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1)Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'Hence OA = OB = 5 cmO'A = O'B = 3 cmOO' is the perpendicular bisector of chord AB.Therefore, AC = BCGiven OO' = 4 cmLet OC =xHence O'C = 4 −xIn right angled ΔOAC, by Pythagoras theoremOA2= OC2+ AC2⇒ 52= x2+ AC2⇒ AC2= 25 −x2.....(1)In right angled ΔO'AC, by Pythagoras theoremO'A2= AC2+ O'C2⇒ 32= AC2+ (4 –x)2⇒ 9 = AC2+ 16 +x2− 8x⇒ AC2= 8x −x2− 7 ......(2)From (1) and (2), we get25 −x2= 8x −x2− 78x= 32Therefore, x= 4Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.AC2= 25 −x2= 25 − 42= 25 − 16 = 9Therefore, AC = 3 mLength of the common chord, AB = 2AC = 6 m

sin(45+theta)-cos(45-theta)= sin(45)- cos(90-45)= sin(45)-sin(45)=(1/V2)-(1/V2)=1

sin45 is 1/V2, cos45 is 1/V2 1/V2- 1/V2=0

it is correct answer

Steps of Construction:

1.Draw a line segment PQ = 11 cm.( = AB + BC + CA).

2.At P construct an angle of 30° and at Q, an angle of 90°.

3.Bisect these angles. Let the bisectors of these angles intersect at a point A.

4.Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.

5.Join AB and AC. Then, ABC is the required triangle.

For justifying construction of 90 degrees angle first follow all the steps.The below given steps will be followed to construct an angle of 90°.(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.(v) Join PU, which is the required ray making 90° with the given ray PQ.

JUSTIFICATION :-You can justify it by proving∠UPQ = 90°. For that joinPSandPT.

SinA=opposite side to A/hypotenusesinA=BC/CA=15/17cosA=adjacent side to A/hypotenusecosA=AB/CA=8/17tanA=sinA/cosAtanA=BC/AB=15/8

hi

hello sir

First multiply under root than multiply with 5 answer is 170

9.1^2 = X^2 + 8.4^2suppose the length is x cm according to Pythagoras then after solving this x= 3.5cm

Sorry it is wrong answer

this is very hard

Given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=2×22×12.5×12.5/7=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25cost of external bowl=982.14×0.05=Rs.49.1cost of ring=38.5×0.05=Rs.1.925cost of bowl=45.25+49.1+1.9=96.25 rupees approxplease like the solution 👍 ✔️👍✔️👍

AB = 5 cm

BC = 3 cm

CA = 4 cm

This implies at once that angle BCA = 90 degree.

That isC and E coincide.

BE = 3 cm

BD = 6 cm

Let O and O' two circle .which intersect in A and B .so, AB is common chord .

we know,chord is perpendicularly bisected by line joining of center to its .let line meet at T

now,∆ OAT is right angle ∆so,length of OT =√{(5^2 -(x/2)^2 }

where x is length of chord

again ,for ∆ O' AT

length of O'T =√{(3)^2 -(x/2)^2

but here ,length of OT + length of O'T =distance between centre of circles

√(25 - x^2/4) +√(9 -x^2/4 ) =4letx^2/4 =r√(25-t) +√(9-t) =4

if we put t = 9then,√(25 -9) +√(9-9) = √16 +0 =4LHS = RHS

so,

t =9x^2/4 =9x^2 =36

x=6 cm

so,length of chord = 6 cm

Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'Hence OA = OB = 5 cmO'A = O'B = 3 cmOO' is the perpendicular bisector of chord AB.Therefore, AC = BCGiven OO' = 4 cmLet OC =xHence O'C = 4 −xIn right angled ΔOAC, by Pythagoras theoremOA2= OC2+ AC2⇒ 52= x2+ AC2⇒ AC2= 25 −x2.....(1)In right angled ΔO'AC, by Pythagoras theoremO'A2= AC2+ O'C2⇒ 32= AC2+ (4 –x)2⇒ 9 = AC2+ 16 +x2− 8x⇒ AC2= 8x −x2− 7..... (2)From (1) and (2), we get25 −x2= 8x −x2− 78x= 32Therefore, x= 4Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.AC2= 25 −x2= 25 − 42= 25 − 16 = 9Therefore, AC = 3 mLength of the common chord, AB = 2AC = 6 m

thank u

Right answer is B.70°

Given that distance = 63 km.Let original speed of train = x km/hr.time = distance / time = 63/x hrs.

And ittravels a distance of 72 kkm at a average speed of 6 km/hr more than the original speed.

distance = 72 km ; speed = (x + 6) km/hr .time = 72/(x+6) hrs.If it takes 3 hours to complete the whole journey63/x + 72/(x + 6) = 3 hrs

⇒ 63(x + 6) + 72x = 3x(x + 6)

⇒ 21(x + 6) + 24x = x(x+6)

⇒ 45x + 21×6 = x^2+ 6x

⇒ x^2- 39x - 126 = 0

⇒ x^2- 39x - 126 = 0

⇒ (x - 42)(x + 3) = 0

∴ x = 42 km/hr∴the original average speed= 42 km/hr

Let R and r are the Radii of two

circles.

R = 4.8 cm,

r = 3.5 cm

Distance between two centres

= R - r

= 4.8 cm - 3.5 cm

= 1.3 cm

thanks

Answer is not clear, please try again

my name is Raj Singh

aab tum question de sakti ho, please question dena,

question do, kishi bhi subject ka, me Raj

Area of four walls=2h(l+b)=2 x 6(20+12)=12 x 32=384 m^2no.of cans needed=384/96=4cans

the correct answer is C

option A is correct answer

1/12+1/15+1/20=5+4+3/60=12/60ALL DO=12/60×1/2=101/10-1/15=6-4/60=2/60=1/30HENCE 30 is correct ans

correct answer is 30

option (b) 30 is the right answer

option b. 30 is the right answer

as SP > CP therefore it is profit

profit% = 100 x (880-800)/800= 10% profit