1404 cm²

535.71 cm²

From P, draw PQ⊥AB, join PA and PB,
In ΔAPQ and Δ BPQ, we have
AP = BP = 6cm
PQ = PQ
ㄥAQP = ㄥBQP=90⁰
⇒ ΔAPQ ~ Δ BPQ
Therefore, \(AQ = BQ={1\over 2}AB\)
\(⇒\ AQ=BQ=3cm\)
Let ㄥPBQ=θ

∴ In rt. Δ QBP, ㄥQ=90⁰
\(⇒\ cos\theta={QB\over PB}\)
\(cos\theta={3\over 6}={1\over 2}\)
⇒ cos θ = cos60⁰
⇒ θ=60⁰
Also, \({PQ\over PB}=sin60^0\)
\(PQ=6\times{\sqrt3\over2}=5.196cm...(i)\)
Now, area of sector BPA=\({60^0\over360^0}\times{22\over 7}\times6\times6\)
=18.857cm²
Area of the portion APB
= 2 X [Area of sector BPA - Area of ΔBPQ]
= 2 x [18.857 - 7.7941= 22.126cm²
Area of the shaded portion = 2 x [Area of quadrant ABC — Area of the portion APB]
\(=2\times\left[{90^0\over 360^0}\times{22\over 7}\times6\times6-22.126\right]\)
=2 x [28.286-22.126]
=12.32cm²
Hence, the required area of the shaded portion is 12.32cm²

Suppose R and r be the radii of bigger and smaller circles, respectively
⇒ AB - AC = CB
⇒ 2R-2r=9
⇒ R-r=\(9\over 2\)= 4.5 cm ...(i)  

Join AD and CD
ΔAOD ~ ΔDOC
\(⇒\ {OD\over OA}={OC\over OD}\)
⇒ OD² = OA x OC
⇒ (R-5)²=R x (R-9)
⇒R²+25-10R=R²-9R
⇒ R=25cm
From (i), we have R-r = 4.5
=25 - 4.5 = 20.5cm
Now, area of the shaded portion = πR²-π​​​​​​​r²
=π​​​​​​​(R²-r²)
=π​​​​​​​(R+r)(R-r)
=3.14 x (25+20.5)(25-20.5)
=3.14 x 45.5 x 4.​​​​​​​5
= 642.915 cm²
Hence, the required area of the shaded portion is 642.915cm²

Let ABCD be a trapezium with AB 75 m,
DC = 50 m, ㄥABC = 60⁰ and ㄥDAB = 90⁰.
Draw CE ⊥AB.
∴ AE = 50m and EB = 75-50 = 25m
We know that sum of angles of a trapezium is 360⁰.
∴ Area of four sectors = πr²
\(={22\over 7}\times17.5\times17.5\)
=962.5m²
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
=1743.75m²