## LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius7 cm. Find,

(1) Given LMN is an equilateral triangle .LM = 14 cmArea of LMN = (√3/4)a2Here a represents the side of equilateral triangle.a = 14Area of LMN = (√3/4)×142= 49√3= 84.87cm2Hence area of LMN is 84.87cm2 (2) Since LMN is equilateral, L = M = N = 60˚= 60˚Given r = 7Area of sector = (/360)r2Area […]

## In figure 7.35, PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Given PQ = 14cmQR = 21cmQ = = 90˚Area of part x = (/360)r2Area of part x = (90/360)×(22/7)×142= 11×14= 154cm2Consider sector (R-BYA)QR = QB+BRBR = 21-14 = 7cmAR = 7cm [radius of same circle]Area of part y = (/360)r2Area of part y = (90/360)×(22/7)×72= 11×7/2= 38.5cm2Area of rectangle PQRS = length ×breadth= QR×PQ= 21×14= […]

## The area of a minor sector of a circle is 3.85 cm2 and the measure of its central angle is 36°. Find the radius of the circle.

Given area of minor sector = 3.85cm2Measure of central angle, = 36˚Area of sector = (/360)r23.85 = (36/360)×(22/7)×r2r2 = (3.85×360×7)/(36×22)r2 = (3.85×10×7)/22r2 = 12.25r = 3.5cmHence the radius of the circle is 3.5cm

## Radius of a sector of a circle is 7 cm. If measure of arc of the sector is –

Given radius of sector, r = 7cm (1)Measure of arc, = 30˚Area of sector = (/360)r2= (30/360)×(22/7)×72= (1/12)×(22/7)×49= (1/12)×22×7= 12.83cm2Hence the area of sector is 12.83cm2 (2)Measure of arc, = 210˚Area of sector = (/360)r2= (210/360)×(22/7)×72= (7/12)×(22/7)×49= (7/12)×22×7= 89.83cm2Hence the area of sector is 89.83cm2 (3)Measure of arc, = 3 right angles = 3×90 = […]

## In figure 7.34, if A(P-ABC) = 154 cm2 radius of the circle is 14 cm,

Given radius of circle , r = 14cmArea ,A(P-ABC) = 154cm2(1) Let APC =A(P-ABC) = (/360)r2154 = (/360)×(22/7)×142= (154×360×7)/(22×14×14)= 90˚Hence APC is 90˚.(2)length of arc, l(arc ABC) = (/360)2rl(arc ABC) = (90/360)×2×(22/7)×14= (1/4)×2×22×2= 22cm.Hence l(arc ABC) is 22cm.

## In figure 7.33 O is the centre of the sector. ROQ = MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. ( = 22/7 )

Given ROQ = MON ,= 60°.Radius OR, r = 7 cmRadius OM, R = 21 cmLength of arc RXQ = (/360)2r= (60/360)×2×(22/7)×7= (1/6)× 2×22= 22/3= 7.33cmHence Length of arc RXQ is 7.33cm.Length of arc MYN = (/360)2R= (60/360)×2×(22/7)×21= (1/6)× 2×22×3= 22cmHence Length of arc MYN is 22cm.

## In figure 7.32, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).

Given radius , r = 3.4cmPerimeter of sector P-ABC = PA+arc ABC +PC12.8 = 3.4+ length of arc ABC +3.4length of arc ABC , l = 12.8-3.4-3.4 = 6cmArea of sector P-ABC = lr/2= 6×3.4/2= 3×3.4= 10.2cm2Hence A(P-ABC) is 10.2cm2.

## In the figure 7.31, radius of the circle is 7 cm and m(arc MBN) = 60°, find

Given radius of circle , r = 7cm(1) Area of the circle , A = r2= (22/7)×72=( 22/7)×49= 154cm2Hence the area of the circle is 154cm2 (2)Given m(arc MBN), = 60°Area of sector = (/360)r2A(O-MBN) = (60/360)×(22/7)×72= (1/6)×22×7= 25.67cm2Hence area of sector O-MBN is 25.67cm2 (3) Area of major sector = Area of circle -Area […]

## Area of a sector of a circle of radius 15 cm is 30 cm2 . Find the length of the arc of the sector.

Given radius of circle , r = 15cmArea of the sector of circle , A = 30cm2Area of sector = length of arc ×radius/2 = lr/2Length of arc , l = 2A/r = 2×30/15 = 4cmHence the length of the arc is 4cm.

## Radius of a circle is 10 cm. Area of a sector of the sector is 100 cm2 . Find the area of its corresponding major sector. ( = 3.14 )

Given radius of circle , r = 10cmArea of minor sector , A = 100cm2Area of circle = r2 = 3.14×102 = 3.14×100 = 314cm2Area of major sector = Area of circle – Area of minor sector= 314-100= 214cm2Hence the area of the major sector is 214cm2