Ans. Diameter of Well = 10 m

Radius of Well = 10/2 = 5m

Depth (height) of well = 8.4 m

Volume of Well = \({22\over7} \times({5})^2 \times8.4 = {22\over7} \times25\times8.4 \)

= 660 m³

Volume of Well = Volume of Earth Taken Out 

=> Volume of Earth Taken Out = 660m³ .....(1)

This earth spread around well, so it ll be in form of hollow cylinder.

Let Height of Embankment = h

Inner Radius  (r) = 5m

Width of Embankment (t) = 7 m

Outer Radius (R) = r + t = 5+7 =12 m

Volume of Embankment = Volume of outer cylinder - volume of inner cylinder

=> \([{{22\over 7} \times ({R})^2\times h }] - [{{22\over7}\times ({r})^2\times h}]\)

=> \({22\over7}\times h[{144-25}] ={22\over7}\times h \times 119\)

=> 374h .........(2)

From (1) and (2)

=> 374h = 660

=> h = 1.765m

Rate For Digging And making embankment  = Rs. 500 per cubic meter

Actual Amount = 500 * [ Total Volume of Well + Volume Of embankment ] 

= 500*[660+660] = Rs. 660000

(as he consider height 2m)volume of embankment = 22*2*[144-25]/7= 44*119/7 = 748m³

Amount He Charged = 500* [ 660 +748 ]= 550*1408=774400

Extra money Made = 774400-660000 = Rs.114400

But madam my book answer is not match your answer.

After digging and spreading the earth, we consider the shape of hollow cylinder.

Here, Radius of well = 5 m and Outer Radius of well = 5 + 7 = 12 

Let height of embankment be h meters.

Now, Volume of Earth spread = Volume of Well

π(12)² × h = π(5)² × 8.4

h = 25 × 8.4 ÷ 144 = 1.46 m

But charge taken by contractor for 2 m height = Rs.500

Actual charge for 1.46 m height = 500/2 × 1.46 = Rs.365

Contractor charged more by Rs.500 - Rs.365 = Rs.135.

\(CSA = {2\pi rh}\)

\(TSA = {2 \pi r(h+r) }\)

Their ratio is 1:2

i.e. \( {h\over h+r}= 1/2\)

2h = h+r

h=r

h :r = 1:1

Let the two numbers be x and y

x>y

5x = 2y+9     {  Dividend = Divisor X Quotient + Remainder}

Required linear equation is : 5x-2y-9=0

Solutions:

Put x=1

5=2y+9

2y=-4

y=-2

Also,

take x= 3

15=2y+9

2y=6

y=3

So, solutions are (1,-2) and (3,3) 

Ans. Thickness of Wooden Cylinder (t) = 2 cm 

Height of Cylinder (h)= 35 cm

Inner radius (r)= 12 cm 

Outer radius (R) = r + t = 12 +2 = 14 cm

Volume of Wood Required = Outer Volume of Cylinder - Inner Volume of Cylinder

=> \(({\pi \times R^2 \times h } ) - ({\pi \times r^2 \times h}) = {\pi \times h (R^2 - r^2)}\)

=> \({22 \over 7 }\times 35 \times ({196-144}) = {22 \over 7 }\times 35 \times52 = 5720 cm^3\)

=> Volume of Wood Required  = 5720 cm³

Wooden cylinder

Thickness = 2cm

Inner radius = 12cm

Outer radius = Inner radius + thickness

= 12+2 = 14cm

Height of cylinder = 35cm

Volume of wood required in making the cylinder = \(V = {\pi (R^2-r^2)h }\) 

= [22(14²-12²) X 35 ]/7

= 5720 cm³

Let the length,breadth and height of the cuboid be l,b and h respectively.

Given  l+b+h= 19cm

Squaring both sides

(l+b+h)²= 361

l²+b²+h²+2(lb+bh+hl)= 361      -(1)

Also, Length of the diagonal = \(d = { \sqrt{l^2+b^2+h^2}}\)= 11 cm

Squaring both sides , d²= l²+b²+h²

l²+b²+h²⁼ 121

Putting it in eqn (1)

121 + 2(lb+bh+hl)= 361

2(lb+bh+hl)= 361-121

Surface area of cuboid = 240 cm²

Ans. let length of cuboid = l

breadth of cuboid = b

height of cuboid = h

According to question

l + b + h = 19   ............ (1)

We Know 

Length of Diagonal of Cuboid = \({ \sqrt{ l^2+ b^2 + h^2}}\)

=> \({ \sqrt{ l^2+ b^2 + h^2}} = 11\)

Squaring both sides, we get 

=> \({ l^2+ b^2 + h^2} = 121\)    .......(2)

Also 

=> \(({ l+ b + h})^2 = {l^2 + b^2 +h^2 + 2lb + 2lh +2bh }\)

=> \( { 2(lb + lh +bh) } = {({ l+ b + h})^2 - {(l^2 + b^2 +h^2)}}\)

put values from (1) and (2), 

=> \( { 2(lb + lh +bh) } = 361 - 121 = 240\)   ....... (3)

Surface Area of Cuboid = \( { 2(lb + lh +bh) } \)

=> So Surface Area of Cuboid = 240 cm²

x + y = 1600

Ans.

