In the given figure, ∠AEF=∠AFE and E is the mid point of . . .

Question: In the given figure, ∠AEF=∠AFE and E is the mid point of CA. Prove that BD/CD = BF/CE.

Doubt by Suraj 

Solution : 

Given : ∠AEF=∠AFE i.e. ∠1=∠2

E is the mid point of CA i.e AE=CE

To Prove : BD/CD = BF/CE

Construction : Take point G on AB such that CG||DF

Proof : 

∠1=∠2 (Given) 
AE=AF (Sides opposite to equal angles of a triangle are equal) — (1) 

Also, 
AE=CE (Given) — (2) 

In ΔAGC
E is the mid point of AC (Given)
CG||EF [∵CG||DF]
⇒F is the mid point of GA
(By Mid Point Theorem)
 AF=GF — (3)

In ΔBDF
CG||DF (By Const.) 
BC/CD = BG/GF (By BPT)
Adding 1 both sides
[BC/CD] + 1 = [BG/GF] +1
[BC+CD]/CD = [BG+GF]/GF
BD/CD = BF/GF
BD/CD = BF/CE

[Using equation (1), (2) and (3)
GF=AF=AE=CE]

Hence Proved.

A solid iron pole consists of a solid cylinder of height 200 cm . . .

Question : A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm, which is surmounted by another cylinder of height 50 cm and radius 7cm. Find the mass of the pole, given that 1cm³ of iron has approximately 8g mass.

CBSE 2024

Solution : 

Dimensions of Bigger Cylinder 

H = 200 cm 
D = 28 cm
R = D/2 = 28/2 = 14 cm

Dimensions of smaller Cylinder
h = 50 cm 
r = 7 cm 

Volume of the Iron pole  = πR²H+r²h
= π[R²H+r²h]
= 22/7 [(14)²×200+(7)²×50]
= (22/7)×(7)²[2²×200+1²×50]
= 22×7[4×200+50]
= 154×[800+50]
= 154×850
= 130900 cm³

Mass of 1cm³ iron = 8g
Mass of 130900 cm³ iron = 8×130900
= 1047200 g
= [1047200/1000] kg
= 1047.2 kg

Tamper-proof tetra-packed milk guarantees . . .

Case Study Based Question on Surface Area and Volume

Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals.

500 mL milk is packed in a cuboidal container of dimensions 15cm×8cm×5cm. These milk packets are then packed in cuboidal cartons of dimensions 30cm×32cm×15cm.

Based on the above given information, answer the following questions : 

(i) Find the volume of the cuboidal carton. [1 Mark]

(ii) (a) Find the total surface area of a milk packet. [2 Marks]

OR

(b) How many milk packets can be filled in a carton? [2 Marks]

(iii) How much milk can the cup (as shown in figure) hold? [1 Mark]

CBSE 2024

Solution : 

Dimensions of cuboidal container 
l = 15 cm 
b = 8 cm 
h = 5 cm

Dimensions of cuboidal cartons 
L = 30 cm 
B = 32 cm 
H = 15 cm

Dimensions of cylindrical cup 
r = 5 cm
h'= 7cm

(i) Volume of the cuboidal carton
= LBH
= 30×32×15
= 14400 cm³

(ii)
(a) Total Surface Area of a milk packet 
=2(lb+bh+hl)
=2[(15×8)+(8×5)+(5×15)]
=2[120+40+75]
=2[235]
=470 cm²

OR

(b) No. of milk packets filled in a carton 
= Volume of carton / Volume of milk packet 
= LBH/lbh
= (30×32×15) / (15×8×5)
= 2×4×3
= 24

(iii) Capacity of the cup = Volume of the cylinder
= πr²h
=(22/7)×5×5×7
= 22×25
= 550 cm³

A solid toy is in the form of a hemisphere surmounted by a . . .

Question : A solid toy is in the form of a hemisphere surmounted by a right circular cone. Ratio of the radius of the cone to its slant height is 3:5. If the volume of the toy is 240π cm³, then find the total height of the toy.

CBSE 2024 (July Session)

Solution : 
For Right Circular Cone
r:l=3:5
Let
r=3x
l=5x

Height of the cone
h=√[l²-r²]
h=√[(5x)²-(3x)²]
h=√[25x²-9x²]
h=√[16x²]
h=4x

Also the radius of hemisphere would be equal to that of cone. 

Volume of the toy= Volume of cone + Volume
f hemisphere

240π = (1/3)πr²h + (2/3)πr³
240π = (1/3)πr²[h+2r]
240=(1/3)(3x)²[4x+2(3x)]
240=3x²[4x+6x]
240=3x²[10x]
240=30x³
240=30x³
240/30=x³
8=x³
x³=2³
x=2

Total height of the toy
= h+r
= 4x+3x
=7x
=7(2) [∵x=2]
=14 cm

Amita buys some books for Rs 1920 . . .

Question : Amita buys some books for ₹1920. If she had bought 4 more books for the same

amount each book would cost her ₹ 24 less. How many books did she buy? What was the initial price of one book?

