By triangle inequality property,
In \(\triangle\) ABC, AB + BC>AC   ...(1)
In \(\triangle\)BCD, BC + CD > BD  ...(2)
In \(\triangle\)CDA, CD + DA >AC   ...(3)
In \(\triangle\)DAB, DA +AB > BD   ...(4)
using the fact that sum of two sides of a triangle  is greater than the third side​​​​​​​

Adding (1), (2), (3) and (4), we get
AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD
\(\Rightarrow\) 2(AB + BC + CD + DA) > 2(AC + BD)
Hence, Perimeter > Sum of its diagonals.

\(x+\frac { 1 }{ x } =5\)
On Squaring both sides, we get
\({ \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ 5 }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+{ \left( \frac { 1 }{ x } \right) }^{ 2 }+2\times x+\frac { 1 }{ x } =25\)
[\(\because\) (a+b)²=a²+b²+2ab)]
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=25\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =25-2\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =23\) 

Total number of families = 1500
(i) at most one girl means 0 girl or 1 girl.
Number of families which have at most one girl
= Number of families which have 0 girl + Number of families which have 1 girl.
= 211 + 814 = 1015
Probability that it has at most one girl=\(\frac { 1015 }{ 1500 } =\frac { 203 }{ 300 } \)
(ii) at least one girl means 1 girls or 2 girls.
 Number of families which have at least one girl
= Number of families which have I girl + Number of families which have 2 girls.
= 814 + 475 = 1289
 Probability that it has at least one girl=\(\frac { 1289 }{ 1500 } \) 

Mean
\(=\frac {76+44+45+87+71+72+82+83+41+32+75+32+46+78+17+70+84+12+77+74} {20}\)
\(=\frac {1198}{20}=59.9\)

For \(\Delta \) ABC
a = 34 cm, b = 42 cm, c = 20 cm

\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 34+42+20 }{ 2 } =48\) cm
\(\therefore \)  Area of  \(\Delta \) ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 48(48-34)(48-42)(48-20) } \)
\(=\sqrt { 48(14)(6)(28) } =\quad 336\) cm²
\(\therefore \) Area of parallelogram ABCD = 2 area of triangle ABC
= 2\(\times \) 336 cm² = 672 cm²

(i) Steps of Construction
1. Draw the base AB = 5.8 ern.
2. At the point A, make an angle, say XAB=60°.
3. Cut a line segment AD equal to BC + CA = 8.4 cm from the ray BX.
4.Join DB
5.Make an angle DBY equal to ADB.
6.Let BY intersect AD at C. Then, ABC is the required triangle.

(ii) By measurement, AC = 3.4 cm
(iii) By measurement, BC = 5cm
(iv) Yes! AC + B C =3.4 + 5 = 8.4cm
(v) By measurement, ㄥACB = 84°
Meenu is right.
The value 'exactness' is depicted by Meenu's statement.

Given: ABC is an isosceles triangles in which AB = AC. A circle has been drawn with AB as diameter to intersect the third side BC at D.
To Prove: This circle bisects the third side, i.e., BD=DC
Construction: Join AD.

Proof: In \(\Delta ADB\) and \(\Delta ADC\)
AB=AC | Given    ............(1)
\(\angle ADB=90°\)     I Angle in a semicircle is 90°
\(\angle ADB+\angle ADC=180°\)
| Linear pair axiom
∴ \(\angle ADB=\angle ADC=90°\)  .........(2)
AD=AD     ..........(3) | Common
In view of (1), (2), and (3),
 \(\Delta ADB\cong \Delta ADC\) | R.H.S congruence rule
∴BD=CD | CPCT
⇒ D is the mid-point of BC.