Given : x = 3k - 2 and y = 2k is solution of eq. 4x -7y + 12 =0

To Find: value of k

Solution : 

put value of x and y in eq., we get

4(3k-2) - 7(2k) + 12 = 0

=> 12k - 8 - 14k + 12 =0

=> -2k + 4 = 0

=> 2k = 4

=> k = 2

Value of k = 2

Putting the given values of x and y in equation

4x + 7y + 12 = 0

4 + 3k + 7 × 2k + 12 = 0

12k + 14k + 12 = 0

26k = -12

k = -6/13

Putting the coordinates of point A in given equation,

4a + 6y = 8.   

2a + 3y = 4.        ..........(i)

Again putting the coordinates of point B in given equation

a + 3b = 8.         ..........(i)

Subtracting eq.(ii) from eq.(i)

a = -4

Putting the value of a in eq.(ii)

-4 + 3b = 8

3b = 12

b = 4

Therefore values of a and b are -4 and 4 respectively.

Volume of tank = 5040 litres = 5040000 cu.cm

Dimensions of tank = 2.2 m x 1.7 m x 1.7 m = 220 cm x 170 cm x 170 cm

Walls of tank are 5 cm thick.

Therefore, length = 220 - 5 - 5 = 210 cm

Breadth = 170 - 5 - 5 = 160

According to question,

l x b x h = 5040000

210 x 160 x h = 5040000

h = 150 cm

Then thickness of bottom = 150 - 5 - 5 = 140 cm = 1.4 m

Ans. Let Water raise by h meter height. 

Length of Rectangular Reservoir = 10.8 m

Width of Rectangular Reservoir = 3.75 m 

Volume of Rectangular Reservoir =  \(10.8 \times 3.75 \times h = ({40.5 \times h } )m^3\)    (1)

Cross section of Pipe = 7.5 * 4.5  = 33.75 cm² = 0.003375 m²

Speed of water = 18m/s

Distace covered in 30m ( 1800s) = 18 * 1800 = 32400m 

Volume of Water flowed in 30 m = cross section of pipe * distance covere = 0.003375 * 32400 = 109.35 m³    (2)

From (1) and (2)

40.5 * h = 109.35 

=> h = 2.7 m

According to question

4πr² = 6a²

r²/a² = 6/4π.         

r/a = √3/√2π

Now ratio of their volumes

(4/3)πr³ / a³

= 4π/3 × (√3/√2π)³

= (4π/3)(3/2π)(√3/√2π)

= 2(3×7/2×22)

= √21/√11

Therefore ratio of their volumes is √21 : √11

Ans. Suppose radius of sphere is r and side of cube is x.

now,

Surface Area of Sphere = Surface Area of Cube

=> \(4 \times \pi \times r^2 = 6 \times x^2\)

=> \({ r^2 \over x^2} = {3 \over 2\pi}\)

=> \({r \over x } = {\sqrt{3 \over 2\pi}}\)

Let volume of Sphere = V_s 

and volume of Cube = V_c

Then

=> \({V_s \over V_c} ={{{ 4 \over 3 }{ \pi r^3} } \over x^3}\)

=> \( {{4 \over 3 }\pi {({r \over x})^3}}\)

=> \( {{4 \pi \over 3} \times {3 \over 2\pi } \times \sqrt{3 \over 2\pi }}\)

=> \( \sqrt{12\over 2\pi}\)

=> \({{ \sqrt 6} \over {\sqrt \pi}}\)

So Required Ratio is = \(\sqrt 6 : \sqrt \pi \)

Ans. Let the length of each edge of the cube is a.
Volume of cube = a³  (1)
 

Since the Cylinder is within the Cube and it touches all the vertical faces of Cube.

