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Area of triangle=(1/2)*base*height=(1/2)4.8*3.6=8.64 cm²

Area =length*breadthArea=45*25=1170 cm²

Given,

AP=3cm

PC=8cm

AP=BP (tangents drawn from an external point)

BP=3cm (putting in the values)

now,

PC= BP + BC

8=3+BC (putting in the values)

BC=5cm

BC=CD (tangents drawn from an external point)

CD=5cm. (putting in the values)

Here curved surface area = h* h = h^2= as TSA= 2πr(r+h)

Given:- the height of a cylinder is 77 m and the circumference of the base of the cylinder is 24 m.

∴2πr=24 m=>r=(12/π) m∴Volume of a cylinder=πr²h =π×(12/π)²×77 =3529.42 m³

given,

AP=3cm

PC=8cm

AP=BP (tangents drawn from an external point)

BP=3cm (putting in the values)

now,

PC= BP + BC

8=3+BC (putting in the values)

BC=5cm

BC=CD (tangents drawn from an external point)

CD=5cm.

We know that the lengths of the tangents from an external point to a circle are equal.

AP = PB = 3 cm and CD = BC

Since BC = PC – PB = 8 – 3 = 5 cm, CD = 5 cm

Hence, the length of CD is 5 cm.

Thank you for asking this question. Here is your answer:

AP = PB

= 3 cm

PC = 8 cm

PC = PB +BC

8 = 3 + BC

BC = 5

BC = CD

So CD will be 5 cm

Compound Interest=P( 1+ R/100)^T=20000(1+ 8/100)^1=₹21600simple interest=P×R×T/100=20000×8×1/100=₹1600Difference=21600-1600=₹20000

lateral surface area of cuboid = 2h ( l+ b ) 2*3 ( 5+4 ) 6 ( 9 )54 is the answer

Area of four walls = 2h(length+ breadth)2*3(4+5)6*954m^2

Given that,length = 5 mbreadth = 4 mheight = 3 marea of wall = 2h (L+ b) = 2 * 3 (4 +5 ) = 6*9 = 54 sq m

let alpha = aput x = aand y = -2-asoy² + 4x = 0(2+a)² + 4a = 0a² + 8a + 4 = 0a = [-8 + or - √(64 -16)]/2a = [ -8 + or - √ 48 ] /2a = [-8 + or - 4√3 ] /2a = [-4 + or - 2√3]

Assume that the length of all edges of the cube=x unitsFormula for volume of cube=x³ cube unitDue an occurrence that the cylinder is within the cube &it touches all the vertices faces of the cube.Therefore radius of the base of the cylinder & cone & height of cylinder of a cone=xVolume of the cylinder=π×{x/2}²×x=22/7×x³/411/14x³ cube unitVolume of cone should be=1/3π{x/2)²×x=11/42 cubic unitRequired ratio=volume of cube:volume of cylinder:volume of cone=x³:11/14x³:11/42x³=42:33:11

Length will be now 6+6+6=18cmBreadth=6cm, height=6cmSurface area of cuboid=2(lb+bh+hl)=2(108+36+108)=2(216+36)=2(252)=504sq.cm

Triangle is possible only when sum of angle is 180°in (i), (ii) and (iv) it is possible but in iii) 60°+ 60° + 50° = 170° so triangle is not possible

thank you so much

3360÷ 32= 105

Rs. 15.60 + Rs. 7.80 = 23.4

rs15.60+rs7.80=rs23.4

Area of shaded region= area of square - area of 4 circle= 14*14-(4*22/7*7*7)

196-(615.734)= 419 as it cant be negative

Given radius, OP = OQ = 5 cmLength of chord, PQ = 4 cmOT ⊥ PQ,∴ PM = MQ =4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]In right ΔOPM,OP2= PM2+ OM2⇒ 52= 42+ OM2⇒ OM2= 25 – 16 = 9Hence OM = 3cmIn right ΔPTM,PT2= TM2+ PM2→(1)∠OPT = 90º [Radius is perpendicular to tangent at point of contact]In right ΔOPT,OT2= PT2+ OP2→(2)From equations (1) and (2), we getOT2= (TM2+ PM2) + OP2⇒ (TM + OM)2= (TM2+ PM2) + OP2⇒ TM2+ OM2+ 2 × TM × OM = TM2+ PM2+ OP2⇒ OM2+ 2 × TM × OM = PM2+ OP2⇒ 32+ 2 × TM × 3 = 42+ 52⇒ 9 + 6TM = 16 + 25⇒ 6TM = 32⇒ TM =32/6 = 16/3Equation (1) becomes,PT2= TM2+ PM2 = (16/3)2+ 42 =(256/9) + 16 = (256 + 144)/9 = (400/9) = (20/3)Hence PT = 20/3Thus, the length of tangent PT is(20/3) cm.

thanks

Similar question is already answered.Please go through.

65%of 8065/100*80=52 marks

Let Rohan's age be x Therefore Rohan's mom's age is x+26 Now Roahn age after 3 yrs x+3 and his mothers age after three years x+29 Now we know (x+3)(x+29) = 360 x²+3x+29x+87 = 360 x²+32x+87=360 x²+32x=360-87 = 273 x(x+32)=273 After solving we get x=7 So, age of Rohan = 7 yrs The solution for how to get x is Take one factor of the equation x²+32x-273 = 0 First factorise .... Factors are x-7 and x+32 ...... So take any one factor and solve it and you will get two solutions of the quadratic equation...... X-7.....x=7 And x+32.....x=-32 Now put your logic age can't be in minus so x=7 is correct solution for this question...Hence solved

17th term = 10th term + 7

nth term, an = a+ (n - 1)d

a + 16d = a + 9d + 7 16d - 9d = 7 7d = 7 d = 1

Common difference = 1

let the age of organ be x, rohan mother=x+26 after 3 years, ( x+26+3)( x+3)=( x+29)( x+3)=x^2+3x+29x+87=360= x^2+32x=273;x^2+32x-273=0; x( x+32)-273=0; solve x=7; mother age =7+26=33

Greatest 8-digit number 999999997-digit number 99999996 digit number 999999hence sum =110,999,997

In one toss possible outcome is 2

In 6 toss possible outcome is 6×2 = 12

b) option is correct

number of coin is 1 in one coin two faceses .6 time tossed than 6×2 12

please like my answer if you find it helpful

Tossed two times Then the total number of outcomes{(H,H),(H,T),(T,H),(TT)}

the set of all possible outcome at the toss of two coin isS={HH,HT,TH,TT}

1)head, head2)tail, tail3) head, tail

Probability = possible outcomes / total outcomes

(1) Number of face cards in deck of 52 cards = 12

Probability of getting face card = 12/52 = 3/13

(2) Number of red cards in deck of 52 cards = 26

Probability of getting redcard = 26/52 = 1/2

(3) Number of ace cards in deck of 52 cards = 4

Probability of getting acecard = 4/52 = 1/13

(4) Number of spade cards in deck of 52 cards = 13

Probability of getting spade card = 13/52 = 1/4