3.146

\(\frac{22}{3}\)

1

(a, 3a)

Scottish mathematician

two different ways

↔

40°

40 cm²

60°

45 m

2(l + b)h

5

1

\(\quad a=-\frac { 179 }{ 171 } ,b=\frac { -20 }{ 171 } \)

Comparing the given expression with (x + y + z)², we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

x=4

PQ = QR
QX = QY

If equals are subtracted from equals, the remainders are also equal.
We have PQ - QX = QR - QY
PX = RY

Let x=3k, y=2k
Then, x+y=3k+2k=180^o
(Angles on the same side of transversal)
5k=180^o
k=36⁰
x=3k=108^o
y=2k=72^o
Thus, \(\angle z=\angle x=108^o\) (Alternate interior angles)

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

36, 60

84 cm²

17.6 m³

We draw the bar graph of this data in the following steps. Note that the unit in the second column is thousand rupees. So, ‘4’ against ‘grocery’ means Rs. 4000.
1. We represent the Heads (variable) on the horizontal axis choosing any scale, since the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one Head be represented by one unit.
2. We represent the expenditure (value) on the vertical axis. Since the maximum expenditure is Rs. 5000, we can choose the scale as 1 unit = Rs.1000.
3. To represent our first Head, i.e., grocery, we draw a rectangular bar with width 1 unit and height 4 units.
4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

\((i)\frac { 53 }{ 100 }\) 
\((ii)\frac { 3 }{ 25 } \)

\(b=\frac { 1 }{ 2+\sqrt { 5 } } \)
\(=\frac { 1 }{ 2+\sqrt { 5 } } \times \frac { 2-\sqrt { 5 } }{ 2-\sqrt { 5 } } \)
\(=\frac { 2-\sqrt { 5 } }{ -1 } =-2+\sqrt { 5 } \)
\({ a }^{ 2 }={ \left( 2+\sqrt { 5 } \right) }^{ 2 }=9+4\sqrt { 5 } \)
\({ b }^{ 2 }={ \left( -2+\sqrt { 5 } \right) }^{ 2 }=9-4\sqrt { 5 } \)
\({ a }^{ 2 }+{ b }^{ 2 }=9+4\sqrt { 5 } +9-4\sqrt { 5 } \)
= 18

Term with the highest power of x = 5x³
Exponent of x in this term           = 3
Therefore Degree of this polynomial = 3

(0,0); (-p,0); (5,0); (0,-q)

x=4y
\(\Rightarrow\ \ y={x\over 4}\)
Put x=0, we get \(y={0\over4}=0\)
Put x=4, we get \(y={4\over4}=1\)
Put x=-4, we get \(y={-4\over 4}=-1\)
Put x=2, we get \(y={2\over4}={1\over 2}\)
Four solutions are (0,0), (4, 1), (-4, -1) and \(\left(2,{1\over 2}\right)\)

Here, \(\angle1=\angle3, \angle2=\angle4\ and\ \angle3=\angle4\), Euclid's first axiom says, the things which are equal to same things are equal to one another.

So, \(\angle 1=\angle 2\)

\(OP\bot AB\)
\(\Rightarrow \angle POA=90^o\)
Let \(\angle POQ=a\)
\(\therefore \angle QOR=3a\)
\(\angle ROA=5a\)
\(\Rightarrow\) a+3a+5a=90^o
\(\Rightarrow\) 9a=90^o
\(\Rightarrow\)a=10^o
\(\therefore\)x=10^o
and y=3x10^o=30^o
z=5x10^o=50^o

Construction: Join OA and OB.
∵ OA = OB = AB              I Given
∴ \(\Delta AOB\) is equilateral.
∴ \(\angle AOB=60°\)

\(\angle ACB=\frac { 1 }{ 2 } \angle AOB\)
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
\(=\frac { 1 }{ 2 } \times 60°=30°\)
Now, ∵ ADBC is a cyclic quadrilateral.
∴ \(\angle ADB+\angle ACB=180°\)
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°
\(\Rightarrow \ \angle ADB+30°=180°\)
\(\\ \Rightarrow \angle ADB=180°-30°\)
\(\\ \Rightarrow angle ADB=150°.\) 

Area of paper of shade I \(2\times \left( \frac { 1 }{ 2 } \times 16\times 16 \right) =256\)cm²
Similarly, Area of paper of shade II = 256 cm²
For area of paper of shade III
a = 8 cm, b = 6 cm, c = 6cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 8+6+6 }{ 2 } =10\) cm
\(\therefore \) Area of paper of shade III = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-8)(10-6)(10-6) } \)
 \(=\sqrt { (10)(2)(4)(4) } =8\sqrt { 5 } \) 
= 17.89 cm²