Doubt by Yana

Solution : 

Let the original number of books purchased be x. 

Cost of each book = 1920/x

Number of books when she bought 4 more books = x+4

New cost of each book = 1920/(x+4)

ATQ 
1920/x-1920/(x+4)=24
1920[1/x-1/(x+4)]=24
[(x+4)-x]/[x(x+4)]=24/1920
4/[x²+4x]=1/80
x²+4x=320
x²-4x-320=0
x²+(20-16)x-320=0
x²+20x-16x-320=0
x(x+20)-16(x+20)=0
(x-16)(x+20)=0
x-16=0 OR x+20=0
x=+16 OR x=-20
Number of books can't be negative. 
So, x=-20 (Rejected)
x=16
Hence, she bought 16 Books.

Initial Price of each Book = 1920/x = 1920/16 

=₹120

If the sum of the zeroes of 5x²+(p+q+r)x+pqr is zero . . .

Question : If the sum of the zeroes of 5x²+(p+q+r)x+pqr is zero, then find p³+q³+r³.

Doubt by Suraj

Solution : 

5x²+(p+q+r)x+pqr
Here
a=5
b=p+q+r
c=pqr

α+β=-b/a
α+β=-(p+q+r)/5
But α+β=0 (Given)
0=-(p+q+r)/5
0×5=-(p+q+r)
p+q+r=0

Using Identity
If a+b+c=0 then a³+b³+c³=3abc

Here 
p+q+r=0
So, 
p³+q³+r³ = 3pqr

The word circus has the same root as circle. In a closed circular . . .

Case Study Based Question on Surface Area and Volume

The word circus has the same root as circle. In a closed circular area various entertainment acts including human skill and animal training are presented before the crowd. 

A circus tent is cylindrical upto a height of 8m and conical above it. The diameter of the base is 28 m and total height of the tent is 18.5 m.

Based on the above, answer the following questions : 

(i) Find the slant height of the conical part. (1 Mark)

(ii) Determine the floor area of the tent. (1 Mark)

(iii) (a) Find the area of the cloth used for making tent. 

OR

(b) Find total volume of air inside an empty tent.

CBSE 2024 

Solution : 

Height of cylinder (H) = 8 m
Diameter of the base (d) = 28 m 
Radius of the base (r) = 28/2 = 14 m
Total Height of the tent = 18.5 m
Height of cone (h) = 18.5-8 = 10.5 m

(i) Slant height of the conical part
l=√[r²+h²] 
l=√(14)²+(10.5)²
l=√[196+(21/2)²]
l=√[196+(441/4)]
l=√[784+441]/4
l=√1225/4
l=35/2 m
l=17.5 m

(ii) Floor area of the tent = πr²
= (22/7)×14×14
=44×14
=616 m²

(iii) (a) 
Area of the cloth used for making tent
= CSA of cylinder + CSA of Cone
=2πrH+πrl
=πr[2H+l]
=(22/7)(14)[2(8)+17.5]
=22×2[16+17.5]
=44[33.5]
=1474 m²

OR 

(b) Volume of the air inside the tent
=Volume of the cylinder + Volume of the cone
= πr²H+1/3πr²h
=πr²[H+h/3]
=(22/7)×(14×14)[8+10.5/3]
=22×2×14[8+3.5]
=44×14×11.5
=7084 m²

To keep the lawn green and cool, Sadhna uses . . .

Question : To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area. The diagram below shows the circular areas covered by two sprinklers:

Two circles touch externally. The sum of their areas is 130π sq. m and the distance between their centres is 14 m. 

Based on the above information, answer the following questions :

(i) Obtain a quadratic equation involving R and r from above. [1 Mark]

(ii) Write a quadratic equation involving only r. [1 Mark]

(iii) (a) Find the radius r and the corresponding area irrigated. [2 Marks]

OR

(b) Find the radius R and the corresponding area irrigated. [2 Marks]

Solution :

(i) Circular Area covered by First Sprinkler + Circular Area covered by Second Sprinkler = 130π m² (Given) 
πR²+πr²=130π
π(R²+r²)=130π
R²+r²=130 — (1)

which is the required quadratic equation involving R and r. 

(ii) R+r=14 m (Given)
R=14-r — (2)
Substituting in equation (1) 

 R²+r²=130
(14-r)²+r²=130
(14)²+(r)²-2(14)(r)+r²=130
196+r²-28r+r²=130
2r²-28r+196=130
2r²-28r+196-130=0
2r²-28r+66=0
2[r²-14r+33]=0
r²-14r+33=0/2
r²-14r+33=0

which is the required quadratic equation involving only r.

