CBSE 10th Standard Maths Subject Areas Related to Circles HOT Questions 2 Mark Questions With Solution 2021
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CBSE 10th Standard Maths Subject Areas Related to Circles HOT Questions 2 Mark Questions With Solution 2021
10th Standard CBSE

Reg.No. :
Maths

The figure given shows a kite, in which BCD is in the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and \(\triangle CEF\) is an isosceles rightangled triangle whose equal sides are 6 cm long. Find the area of the shaded region.
(a) 
Four cows are tethered at the four corner of a squares plot of size 50 m. So that they just cannot reach one another. What area will be left ungrazed?
(a) 
In the adjoining figure, ABCD is a square of side 6cm.Find the area of the shaded region.
(a) 
In the figure alongside, crescent is formed by two circles which touch at the point A, O is the centre of the point A, O is the centre of the bigger circle.If CB=9cm and ED=5cm, find the area of the shaded region.[Take \(\pi\)=3.14]
(a) 
ABCD is a field in the shape of a trapezium.ABDC and \(\angle\)ABC=60^{0},\(\angle\)DAB=90^{0}.Four sector are formed with centres A, B, C and D.THe radius of each sector is 17.5m.Find:
(i)the total area of the four sectors
(ii)the area of remaining portion, given that AB=75m and CD=50m.
(a)
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CBSE 10th Standard Maths Subject Areas Related to Circles HOT Questions 2 Mark Questions With Solution 2021 Answer Keys

1404 cm^{2}

535.71 cm^{2}

From P, draw PQ⊥AB, join PA and PB,
In ΔAPQ and Δ BPQ, we have
AP = BP = 6cm
PQ = PQ
ㄥAQP = ㄥBQP=90^{0}
⇒ ΔAPQ ~ Δ BPQ
Therefore, \(AQ = BQ={1\over 2}AB\)
\(⇒\ AQ=BQ=3cm\)
Let ㄥPBQ=θ
∴ In rt. Δ QBP, ㄥQ=90^{0}
\(⇒\ cos\theta={QB\over PB}\)
\(cos\theta={3\over 6}={1\over 2}\)
⇒ cos θ = cos60^{0}
⇒ θ=60^{0}
Also, \({PQ\over PB}=sin60^0\)
\(PQ=6\times{\sqrt3\over2}=5.196cm...(i)\)
Now, area of sector BPA=\({60^0\over360^0}\times{22\over 7}\times6\times6\)
=18.857cm^{2}
Area of the portion APB
= 2 X [Area of sector BPA  Area of ΔBPQ]
= 2 x [18.857  7.7941= 22.126cm^{2}
Area of the shaded portion = 2 x [Area of quadrant ABC — Area of the portion APB]
\(=2\times\left[{90^0\over 360^0}\times{22\over 7}\times6\times622.126\right]\)
=2 x [28.28622.126]
=12.32cm^{2}
Hence, the required area of the shaded portion is 12.32cm^{2} 
Suppose R and r be the radii of bigger and smaller circles, respectively
⇒ AB  AC = CB
⇒ 2R2r=9
⇒ Rr=\(9\over 2\)= 4.5 cm ...(i)
Join AD and CD
ΔAOD ~ ΔDOC
\(⇒\ {OD\over OA}={OC\over OD}\)
⇒ OD^{2 }= OA x OC
⇒ (R5)^{2}=R x (R9)
⇒R^{2}+2510R=R^{2}9R
⇒ R=25cm
From (i), we have Rr = 4.5
=25  4.5 = 20.5cm
Now, area of the shaded portion = πR^{2}πr^{2}
=π(R^{2}r^{2})
=π(R+r)(Rr)
=3.14 x (25+20.5)(2520.5)
=3.14 x 45.5 x 4.5
= 642.915 cm^{2}
Hence, the required area of the shaded portion is 642.915cm^{2} 
Let ABCD be a trapezium with AB 75 m,
DC = 50 m, ㄥABC = 60^{0} and ㄥDAB = 90^{0}.
Draw CE ⊥AB.
∴ AE = 50m and EB = 7550 = 25m
We know that sum of angles of a trapezium is 360^{0}.
∴ Area of four sectors = πr^{2}
\(={22\over 7}\times17.5\times17.5\)
=962.5m^{2}
In rt .ㄥed ΔCEB,
\({EC\over EB}=tan60^0\)
\(⇒ {EC\over 25}=\sqrt{3}\)
⇒ EC=25√3m
Now, area of trapezium\(={1\over 2}(AB+DC)\times EC\)
\(={1\over 2}(75+50)\times25\sqrt3\)
\(={1\over 2}\times125\times25\sqrt3\)
=2706.25
Area of remaining portion = 2706.25  962.5
=1743.75m^{2}