Let original length = x metres and original breadth = y metres.

Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%

Explanation:
Let breadth =x metres.
Then length =(115x/100)metres.

=x∗115×100=460×2=(460×100/115)x2=400x=20=x∗115×100=460×2=(460×100/115)x2=400x=20

Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20
= 88 feet

Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

From the question, 2b+l = 30
=> l = 30-2b

Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0

b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m

Explanation:
Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000

Explanation:
Let Breadth = x cm,
then, Length = 3x cm

x2+(3x)2=(810−−√)2=>10×2=640=>x=8×2+(3x)2=(810)2=>10×2=640=>x=8

Explanation:
In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440

Explanation:
Let original length = x
and original width = y
Decrease in area will be

=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%

Clearly first we need to find the areas of the given squares, for that we need its side.

Side of sqaure = Perimeter/4

So sides are,

(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961−−−√=31cmRequired perimeter =(4×31)=124cm(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961=31cmRequired perimeter =(4×31)=124cm

Explanation:
Area of each slab =

7250m2=1.44m2Length of each slab =1.44−−−−√=1.2m=120cm7250m2=1.44m2Length of each slab =1.44=1.2m=120cm

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =

Area of floorArea of tile=(1517×90241×41)=814Area of floorArea of tile=(1517×90241×41)=814

If area is given then we can easily find side of a square as,

Side=69696−−−−−√=264cmwe know diagonal =2√×side=2√×264=1.414×264=373.296cmSide=69696=264cmwe know diagonal =2×side=2×264=1.414×264=373.296cm

Explanation:
a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16

Explanation:
Let the lengths of the line segments be x and x+2 cm
then,

(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm

Explanation:
Area of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cmArea of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cm

Ratio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cmRatio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cm
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 cm

Explanation:
Let each side be a cm, then

(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=3√4∗a2=(3√4∗4003)cm2=1003√cm2(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=34∗a2=(34∗4003)cm2=1003cm2

Explanation:
We know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2aWe know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2a

Explanation:
As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.

Lets solve this,

Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1

Option D
Explanation:
We know the product of diagonals is 1/2*(product of diagonals)

Let one diagonal be d1 and d2
So as per question

12∗d1∗d2=15012∗10∗d2=150d2=1505=3012∗d1∗d2=15012∗10∗d2=150d2=1505=30

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then

Area of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200mArea of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200m

Explanation:
Area of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84−−−−√=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60mArea of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60m

Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference

So we will be having distance and time, from which we can calculate the speed. So let solve.

Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be

40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr

Let the original radius be R cm. New radius = 2R

Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%

2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π

Explanation:

Perimeter = Distance covered in 8 min. =		12000	x 8	m = 1600 m.
		60

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m² = 153600 m².

Explanation:

Perimeter = Distance covered in 8 min. =		12000	x 8	m = 1600 m.
		60

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m² = 153600 m².

Explanation:

100 cm is read as 102 cm.

A₁ = (100 x 100) cm² and A₂ (102 x 102) cm².

(A₂ - A₁) = [(102)² - (100)²]

Percentage error =		404	x 100	%	= 4.04%
		100 x 100

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = (xy) m².

New length =		120	x	m	=		6	x	m.
		100					5

New breadth =		120	y	m	=		6	y	m.
		100					5

New Area =		6	x x	6	y	m2	=		36	xy	m2.
		5		5					25

The difference between the original area = xy and new-area 36/25 xy is

= (36/25)xy - xy

= xy(36/25 - 1)

= xy(11/25) or (11/25)xy

Increase % =		11	xy x	1	x 100	%	= 44%.
		25		xy