Diameter of Cylinder = a 
Radius of base of the cylinder = \(a\over 2\)     
Height of Cylinder = a

Volume of Cylinder = \({ {22 \over 7} \times ({a \over 2})^2} \times a = {11 \over 14 }a^3\)      (2)

As Cone and Cylinder has same base

Radius of the Cone = \(x \over a\)
Height of Cone =  a

Volume of Cone = \({1 \over 3} \times { {22 \over 7} \times ({a \over 2})^2} \times a = {11 \over 42 }a^3\)    (3)

Ratio of volumes, From (1) (2) and (3)

=> \( a^3 :{11 \over 14 }a^3:{11 \over 42 }a^3\)

Multiply by 42 and divide by a³, We get 

=> 42 : 33 : 11

Ans. Radius of Cylinderical Tub = 12 cm

Raised Height of Water in cylindrical tub = 6.75 cm  

Suppose Radius of sphereical Ball = r

Then volume of raised water in tub = volume of ball

=> \(\pi\times (12)^2 \times 6.75 = {{4 \over 3} \times \pi\times r^3 }\)

=> \(4 \times r^3 = {3 \times 144 \times 6.75}\)

=> \(r^3 = {2916\over 4}\)

=> \(r^3 = 729\)

=> r = 9

Radius of Spherical Ball = 9 cm

Volume of sphere = 4851

(4/3)πr³ = 4851

r³= (4851 × 3 × 7)/(4 × 22)

r = 10.5 cm

Again,

Reduced volume of sphere = 4312/2 

(4/3)πr³= 4312/2

r³= (4312 × 3 × 7)/(3 × 4 × 22)

r = 7 cm

There for radius should be reduced to 10.5 - 7 = 3.5 cm.

Ans. Let Initial Radius of Sphere is R and FInal Radius of Sphere is r.

Initial Volume of the sphere = 4851 cm³

^=> \({4 \over 3} \times {22 \over 7} \times R^3 = 4851\)

^=> \(R^3 = {{4851 \times 21 } \over 88}\)

^=> \(R^3 = 1157.625 \)

=> R  = 10.5             (1)

Final Volume of Sphere = \({4312 \over 3} cm^3\)

=> \({4 \over 3} \times {22 \over 7} \times r^3 = {4312\over3}\)

=> \(r^3 = {{4312 \times 7 } \over 88}\)

=> \(r^3 = 343\)

=> r = 7

Radius to be Reduced by = 10.5 - 7 = 3.5 cm

Ans.

Given : An isosceles Right Triangle ABC. A square CMPN is Inscribed in it .

To Prove : CP bisects the hypotenuse AB. i.e. AP = PB
Proof : CMPN is square.
So, 
CM = MP = PN = NC      [ all sides are equal]

Also Triangle ABC is Isosceles.
So, AC = BC 

=> AN + NC = CM + MB 

=> AN = MB   [as NC = CM]

Now, Consider Traingles ANP and PMB
AN = MB  [Proved Above]

∠ANP  =  ∠ PMB  [Both are 90]

PN = PM    [Sides of square]

So, By SAS  Rule 
Triangle ANP = Triangle PMB 

=> AP = PB         [By CPCT]
Hence Proved
 

Ans.

Given : In ΔABC, P, Q and R are the mid points of sides BC, CA and AB respectively. AD ⊥ BC.
To prove : P, Q, R and D are concyclic.

Proof : In ΔABC, R and Q are mid points of AB and CA respectively.

∴ RQ || BC (Mid point theorem)

Similarly, PQ || AB and PR || CA

In quadrilateral BPQR,

BP || RQ and PQ || BR        [RQ || BC and PQ || AB]

∴ Quadrilateral BPQR is a parallelogram.

Similarly, quadrilateral ARPQ is a parallelogram

∴ ∠A = ∠RPQ (Opposite sides of parallelogram are equal)

PR || AC and PC is the transversal

∴ ∠BPR = ∠C (Corresponding angles)

∠DPQ = ∠DPR + ∠RPQ = ∠A + ∠C ...           (1)

RQ || BC and BR is the transversal,

∴ ∠ARO = ∠B (Corresponding angles) ...          (2)

In ΔABD, R is the mid point of AB and OR || BD.

∴ O is the mid point of AD (Converse of mid point theorem)

⇒ OA = OD

In ΔAOR and ΔDOR

OA = OD (Proved)

∠AOR = ∠DOR           (90°) [∠ROD = ∠ODP (Alternate angles)  & ∠AOR = ∠ROD = 90° (linear pair)]

OR = OR (Common)

∴ ΔAOR  ≅ ΔDOR (SAS congruence criterion)

⇒ ∠ARO = ∠DRO (CPCT)

⇒ ∠DRO = ∠B (Using (2))

In quadrilateral PRQD,

∠DRO + ∠DPQ = ∠B + ( ∠A + ∠C) = ∠A + ∠B + ∠C (Using (1))

⇒ ∠DRO + ∠DPQ = 180° ( ∠A + ∠B + ∠C = 180°)

Hence, quadrilateral PRQD is a cyclic quadrilateral.

