A cyclic quadrilateral  is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic.

Refer this image for your question.

Ans. Let Edge of Cube = a

Area of Cube = a³

Surface Area of Cube = 6a²

According to Question, 

\(=> a^3 = 6a^2\)

 \(=> {a^3 \over a^2} = 6\)

\(=> a = 6 \)

Edge of Cube = 6 unit

NO EB will not be a median. Because:-

Intersection of medians divides any median in the ratio of 2:1 which means that E should divide AD in 2:1 but here it divided in 1:1. Hence E cannot be the intersection point which implies that EB is not a median

Join OA and OC

As perpendicular drawn from the center, bisects the chord AB and CD at P and Q respectively.
AP = PB = ½ AB = 3 cm
And CQ = QD = ½ CD = 4 cm
In right triangle OAP, by Pythagorean theorem
OA ² = OP ² + AP ² 
5 ² = OP ² + 3 ² 
⇒ OP ² = 5 ² - 3 ² 
⇒ OP ² = 25 – 9 = 16
⇒ OP = 4 cm 
In right triangle OCQ, by Pythagorean theorem
OC ² = OQ ² + CQ ² 
⇒ 5 ² = OQ ² - 4 ² = 9
⇒ OQ = 3 
∴ PQ = PO – QO 
PQ = 4 -3 
PQ = 1 cm.

In an equilateral triangle, all sides are equal. Let the lenght of side be a unit.

Then, Semi Perimeter (s) = (a + a + a)/2 = 3a/2

Usinf Heron's formula,

Area of equilateral triangle = [s{s - a}{s - b}{s - c}] ^1/2

                                           = [3a/2{(3a/2) - a}{(3a/2) - a}{(3a/2) - a}] ^1/2

                                           = [3a/2{a/2}{a/2{{a/2}] ^1/2

                                           = [3a⁴/16] ^1/2

                                           = (3)^1/2a²/4

Let take all sides be a

a=a

b=a

c=a

Put it in heron's formula

90°

Total no. Of balls =35

No.of times he hits a boundary =7

No.of times he does not hit a boundary = 35-7 =28

P(E) =28/35 = 4/5

Ans. x + y = 21 ......(1)

xy = 80   .....(2)

From (1)

x = 21-y

(21-y)×y = 80

=> 21y - y² = 80

=> y² -21y +80 = 0

=> y² -16y -5y +80=0

=> y(y-16)-5(y-16) = 0

=> (y-16)(y-5) = 0

=> y = 16 or y = 5

If y =16 then x = 5

If y =5 then x =16

Ans. Side of Cube (a) = 4 cm

Volume of Cube = \(a^3 = 4 \times 4 \times 4 = 64 cm^3\)

As the Sphere touch the side of Cube

So, Diameter of Sphere = 4 cm

Radius of Sphere(r) = 4/2 = 2 cm

Volume of Sphere = \({4\over 3} \times \pi \times r^3 = {4\over 3} \times {22\over 7}\ \times 8 = {704\over 21} = 33.5 cm^3\)

Volume of Gap = Volume of Cube - Volume of Sphere 

=> 64 - 33.5 = 30.5cm³

there is two process finding the cube root

1. prime factorisation

2. estimatiosation 

for estimation

A. divide the nuber into two group, from right most choose only three number,remaining number should other group

for example 17576,,group1=17 and group=576

B. look the group2 last number is 6.so its one digit of cuberoot should 6

C. 17 lie between 2³ and 3³,for the estimation we choose smallest number i.e 2

thus cuberoot of 17576 is 26 

similarly, cube root of 85375 is 95

sorry for text problem in previous answer

Let a and b represent the lower and upper limits of a class interval respectively

Class mark = 10 

Class size = 8

Lower limit = Class mark - Half of class size

= 10 -4 = 6

Upper limit = Class mark + Half of class size

= 10+4 = 14 

Abcd is a llgm and E and F are the midpointsTherefore, CE=AF also ce ll AF

Therefore aecf is aparallelogram.

