10)+(106×10)+2×\dfrac{π}{2}(r^2-R^2)$ $m^2$
= $2120+\dfrac{22}{7}×70×10$ $m^2$
= 4320 $m^2$
The radius of larger circle, R = 7 cm
The radius of smaller circle, r = $\dfrac{7}{2}$ cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = $\dfrac{1}{2} × AB × OC = \dfrac{1}{2}×7×14 = 49$ $cm^2$
Area of larger circle = $πR^2 = \dfrac{22}{7}×7^2 = 154$ $cm^2$
Area of larger semicircle = $\dfrac{154}{2}$ $cm^2$ = 77 $cm^2$
Area of smaller circle = $πr^2 = \dfrac{22}{7}×\dfrac{7}{2}×\dfrac{7}{2} = \dfrac{77}{2}$ $cm^2$
Area of the shaded region = Area of the larger circle – Area of the triangle – Area of the larger semicircle + Area of the smaller circle
Area of the shaded region = $154-49-77+\dfrac{77}{2}$ $cm^2$
= $\dfrac{133}{2}$ $cm^2$ = 66.5 $cm^2$
ABC is an equilateral triangle.
$\angle A = \angle B = \angle C = 60°$
There are three sectors, each making 60°.
Area of ΔABC = 17320.5 $cm^2$
$\dfrac{\sqrt{3}}{4} ×(side)^2 = 17320.5$
$(side)^2 =\dfrac{17320.5×4}{1.73205}$
$(side)^2 = 4×10^4$
side = 200 cm
Radius of the circles = $\dfrac{200}{2}$ cm = 100 cm
Area of the sector = $\dfrac{60°}{360°}×π r^2$ $cm^2$
= $\dfrac{1}{6}×3.14×(100)^2$ $cm^2$
= $\dfrac{15700}{3}$ $cm^2$
Area of 3 sectors = $\dfrac{3×15700}{3} = 15700$ $cm^2$
Thus, the area of the shaded region = Area of an equilateral triangle ABC – Area of 3 sectors
= 17320.5-15700 $cm^2$ = 1620.5 $cm^2$