Let the radius of the base of the cylinder be r cm.
= 5 cm
Lateral surface =94.2 cm²
\(\Rightarrow 2\pi rh=94.2\)
\(\Rightarrow 2\times 3.14 \times r\times 5=94.2\)
\(\Rightarrow r=\frac{94.2}{2\times3.14\times5}\)
\(\Rightarrow r=\frac{94.2}{31.4}\)
\(\Rightarrow r=3cm\)
(ii) r = 3cm,  h = 5 cm
ஃ Volume of the cylinder =\(\pi r^2h\)
= 3.14 x (3)² x 5 =141.3 cm³

Modified Continues Distribution

(ii) Frequency Polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the original interval 145-153 (or modified interval 144.5-153.5).

\((i)\frac { 21 }{ 40 }\) 
\((ii)\frac { 29 }{ 40 }\) 
\((iii)\frac { 29 }{ 40 } \)

\(\frac { 1 }{ 3+\sqrt { 7 } } =\frac { 1 }{ 3+\sqrt { 7 } } \times \frac { 3-\sqrt { 7 } }{ 3-\sqrt { 7 } } =\frac { 3-\sqrt { 7 } }{ 9-7 } =\frac { 3-\sqrt { 7 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } =\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } \times \frac { \sqrt { 7 } -\sqrt { 5 } }{ \sqrt { 7 } -\sqrt { 5 } } =\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } \times \frac { \sqrt { 5 } -\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } =\frac { \sqrt { 5 } -\sqrt { 3 } }{ 5-3 } \)
\(=\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 3 } +1 } =\frac { 1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } =\frac { \sqrt { 3 } -1 }{ 3-1 } =\frac { \sqrt { 3 } -1 }{ 2 } \)
\(LHS=\frac { 3-\sqrt { 7 } }{ 2 } +\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } +\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } +\frac { \sqrt { 3 } -1 }{ 2 } \)
\(=\frac { 3-1 }{ 2 } =\frac { 2 }{ 2 } =1=RHS\)

Factor theorem - statement
Let p(x)=x³-3x²-x+3
the factors of the constant term 3 are \(\pm \)1, \(\pm \)3
p(1)=1³-3(1)²-1+3=0
(x-1) is a factor
p(-1)=(-1)³-3(-1)²-(-1)+3=0
(x+1) is a factor
p(3) = 3³-3(3)²-3+3 = 0
(x-3) is a factor
(x-1)(x+1)(x-3) are the factors of p(x)

The figure formed is a square.
Its all the sides are of equal length.

Let us represent time (in hour) by x and distance (in km) by y. Then, we have y =80x.
Table of solution

We plot the points (1, 80), (2, 160), (3, 240) and (4, 320) on a graph paper and join these points by a ruler to get the line which is the graph of the equation y = 80x.

We have
AB = BC
\(\Rightarrow \) AB - BX = BC - BX
|If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
AB - BX = BC - BY  \(|\)\( \because\)  BX = BY
AB - BX coincides with AX;
BC - BY coincides with CY
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

Given l,m,n are three lines such that l || m and  n \(\bot \)m

\(\therefore \)l || m and n is a transversal
\(\therefore \) \(\angle \)1=\(\angle \)2 |Corresponding angles
But \(\angle \)1=\(90^{ 0 }\)
\(\therefore \) \(\angle \) 2= \(90^{ 0 }\)
\(\Rightarrow \) n \(\bot \) m

In \(\triangle\)AMB and \(\triangle\)PNQ,
AB = PQ   (Given)
AM = PN   (Given)
\(\angle\)1 = \(\angle\)2 = 90°
(AM\(\bot \)BC & PN \(\bot \)QR)
\(\Rightarrow\) \( \triangle AMP\cong \triangle PNQ\) (By RHS)
\(\Rightarrow\) \(\angle\)3 =\(\angle\)4 (By c.p.c.t.) 
Similarly, \(\triangle AMC\cong \triangle PNR\)
\(\Rightarrow\) \(\angle\)5 = \(\angle\)6
(In congruent triangles, corresponding angles are equal)
\(\therefore\) In \(\triangle\)ABC and \(\triangle\)PQR,
AB = PQ  (Given)
AC = PR (Given)
\(\angle\)A =\(\angle\)P
(\(\angle\)3 + \(\angle\)5 = \(\angle\)4 + \(\angle\)6) (proved)
\(\triangle ABC\cong \triangle PQR\)     (By SAS)

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA = 90°
⇒ ㄥLON = 90°
Similarly, ㄥOLM =ㄥLMN =ㄥMNO = 90°

ஃ A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is rectangle.