(iii) 

(a) r²-14r+33=0

Solving by Factorisation Method (Splitting the Middle Term)

r²-(11+3)r+33=0
r²-11r-3r+33=0
r(r-11)-3(r-11)=0
(r-11)(r-3)=0
r-11=0
r=11 

but we know R+r=14 m and r is smaller radius so r≠11

r-3=0
r=3 m

Hence, the required value of r is 3 m 
Corresponding Area irrigated
= πr²
= π(3)²
=9π m²

(b) We know, 
r = 3 m 
Also
R+r=14
R+3=14
R=14-3
R=11 m
Corresponding Area Irrigated 
=πR²
=π(11)²
=π(121)
=121π m²

A backyard is in the shape of a triangle ABC with right angle . . .

Question : A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = x m. 

Based on the above information, answer the following questions : 

(i) Find the length of AR in terms of x. [1 Marks]

(ii) Write the type of quadrilateral BQOR. [1 Marks]

(iii) 

(a) Find the length PC in terms of x and hence find the value of x. [2 Marks]

OR

(b) Find x and hence find the radius r of the circle. [2 Marks]

CBSE Delhi 2024 (30/1/1)

Solution : 

This is a case study based question from circles.

In Δ ABC
∠ABC=90°
AB=7 m
BC=15 m

AP=x m

(i) Find the length of AR in terms of x. 
Ans : 

AP=x m (Given)

AP=AR (Length of tangent drawn from an external point to a circle are equal in length)
∴ AR = x m

(ii) Write the type of quadrilateral BQOR.

Ans :

In BQOR
BR=BQ (Length of tangent drawn from an external point to a circle are equal in length)

OR=OQ (Radii of the same circle)

∠ABC=90° (Given)
∠BRO=∠BQO=90° (Tangent at any point of the circle is perpendicular to the radius through the point of contact)

∠ROQ=90° (By ASP of a quadrilateral)

∵ All sides are equal and all angles are of 90°.

∴ BQOR must be a square.

(iii) 

(a) Find the length PC in terms of x and hence find the value of x.

Ans : 
We know, length of tangent drawn from an external point to a circle are equal in length

AP=AR = x 
BR=BQ = (7-x) cm
PC=QC
= 15-(7-x)
= (15-7+x) cm
= (8+x) cm

∴ PC=(8+x) cm

In Rt. ΔABC

AC²=AB²+BC² (By Pythagoras Theorem)
AC²=(7)²+(15)²
AC²=49+225
AC²=274
AC=√274
AC=√274 cm 
AP+PC=√274
x+8+x=√274
2x+8=√274
2x=(√274)-8
x=[(√274)-8]/2 cm

OR

(b) Find x and hence find the radius r of the circle. 

Ans :
x=[(√274)-8]/2 cm

Radius (r)
= BQ=BR 

= (7-x) cm 
= 7-{(√274)-8]/2} cm

A rectangular floor area can be completely tiled with . . .

Question : A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.

(i) Assuming the original length of each side be x units, make a quadratic equation from the above information. [1 Mark]

(ii) Write the corresponding quadratic equation in standard form. [1 Mark]

(iii) (a) Find the value of x, the length of side of a tile by factorisation. [2 Marks]

OR

(b) Solve the quadratic equation for x, using quadratic formula. [2 Marks]

CBSE Outside Delhi 2024 (30/3/1)

Solution : 

This is a case study question from Quadratic Equations. 

Here, it should be understood that the area covered by 200 square shaped tiles of each side 'x' units must be equal to the area covered by 128 different square shaped tiles of each area (x+1) units.

(i) Assuming the original length of each side be x units, make a quadratic equation from the above information.

Ans : 

200×Area of each tiles of side x units = 128×Area of each tiles of side (x+1) units.
200×(x)²=128(x+1)²

(ii) Write the corresponding quadratic equation in standard form.

Ans : 

200×(x)²=128(x+1)²
Dividing both sides by 8
25x²=16(x²+2x+1)
25x²=16x²+32x+16
25x²-16x²-32x-16=0
9x²-32x-16=0
which is the required quadratic equation. 

(iii) (a) Find the value of x, the length of side of a tile by factorisation. 
Ans : 

9x²-32x-16=0
9x²-(36-4)x-16=0
[By Splitting the middle term]
9x²-36x+4x-16=0
9x(x-4)+4(x-4)=0
(x-4)=(9x+4)=0
(x-4)=0 or (9x+4)=0
x=4 or 9x=-4
x=4 or x=-4/9
But Side of a tile can't be negative. So, x≠0.
Hence, x=4

OR

(b) Solve the quadratic equation for x, using quadratic formula.

Ans : 

9x²-32x-16=0

Here

a=9
b=-32
c=-16
D=b²-4ac
D=(-32)²-4(9)(-16)
D=1024+576
D=1600

D>0
∴ Roots are real and equal. 

Using Quadratic Formula

x=[-b±√D]/2a
x=[-(-32)±√1600]/2(9)
x=[32±40]/18
x=[32±40]/18
x=[32+40]/18 OR x=[32-40]/18
x=[72]/18 OR x=[-8]/18
x=4 OR x=-4/9

But Side of a tile can't be negative. So, x≠0.
Hence, x=4