I hope that this answer may helped you!!!!!!!

hllo, 

 we know that the opposite angles are always equal in the case of parallelogram and also the sum of all angles is equal to 360 degree

by these properties of palrallelogram <i>L</i>s=Lq

so      <i> </i>Lq=70 degree

so,     by the second property 

          Lp=Lr,

so      we can write Lp in the place of Lr

so     Lp+Lq+Lr+Ls= 360

        2 Lp +70+70=360

        2 Lp=360-140

        Lp=220/2

        Lp=110

         hence Lr=110

Ans. Follow these steps to construct a histogram if only mid-values of class interval are given:

• Subtract the first mid-value from the second mid-value, let the difference is h
• divide the difference h by 2 = h/2
• Subtract h/2 from first mid-value to get lower limit of first  class-interval and add h/2 in first mid-value to get upper limit of class interval.
• repeat all the above stated steps for all remaining mid-values to get all class interval.
• Now you get all class interval.
• draw the histogram using these class interval

Practice all the imp. theorems from NCERT and egs. 

1) Equal chords subtend equal angles at their Centres. 

2) The perpendicular from the Centre of a circle to a chord bisects the chord. 

3) The angle subtended at the centre of a circle is double the angle subtended at any other point of the circle. 

4) Prove that the opposite angles of a cyclic quadrilateral are supplementary. 

5) Practice all the examples in NCERT.

Let water level rises by h meters

Since displacement of water by a person = 4 cu.m

Therefore displacement of water by 500 persons = 500 × 4 = 2000 cu.m

Now,

Volume of tank = Volume of displaced water

80 x 50 x h = 2000

h = 2000/4000 = 0.5 m = 50 cm

Ans. Suppose water level rise by h meter.

Volume of water displaced by a Man = 4m³

Total Volume Displaced By 500 Men = 4 * 500 = 2000m³ 

As Tank is in cuboid shape:

Volume of tank = Volume of water displaced

=> 80 * 40 * h = 2000

=> h = 2000/3200 = 5/8m = 62.5cm

abhi hope u recognised me 

in chapter u need to just understand that: 

two IIgm lie on same base are equal

same with triangle

triangle and IIgma= half of the area of IIgm

Ans.

External faces to be polished

= Area of six faces of cuboidal bookshelf – 3 (Area of open portion ABCD)

= 2 (110 x 25 + 25 x 85 + 85 x 110) – 3 (75 x 30)

[AB = 85 – 5 – 5 = 75 cm and AD = (1/3 x 110) – 5 – 5 – 5 – 5 = 30 cm]

= 2 (2750 + 2125 + 9350) – 3 x 2250

= 2 x 14225 – 6750

= 28450 – 6750

= 21700 cm²

Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.

= Rs. 0.20 per cm² = Rs. 0.20 x 21700 = Rs. 4340

Here, three equal five sides inner faces.

Therefore total surface area

= 3 [ 2 (30 + 75) 20 + 30 x 75] [Depth = 25 – 5 = 20 cm]

= 3 [ 2 x 105 x 20 + 2250] = 3 [ 4200 + 2250]

= 3 x 6450 = 19350 cm²

Now, cost of painting inner faces at the rate of 10 paise i.e. Rs. 0.10 per cm².

= Rs. 0.10 x 19350 = Rs. 1935

Total expenses required = Rs. 4340 + Rs. 1935 = Rs. 6275

Ans. Volume of cubical Tank = 15.625m³

Sides of cubical Tank = (15.625)^1/3 = 2.5m

As now water level is now 1.3m

Volume of water remained = area of base*current water level= 2.5*2.5*1.3= 8.125m³

Volume of water used = total volume - remaining volume = 15.625 - 8.125 = 7.500m³

Ans. Diameter of Sphere = 4.2 cm

Radius of Sphere = 4.2/2 = 2.1cm

Water displaced by sphere = volume of sphere

=> 4*22*2.1*2.1*2.1/3*7 = 38.808cm³

And. if it is asked to justify your answer or construction then you must give reason in support otherwise it is not compulsory. 

Since they are right triangles, it is a cyclic quadrilateral and the circle with diameter AB passes through all 4 points. Since ∠BAC and ∠BDC both subtend the same arc, BC, of this circle, and since the vertices of these angles are both on the circle, the two angles are equal by the theorem that says all angles subtending the same arc with vertices on the circle are equal.