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A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

\frac{60}{\pi}cm 

Solution

Given \theta=60^{\circ}

Length of arc = 20 cm

We know that

Length of arc =\frac{\theta}{360}\times 2 \pi r

 \\20=\frac{\theta}{360}\times 2 \pi r\\\\ \frac{20 \times 360}{60 \times 2 \pi}=r\\ \\r=\frac{60}{\pi}cm

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In Figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

[1386cm2]

Solution

Area of sector =\frac{\pi r^{2}\theta}{360^{\circ}}

Here    \theta=90^{\circ}

Radius = 21 cm

There are four sectors in the figure

Area of sector =\frac{\pi \times (21)^{2} \times 90}{360}

                        =\frac{\frac{22}{7} \times 441}{4}=346.5 cm^{2}

Area of shaded region = 4 × Area of one sector

                                     = 4 × 346.5

                                     = 1386 cm2

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A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

15246 m2

Solution

Area of circle =\pi r^{2}

        

Given that AB = 105m, BC = 21m

Where AB is radius of park and BC is wide of road

 AC=AB+BC

AC=105+21=126 m

Area of big circle=\pi r^{2}

                    \\=\pi (126)^{2}\\ =49896 m^{2}

Area of small circle =\pi r^{2}

                        \\=\pi (105)^{2}\\ =34650 m^{2}

Area of road =Area of big circle - Area of small circle

                 =49896-34650=15246 m2

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In Figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

  [ 308 cm2]

Solution

Area of sector with angle \theta = \frac{\pi r^{2} \theta}{360^{\circ}}

Here \angle p, \angle Q, \angle R=60^{\circ}

Radius of each circle = 14 cm

There are three sectors

Area of each sector = \frac{\pi \times (14)^{2} \times 60}{360}\\

                               \\=\frac{\frac{22}{7} \times 196}{6}\\ =\frac{616}{6}cm^{2}

Area of shaded region = 3 x (Area of one sector)  

                                    \\=3 \times \frac{616}{6}\\ =308cm^{2}                                              

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In Figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. to intesect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region(Use \pi = 3.14).

39.25 cm2

Solution

Angle made by vertices A, B and C = 60°     { In equilateral triangle all angles = 60°}

Diameter of circle = 10

Radius =\frac{10}{2}=5 cm

Area of shaded region = 3 × Area of sector       

                                    \\=3 \times \frac{\pi r ^{2} \theta}{360}\\ =\frac{3 \times (5)^{2} \times 3.14 \times 60 }{360}\\ =\frac{25 \times 3.14}{2}\\ =\frac{25 \times 314}{2 \times 100}\\ =\frac{25 \times 157}{100}\\ =39.25 cm^{2}

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Find the area of the shaded region in Figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use \pi = 3.14).

[30.96 cm2]

Solution

Here ABCD is a square of side 12 cm

Area of ABCD= (side)2=(12)2=144 cm2

Area of sector =\frac{\theta }{360^{\circ}} \times \pi r^{2}                                                                  here \theta=90^{\circ}

           

Here PSAP, PQBP, QRCQ, RSDR all sectors are equal

Area of 4 sectors =4 \times \frac{\theta }{360^{\circ}} \times \pi r^{2}

                                        =4 \times \frac{1 }{4} \times \pi r^{2}\\ =3.14 \times 36\\ =113.04 cm^{2}

Area of shaded region = Area of square – Area of 4 sectors

                                     = 144-113.04

                                      =30.96 cm2

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Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

\left [ \frac{308-147\sqrt{3}}{3} \right ] 

Solution

Here    \theta=60^{\circ}

                        r=14 cm

Area of segment = \frac{\pi r^{2} \theta}{360}-\frac{1}{2}r^{2}\sin \theta

                       

                   \\=\frac{\frac{22}{7}\times 14 \times 14 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \sin 60\\ =\frac{22 \times 28 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \frac{\sqrt{3}}{2}\\ =\frac{308}{3}-49\sqrt{3}\\ =\frac{308-147\sqrt{3}}{3}cm^{2}

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Find the area of the shaded region in Figure.

235.44 m2

Solution

 

There are two semi-circle with diameter (d) 4 cm.

Radius(r) =\frac{d}{2}=\frac{4}{2}=2m

Area of semi-circle =\frac{1}{2}\times \pi \times (r)^{2}=\frac{1}{2}\times \pi \times (2)^{2}=2 \pi

Length and breadth of rectangle ABCD is 16m and 4m respectively

Area of ABCD=16 x 4=64 m2                        (\because Area of rectangle = length× breadth)

Length and breadth of rectangle UVWX is 26m and 12m respectively

Area of UVWX=26 x 12 =312 m2                 (\because Area of rectangle = length× breadth)

Area of shaded region = Area of UVWX – Area of ABCD – 2 × Area of semi-circle

                                   \\=312 -64-2(2 \pi)\;\;\;\;\;\;\;\;\;\;(here \; \pi=3.14)\\ =312-64-12.56\\ =235.44 m^{2}

                                                

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Find the area of the shaded field shown in Figure.

Answer [38.28 m2]

Solution

            

Here length and breadth of rectangle ABCD is 8m and 4m respectively.

Are of rectangle ABCD=l \times b=8 \times 4 = 32 m^{2}

Radius of semi circle = 2m

 Area of semi circle=\frac{1}{2}\pi r^{2}

                              =\frac{1}{2}\times 3.14 \times (2)^{2}=6.28 m^{2}

Area of shaded field = Area of rectangle ABCD + Area of semi circle

                                 = 32+6.28

                                = 38.28 m2

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In Figure, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use p = 3.14).

 [54.5 cm2]

Solution

Given: AC = 6cm and BC = 8cm

In the figure \triangle ABC is a right angle triangle.

Hence using Pythagoras theorem

\\(AB)^{2}=(AC)^{2}+(BC)^{2}\\ =(6)^{2}+(8)^{2}\\ =36+64\\ =100\\ AB=\sqrt{100}=10\\ AB=10 cm

Diameter of circle = AB = 10 cm

Radius =\frac{10}{2}=5 cm

Area of circle =\pi r^{2}

                 =3.14 \times (5)^{2}=78.5 cm^{2}

Area of \triangle ABC=\frac{1}{2}\times AC \times BC\\

                            =\frac{1}{2} \times 6 \times 8=24m^{2}

Area of shaded region = Area of circle – Area of DABC

                                               =78.5-24=54.5 cm2

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