AREA

If the radius of a circle is diminished by 10 %, then its area is diminished by :

Let the original radius be R cm,

New radius = (90% of R) cm = (90/100*R) cm

=9R /10 cm

Original area =πR²

Diminished area = [ πR²-π(9R/10)²] cm²

= [(1-81/100)πR²] cm²

(19/100 πR² )cm²

Decrease% = (19πR²/100 *1/πR²*100)%

=19%

If the radius of a circle is increased by 6%, then the area is increased by:

Let the original radius be R cm. New radius= (106/100 R)cm

=(53R/50) cm

Original area =πR²

Increase in area = π(53R/50)² – πR²

= πR²[(53/50)²-1]

= πR²[(53)² – (50)²]/2500

= πR²(103*3)/2500 m²

Increase %= πR²*309/2500*1/πR²*100)%

=12.36%

The area of the largest circle, that can be drawn inside a rectangle with sides 18 cm by 14 cm, is :

Radius of the required circle= (1/2*14) cm

= 7 cm

Area of the circle =(22/7*7*7) cm²

=154 cm²

The perimeter of a triangle is 30 cm and the circumference of its in circle is 88 cm. The area of the triangle is :

Let  the radius of in circle be r cm

then 2πr = 88

r= (88*7/22*1/2)

=14

semi-perimeter, s=(30/2) cm =15 cm

Area of the triangle =r *s

= (14*15 ) cm²

A semi- circular shaped window has diameter of 63 cm. Its perimeter equals:

Perimeter of window = πR plus 2R

=(22/7*63/2 plus 63) cm

=(99 plus 63)cm

=162 cm

What will be the area of a semi- circle whose perimeter is 36 cm?

Given : πR plus 2R=36

(π plus 2 )R= 36

R=36/(22/7 plus 2) cm

(36*7/36) cm

7 cm

Required area = πR²= (22/7 * 7*7) cm²

= 154 cm²

The radius of a wheels is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be :

Distance covered in 1 revolution= 2πR

= (2* 22/7*25/100) m

= 11/7 m

required number of revolutions = (11000*7 /11)

= 7000

A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratio of 6 : 5. The smaller side of the rectangle is :

Length of wire = 2πR

(2*22/7*42) cm

=24 cm

perimeter of rectangle = 2*(6x plus 5x) cm

22x cm

22x=264

x=12

smaller side =    (5*12) cm

=60 cm

Between a square of perimeter 44 cm and a circle of circumference 44 cm, which figure has larger area and by how much ?

Side of the square = 44/4 cm

=11 cm

Area of the square = (11* 11 ) cm²

= 121 cm²

2πR = 44

R=7 cm

Area of circle =πR²

=154 cm²

Area of circle is larger by 33 cm²

The length of a row is 5.5 meter and width in 3.75 meter. Find the cost of paving the floor by slabs at the rate of Rs 800 per square. meter

Area of the floor = (5.5 * 3.75) m²

= 20.625 m²

Cost of paving = Rs (800*20.625)

= RS. 16500

The difference between the length and breath of a rectangle is 23 m, If its perimeter is 206 m, then its area is :

we have: ( l – b) = 23 and 2(l plus B ) = 206

or l plus B = 103

We get L = 63 and b= 40

Area = (l* b) = ( 63 * 40 )m²

= 2520 m²

The length of a rectangular hall is 5 meter more than its breath the area of the hall is 750 m². The length of the hall is :

Let breath = x m

then length = (x plus 5 ) m

then x (x plus 5 ) = 750

x² plus 5x – 750 =0

(x plus 30 )(x – 25) =0

x=25

length = (x plus 5 )

= 30

A rectangular field is to be fenced on three sides living a side of 20 ft uncovered if the area of the field is 680 square ft how many feet of fencing will be required

we have L = 20  ft

and LB = 680 square ft

so B = 34 Ft

length of fencing = (l plus 2b )

= ( 20 plus  68)ft

= 88 ft

The ratio between the perimeter and the breadth of a rectangle is 5:1 . If the area of the rectangle is 216 sq cm , what is the length of the rectangle ?

2 (l plus b )/b =5/1
2l plus 2b = 5b
3b = 21
b=2/3 l

Then, Area = 216 cm²

L * b =216

l * 2/3l=216

l²=324

l=18 cm

A courtyard 25 m long and 16 m broad is to be paved with bricks of dimensions 20 cm by 10 cm . The total number of bricks required is :

number of bricks = (Area of courtyard /Area of 1 brick )

(2500* 1600/20*10)
=20000

The diagonal of a rectangle is root 41 cm and its area is 20 sq. cm. The perimeter of the rectangles must be :

√l² plus b² = √41

or l² plus b²= 41

also lb = 20

(l plus b )²= (l² plus b² ) plus 2lb

= 41 plus 40

= 81

(l plus b)=9

perimeter = 2(l plus b )

= 18 cm

The area of a sector a circle of radius 5 cm, formed by an are of length 3.5 cm is:

Area of the sector = (1/2*arc*R)

=(1/2*3.5*5) cm²

8.75 cm²

In a circle of radius 7 cm an arc subtends an angle of 108 at the center. The area of the sector is :

Area of the sector= πR²Θ/360

=(22/7*7*7*108/360)cm²

=46.2 cm²

The area of the largest triangle that can be inscribed in a semi circle of radius R, is :

Required area = 1/2 * base* height

= (1/2 * 2r *r)

= r²

The radius of the circumcircle of an equilateral triangle of side 12 cm is :

radius of circumcircle = a / √3

= 12 / √3 cm

= 4√3