We know in aparallelogram diagonals bisect it into 2 congruent triangles 

Therefore ar(AEF)=ar(CFE)

Plz give thump

Let AB be the chord of length 4 cm. C be its mid-point

. O be the center of the circle. Join OA [OA becomes the radius]

Given OC = 5 cm

AC = BC = 4/2 = 2 cm [The perpendicular drawn from the center to a chord bisect the chord]

OAC is a right angled triangle

OA²=OC²+AC²[byPythagorasTheorem]

OA²= 5²+2²

= 25+4

=29

OA= \(\sqrt29\)cm

You can refer this image. Just make a few changes according to the question.

First of all practice all the examples and important questions from NCERT. After that start practicing unsolved papers from U-like and assess your answers. 

Since it is the final time practice as many questions as you can. 

All the best! 

go buy it yourself

No, the question is only that what' s it written and its answer is about 102 g

Ans. In the sentence above 5 g per cm² is supposed to be the density of the material. The units of density are g/cm³. So the density is 5 g/cm³.

Mass = Density × Volume 

Radius of cylinder = 10.5 cm

Height of cylinder = 60 cm

Volume of cylinder = \(\pi r^2 h = {22\over 7} \times10.5 \times 10.5\times 60\)

= 20790cm³

So Mass of Material = 5 × 20790 = 103950g = 103.95Kg

Thanks for trying but it is not the correct answer. I have been searching the solution of this ques. for a week  and not getting sth correct.

First of all material should be 5g per cubic cm.

So now first of all we need to find the volume which is:-

pi*r²*h = (22/7)*10.5*10.5*60

= 3465 cubic cm

So total mass = 5 * total volume

= 5*3465

=17325 g

Ans.

Let X and Y be the centers of circles having radius AX = 10 cm and AY = 8 cm respectively.

Length of the chord AB = 12 cm

Suppose these circles intersect in A and B. XY intersect AB in P.

\(In \space \triangle XAY \space and \space \triangle XBY,\)

AX = BX (Radius of the circle having centre X)

XY = XY (Common)

AY = BY (Radius of the circle having centre Y)

\(\triangle XAY \cong \triangle XBY \space .......[ SSS \space criterion]\)

=> ∠AXY = ∠BXY (CPCT)

\(In \triangle XAP \space and \space \triangle XBP,\)

AX = BX (Radius of the circle)

∠AXP = ∠BXP (∠AXY = ∠BXY)

XP = XP (Common)

ΔXAP  ΔXBP (SAS congruence axiom)

=>  ∠APX = ∠BPX (CPCT)

∠APX + ∠BPX = 180° (Linear pair)

2∠APX = 180° (∠APX = ∠BPX)

=> ∠APX = 90°

=>  XP ⊥ AB

=> P is the midpoint of AB (Perpendicular from the centre to the chord, bisect the chord)

=> AP = PB = 6 cm

Let XP = x and PY = y.

In triangle AXP,

AX² = AP² + XP²

100 = 36 + x²

x²  = 64

x = 8.

XP = 8 cm.  ………..(i)

In triangle APY,

AY² = AP² + PY²

y² = 64 – 36

x²  = 28

x = √28 = 2√7.

PY = 2√7 cm.

Therefore, distance between the center of the circles = XP + PY = (8 + 2√7) cm

Ans.

 \(GIven : \space \angle EAD = 35 \)  \(and \space \angle AED = 40\)

\(To \space Find : \space \angle CBE \)

\(Proof : \space in \space \triangle EAD \)

=> \(\angle AED + \angle EAD + \angle ADE = 180 \space [angle \space sum \space property ]\)

=> \(\angle ADE = 180 -(35 +40) = 105 \)

=> \( \angle ADE + \angle ADC =180 \space [Linear \space Pair]\)

=> \(\angle ADC = 180 - 105 = 75 \)

\(Now \space \angle ADC = \angle ABC [angle\space made\space by\space same\space arc\space AC ]\)

=> \(\angle ABC = 75 \)

=> \( \angle ABC + \angle CBE =180 \space [Linear \space Pair]\)

=> \(\angle CBE = 180 - 75 = 105 \)

Refer this image for your answer.

Ans.

Given: ABCD is a quadrilateral. O is a point inside the quadrilateral ABCD.

To prove : OA + OB + OC + OD > AC + BD

Construction:  Join OA, OB, OC and OD. Also, join AC and BD

Proof:  By triangle in equality the sum of any two sides of a triangle is greater than the third side.