Given: BD is one of the diagonals of a quadrilateral ABCD. AM and CN are perpendiculars from A and C respectively on BD.
To Prove: ar(quad. ABCD)
 \(\frac { 1 }{ 2 } BD.(AM+CN)\)
Proof: ar(quad. ABCD)
 
\(ar(\Delta ABD)+ar(\Delta BCD)\)
\(\\ =\frac { BD.AM }{ 2 } +\frac { BD.CN }{ 2 } \)
Area of a triangle = \(\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
 \(\frac { 1 }{ 2 } BD(AM+CN)\)

\(\because \) ED is the line
\(\therefore \angle 4+\angle 2=180^{ 0 }\)
|Linear Pair Axiom
and \(\angle 5+\angle 3=180^{ 0 }\)
| Linear Pair Axiom

\(\therefore \angle 4+\angle 5+\angle 2\angle 3=360^{ 0 }\)
Now, \(\therefore \angle 1+\angle 2+\angle 3=180^{ 0 }\)
|Sum of the angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow \angle 2+\angle 3=180^{ 0 }-\angle 1\)   .....(5)
From (3) and (5)
\(\Rightarrow \angle 4+\angle 5=(180^{ 0 }-\angle 1)=360^{ 0 }\)
\(\Rightarrow \angle 4+\angle 5=180^{ 0 }-\angle 1\)
\(\Rightarrow \angle 4+\angle 5>180^{ 0 }\) 

We have,
x - 15 = 25
\(\Rightarrow \)  x - I5 + 15 = 25 + 15  | If equals are added to equals, the wholesare equal (Euclid's Axiom (ii))
\(\Rightarrow \) x = 40

if(-m, 3) is a solution of the equation
4x+9y-3=0, then
4(-m)+9(3)-3=0
⇒  -4m+27-3=0
⇒ -4m+24=0
⇒  4m=24
⇒ \(m={24\over 4}=6\)

The graph we get is a straight line.

\(\frac { 3 }{ 7 } =\frac { 30 }{ 70 } \)
\(\\ \frac { 5 }{ 7 } =\frac { 50 }{ 70 } \)
30<31<32<33<34<35
\(\frac { 30 }{ 70 } <\frac { 31 }{ 70 } <\frac { 32 }{ 70 } <\frac { 33 }{ 70 } <\frac { 34 }{ 70 } <\frac { 50 }{ 70 } \)
So four rational numbers between \(\frac { 3 }{ 7 } \) and \(\frac { 5 }{ 7 } \)
\(\frac { 31 }{ 70 } ,\frac { 32 }{ 70 } ,\frac { 33 }{ 70 } \) and \(\frac { 34 }{ 70 } \)
\(\frac { 31 }{ 70 } ,\frac { 16 }{ 35 } ,\frac { 33 }{ 70 } \)and \(\frac { 17 }{ 35 } \)

In \(\Delta\)PCB,
\(\angle\)PCB + \(\angle\)PBC = \(\angle\)APB (exterior angle of a \(\Delta\) is equal to the sum of two opposite angles)
\(\angle\)PCB + 15° = 120°
\(\therefore\) \(\angle\)PCB = 105°
or, \(\angle\)ACB = 105°  
\(\Rightarrow\)\(\angle\)ADB = \(\angle\)ACB = 105°  [Angle in same sagment]

Let the probability of winning a game = p
and probability of losing a game = q
we know that p + q = 1...(i)
According to questions,
\(p=2q-\frac{1}{3}\)
\(\Rightarrow \) 6q - 3p = 1...(ii)
On solving (i) and (ii), we get
\(q=\frac{4}{9}\ and\ p=\frac{5}{9}\)
\(\therefore \) Probability of winning game \(=\frac{5}{9}\)

\(\because \angle A=\angle B+\angle C\)
\(\therefore \angle A+\angle B+\angle C=180^o\Rightarrow 2\angle A=180^o\)
\(\Rightarrow \angle A=90^o\)

A diagonal of a parallelogram divides it into two triangles of equal area
\(\therefore\) ar(\(\Delta\)ABD)=  ar(\(\Delta\)BCD)
Since, given, ar(\(\Delta\)ABD) = 8 cm²
\(\therefore\) ar(\(\Delta\)BCD) = 8 cm².