Given: Diameter AB and a chord AC have a common end point A. AB = 20 cm and AC = 12 cm.
To determine: OD
Determination: ∵ OD丄AC
∴ \(AD=DC=\frac { 1 }{ 2 } AC=\frac { 1 }{ 2 } \times 12=6\quad cm\)
| ∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
\(OA=OB=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } \times 20=10\quad cm\)
In right triangle ODA,
OA² = OD² + AD² I By Pythagoras Theorem
⇒ (10)²=OD²+(6)²
⇒ OD=8 cm
Hence, AC is 8 cm far from the centre of the circle.

Area =  \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)
 \(\Rightarrow 60=\frac { 24 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ (24) }^{ 2 } } \)
\(\Rightarrow 10=\sqrt { 4{ b }^{ 2 }-576 } \)
\(\Rightarrow 100=4{ b }^{ 2 }-576\)      | Squaring
\(\Rightarrow 4{ b }^{ 2 }=676\)
\(\\ \Rightarrow { b }^{ 2 }=\frac { 676 }{ 4 } =169\)
\(\Rightarrow b=\sqrt { 169 } \) = 13 cm
\(\therefore\) Perimeter = a + b + b
= 24 + 13 + 13 = 50 cm

Total number of subjects = 5
(i) Number of subjects in which the student gets a first class = 4
Probability that the students gets a first class \(=\frac { 4 }{ 5 } \)
(ii) Number of subjects in which the student gets a distinction = 2
Probability that the student gets a distinction  \(=\frac { 2 }{ 5 } \)
(iii) Number of subjects in which the student gets marks between 70% and 80% = 1
Probability that the students gets marks between 70% and 80% \(=\frac { 1 }{ 5 } \) 

\(x=3-2\sqrt { 2 } \Rightarrow \frac { 1 }{ x } =3+2\sqrt { 2 } \)
\({ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) }^{ 2 }=8\)
\(\Rightarrow \sqrt { x } +\frac { 1 }{ \sqrt { x } } =\pm 2\sqrt { 2 } \)

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio.

(i) As EB II DL and ED II BL.
Therefore EBLD is a parallelogram.
ஃ BL = ED
=\(\frac{1}{2}\)BC=CL ...(i)
Now in triangles DCL and FBL, we have
CL=BL from (i)
ㄥDLC= ㄥFLB 
ㄥCDL=ㄥBFL
ΔCDL≡ΔBFL
CD= BF
and DL= FL
Now, BF = DC = AB
⇒ 2AB = 2DC
⇒ AF = 2DC
(ii) ∵ DL=FL
⇒ DF = 2DL

(i) Let us assume that A, Band C are the position of Priyanka, Sania and David respectively on the boundary of circular park with centre O.
Draw AD\(\bot\) BC
Since the centre of the circle coincides with the centroid of the equilateral \(\triangle\) ABC
\(\therefore\) Radius of circumscribed circle = \(\frac{2}{3}\) AD
\(\Rightarrow 20=\frac { 2 }{ 3 } AD\)
\(\Rightarrow AD=20\times \frac { 2 }{ 3 } \)
\(\Rightarrow AD=30m\)
Now, AD\(\bot\)BC, and let AB=BC=CA=x
\(\Rightarrow BD=CD=\frac { 1 }{ 2 } BC=\frac { x }{ 2 } \)

In rt. \(\triangle\)BDA, D=90⁰
By Pythagoras Theorem, we have
AB²=BD²+AD²
\(\Rightarrow { x }^{ 2 }={ \left( \frac { x }{ 2 } \right) }^{ 2 }+{ (30) }^{ 2 }\)
\(\Rightarrow{ x }^{ 2 }-{ \frac { { x }^{ 2 } }{ 4 } }=90\)
\(\Rightarrow \frac { 3 }{ 4 } { x }^{ 2 }=90\)
\(\Rightarrow { x }^{ 2 }=900\times \frac { 4 }{ 3 } \)
\(\Rightarrow { x }^{ 2 }=1200\)
\(\therefore x=\sqrt { 1200 } =20\sqrt { 3 } \)
Hence, the distance between each of them is \(20\sqrt { 3 } .\)
(ii) Properties of the circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.