\(In \triangle BOD, \)

OB + OD > BD …........(1)

Similarly \(In \triangle AOC, \)

OA + OC > AC ….........(2)

Adding (1) and (2), we obtain

OB + OD + OA + OC > BD + AC

Therefore; OA + OB + OC + OD > AC + BD

Hence Proved

Ans.

Given : ABCD is a trapezium with AB || DC. AB = 50 cm and CD = 30 cm.  X and Y are Mid-point of AD and BC respectively.

Construction : Join DY and produce it to meet AB produced at P.

Proof: In \(\triangle BYP \) and \( \triangle CYD \)

\(\angle BYP = \angle CYD \)       [vertically opposite angles]

\(\angle DCY = \angle PBY\)     [Alternate opposite angles as DC || AP and BC is the transversal]

 BY = CY              [Y is the mid point of BC]

 Thus \( \triangle BYP\)  \(\cong\)  \(\triangle CYD \)    [by ASA cogence criterion]

=> DY = YP and DC = BP  [By CPCT]

Also, X is the mid-point of AD

=> \(XY \parallel AP\)

=> XY  = \({1\over 2} AP \)  [By Mid-point theorem ]

=> \(XY = {1\over 2} [AB + BP]\)

=> \(XY = {1\over 2} [AB + CD ] \)

=> \(XY = {1\over 2} [50+30] = {1\over2} \times 80 = 40cm\)

As X and Y are the mid points of AD and BC respectively.

Trapezium DCYX and ABYX are of same height, say K cm.

Now, 

\({ar(DCYX) \over ar(ABYX) } = {{{1\over2} [DC+XY]\times K} \over {{1\over2} [AB+XY]\times K}} = {[30+40] \over [50+40]} = {70\over90} = {7\over9}\)

=> ar(DCYX) = 7/9 ar(ABYX)

Hence Proved 

RD Sharma,RS agarwal

First of all practice the NCERT thoroughly. Refer to all the examples of every chapter from NCERT as they make up the most of the paper.

After that, if you want to revise the syllabus practice the model test papers from U-like. Practice as many unsolved papers you can.

All the best!

Volume of rectangular box= l X b X h

= 16 X 8 X 8 = 1024 cm³

Volume of a sphere = \( {4 \pi r^3 \over 3}\)

= \( {4 \times 22 \times2\times 2\times2 \over 3\times7}\)

= 33.52 cm³

Volume of 16 spheres = 16 X 33.52

= 536.38 cm³

Volume of liquid in box = Volume of box- Volume of 16 spheres

= 1024-536.38 = 487.62 cm³ = 488 cm³ (to the nearest integer)

Ans. Radius of Sphere Ball = 2cm

Total Nunber of Balls = 16

Volume of 16 sphere Balls = 16 *(22/7)*(4/3)*(2)³= (16 * 88 * 8)/21 

=> 536.38 cm³

Volume of Rectanglular Box = \({16\times 8\times8} = 1024cm^3\)

Volume of Preservative Liquid = Volume of Box -Volume of Sphere Balls

=> 1024 - 536.38 = 487.62cm³

1:1

I am sorry that I posted the half question.

After taking out the voulme of the bowl with outer radius,

we will  calculate the volume of the bowl with inner radius and subtract the both.

Inner radius = 4cm

Volume = \( {2 \pi R^3\over 3}\)

=( 2 X 22 X (4)³)/21

= 134 cm³

So, volume of steel used = 160.842-134

= 26.842 cm³

Ans. Inner Radius ( r) = 4 cm

Thickness of Bowl (t) = 0.25 cm 

Outer Radius (R) = r + t = 4 + 0.25 = 4.25 cm

Volume of Steel Used = Outer Volume of Bowl - Inner Volume of Bowl

=> \({2\over 3} \pi R^3 - {2\over3} \pi r^3 = {2\over 3}\pi [{R^3-r^3}]\)

=> \({2\over 3} \times {22\over 7} \times [{77-64}] = {2\over 3} \times {22\over 7} \times 13\)

=> 27.23 cm³

Thickness of hemispherical bowl = 0.25 cm 

Inner radius = 4cm

Outer radius = 4+0.25=4.25 cm 

Volume of steel used in making the bowl=

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

= (2*22*4.25*4.25*4.25)/3*7

= 160.842 cm³

=