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in a circle a chord 1 centimetres away

answer is 3cm. because distance between chord and centre of the circle and length of chord are inversely proportional to each other

hope it will help you

3 is the right answer

3 is the most valuable and correct answer

it is away from2 cmso ,6-15 is the right.answer....

O is centre of the circle.

DE is the chord 1 cm away from centre. CH is perpendicular on DE.

∴CH = 1 cm and DE = 6 cm

FG is the chord 2 cm away from centre. CI is perpendicular on FG.

∴CI = 2 cm

From centre C, CE and CG are joined.

CE and CG both are radius of the circle.

In ΔCHE we have,

∠CHE = 90° [∵CH is perpendicular on DE]

CH = 1 cm

HE = DE/2 = 3 cm [∵perpendicular drawn from centre bisects chord]

In ΔCGI we have,

∠CIG = 90° [∵CI is perpendicular on FG]

CI = 2 cm

CG = CE = √10 cm

∴FG = 2 × IG = 2√6 cm [∵perpendicular drawn from centre bisects chord]

∴The length of the chord is = 2√6 cm

l metre contains 0.01 centimetres

Answer:0.01 m is 1 cm

1 acre=4046.856m²25 acre=101171m²

Area =(1/2)base*height101171=(1/2)*20*heightheight=10117.1m

We know that circumradius =abc/4∆

by herons law ∆=✓(s(s-a)(s-b)(s-c))

where s=13+14+15/2=21

∆=√(21(8)(7)(6)

∆=84

now ,circum radius,r1 =∆/(s-a)=84/(21-13)=10.5

Option (1) is correct.

Given,Beam length l = 9m, width w = 50 cm or .5m, depth d = 20 cm or .2m

Density of beam Dn = 30 kg/m^3

Volume of beam = l*w*d= 9*.5*.2= 90/100= .9 m^3

Weight of Beam = Volume of beam*density of beam= .9*30= 27 kg

Given: Length = 1.5 m,Breadth = 1.25 m and Depth = 65 cm = 0.65 m

Area of the sheet required for making the box open at the top

= 2(bh+hl)+lb = 2(1.25*0.65+0.65*1.5)+1.5*1.25

= 2(0.8125+0.975)+1.875 = 2*1.7875+1.875

= 3.575 + 1.875

= 5.45 m^2

HEIGHT WILL BE 4 CM

shape of a cuboid is 1.4m × 90 cm × 70 cm or 140 cm × 90 cm × 70 cm

Volume of the pit = lbh cu units.= 140 × 90 × 70 = 882000cu cm

Dimensions of the bricks is 21 cm × 10.5 cm

Let height of the brick be 'H' cmVolume of the brick = 21 × 10.5 × h cu. cmHence

volume of 1000 bricks = 1000 × 21 × 10.5 × H1000 × 21 × 10.5 × h = 882000⇒ H = 4 cm

..........,., ,.,.,..

600 Sq mConvert into areas600 Sq m0.0006 Sq kmThe answer is correctPlz accept my best answer.

600sq m = 0.6 sq km because 1km = 1000 m

a) 600 Sq mConvert into area0.0006 Sq km

b) 1200 Sq mConvert into area 0.0012 Sq km

The answer is correctPlz accept my best answer.

bt it is wrong

give me solution

3.25 =Cancel using 5325/100

65/20

Again cancel using 5

13/4

(A) option is correct

3(i)=(-1)^2-1=1-1=0, (1)^2-1=1-1=0, (ii)p(x)=2x-1=2(-1/2)-1=-1-1=-2, (iii)p(x)=4x+5; x=-5/4, 4(-5/4)+5=0; (iv)=3x^2=3(0)^2=0; (v)p(x)=ax+b; x=-b/a; a(-b/a)+b=0

Thanks for the Help

Density of beam = 30 kg per m^3Volume of beam = 9x.5x.2= 9/10Density = weight/volumeWeight = 30 x 9/10Weight of beam is 27 kg

thanks

4x*x + 9y*y + 25z*z - 12xy - 30yz + 20zx = (2x)(2x) + (-3y)(-3y) + (5z)(5z) + 2(2x)(-3y) + 2(-3y)(5z) + 2(5z)(2x) = (2x-3y+5z)(2x-3y+5z)

Area of floor=(4.8m)(3.6m)=17.28m^2because rate is 100 Rs/squareso, cost of floor =100x17.28 Rs=1728 Rsandtwo wall area =2x (3.6m)(2m)=14.4m^2other two wall area =2x(4.8m)(2m)=19.2m^2now total area of wall =(14.4+19.2)m^2=33.6m^2now cost of for wall's tiles =33.6x100=3360Rs now total cost =(3360+1728)Rs =5088 Rs

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Q91- 26Q92-156This is correct answer of your questions

Anequilateral trianglecannot have anobtuse angle. In fact, the anglesareall "fixed" -they areeach 60 degrees. It is impossible to create an obtuse equilateral triangle.The word 'equilateral' means equal sides. An equilateral triangle is one that has three equal sides and, subsequently, three equal angles. The angles are each 60 degrees and, when added together, add up to 180 degrees.An obtuse angle is an angle greater than 90 degrees and less than 180 degrees. Thus, an equilateral triangle cannot have an obtuse angle because the angles must all be the same degrees and the sum of the angles cannot exceed 180 degrees. If an obtuse angle could be used in an equilateral triangle, this would mean that all the angles would be, at least, 90 degrees. In turn, it would mean that the sum of the angles far exceeds 180 degrees, with the sum being a minimum of 270 degrees. It's just not possible.

yes, we think it is possible to sketch an obtuse angled equilateral triangle.

cos^2 30°(1+tan^2 30°)=(√3/2)^2{1+(1/√3)^2}=3/4 * 4/3=1

Crop only the question that you want a solution for. We will not be able to provide solutions to multiple questions.

give answer of 3rd question

give answer of 5th question

Give 5 question answer

draw an acute angled triangle and an obtuse angle triangle.draw the perpendicular bisectors of their sides.where does the point of concurrent lie?

give answer 1question

It is an A.P.,

in which, a= -5, d= -8-(-5)=-3, l=-230

First calculate the number in series.....

l=a+(n-1)d

-230=-5+(n-1)-3

-225=-3n + 3

3n=228

n=76.

Now,

Sum of the series = n/2 (a+l)

76/2 (-5-230)

38 * -235 = -8930.

So, the sum of your series would be -8930.

d is correct answer, 8:125

Let the smaller number = xbigger number = 5x.

if 21 is added to both; 5x+21 = 2×(x+21)

5x + 21 = 2x + 42

5x - 2x = 42 - 21

3x =. 21

x = 21/3= 7

So the positive number is = 5×7 = 35

answer is 35

Sun of areas of walls=2(length*height)+2(breadth*Height)=2(6.5*6.5)+2(12*6.5)=2(42.25)+2(78)=84.5+156=240.5m²