Let the measure of the angles be x, 2.x, 3x and 4x then,
x + 2x + 3x + 4x = 360°
⇒ x = 36°
ஃ Angles of quadrilateral are 36°, 72°, 108°,144°

Let line segment \(\overline{AB}\) has 2 mid-points, say X and Y
then, \(\frac{AB}{2}=AX\ and\ \frac{AB}{2}=AY\)
\(\therefore\) AX = AY
X and Y coincides

(i) 15
(ii) 30-35
(iii) 47.5

Since the area of the canvas = 551 m² and area of the canvas lost in wastage is 1 m², therefore the area of canvas available for making the tent is (551 – 1) m² = 550 m².
Now, the surface area of the tent = 550 m² and the required base radius of the conical tent = 7 m
Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!).
Therefore, curved surface area of tent = 550 m².
That is, \(\pi\)rl = 550
or, \(\frac{22}{7} \times 7 \times l=550\)
or, \(l=3 \frac{550}{22} \mathrm{~m}=25 \mathrm{~m}\)
Now, l² = r² + h²
Therefore, \(h=\sqrt{l^{2}-r^{2}}=\sqrt{25^{2}-7^{2}} \mathrm{~m}=\sqrt{625-49} \mathrm{~m}=\sqrt{576} \mathrm{~m}\)
= 24 m
So, the volume of the conical tent \(=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \mathrm{~m}^{3}=1232 \mathrm{~m}^{3}\)

12 cm²

0x+5y-2=0

(i)

(ii) Parallelogram

61, -143

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

Required probability=\(\frac { 150 }{ 600 } =\frac { 1 }{ 4 } \)

Mean \(\frac {2+3+5+7+11}{5}=5.6\)

v = 5 \(\times\) (6 \(\times\) 2 \(\times\) 1.5)

See Hero's formula.

Theorem

Evident.

no of angle of a rhombus is 90⁰

ㄥA+ㄥB+ㄥC=180⁰
⇒ 35⁰+65⁰+ㄥC=180⁰
⇒ ㄥC=180⁰
∵ ㄥC > ㄥA
∵ AB > BC
∵ ㄥC > ㄥB
∴ AB > AC
In view of (1) and (2), AB is the longest side.

\(\frac { x }{ y } =\frac { 4 }{ 3 } =k\)

(4, -1) satisfies x-2y=6

\(x+p=0\ \Rightarrow \ x=-p\)
By factor theorem,
\((-p)^2+p(-p)+3-p=0 \ \Rightarrow \ p=3\)

\(\frac { { \left( { x }^{ \frac { 1 }{ 2 } } \right) }^{ \frac { -2 }{ 3 } }{ \left( { y }^{ 4 } \right) }^{ \frac { 1 }{ 2 } } }{ { x }^{ \frac { -1 }{ 4 } }{ y }^{ \frac { -1 }{ 4 } } } ={ x }^{ \frac { -1 }{ 3 } }.{ y }^{ 2 }.{ x }^{ \frac { 1 }{ 4 } }.{ y }^{ \frac { 1 }{ 4 } }\)
\({ x }^{ \frac { -1 }{ 12 } }.{ y }^{ \frac { 9 }{ 4 } }=\frac { { y }^{ \frac { 9 }{ 4 } } }{ { x }^{ \frac { 1 }{ 12 } } } \)

Total number of times the three coins are tossed = 200
Number of times when 2 heads come up = 72
Probability of 2 heads coming up
 =\(\frac { 72 }{ 200 } =\frac { 9 }{ 25 } \) 

6

672

36000 cm²

Given: ABCD is a trapezium whose nonparallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE 11 AD.
Proof: ∵ AB || DE        I Given
and     AD || BE           I By construction
∴  Quadrilateral ABCD is a parallelogram.

∴ \(\angle BAD=\angle BED\)                        ....(1)
| Opp.\(\angle \) s of a || gm are equal
and    AD = BE                                 ...(2)
Opp. sides of a || gm are equal
But     AD = BC            ...(3) I Given
From (2) and (3),
BE = BC
∴ \(\angle BEC=\angle BCE\)        ....(4)
| Angles opposite to equal sides of a triangle are equal
\(\angle BEC+\angle BED=180°\)       | Linear Pair Axiom
⇒ \(\angle BCE+\angle BAD=180°\) I From (4) and (1)
⇒ Trapezium ABCD is cyclic.
I ∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD)
⇒ Area (ΔABD) = 1/2Area (ΔABC)... (1)
In ΔABD, E is the mid-point of AD. Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
⇒ Area (ΔBED) = 1/2Area (ΔABD)
⇒ Area (ΔBED) = 1/2 x 1/2Area (ΔABC) [From equation (1)]
⇒ Area (ΔBED) = 1/4Area (ΔABC)

40⁰

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

\(P\rightarrow \left( II \right) \)
\(\\ Q\rightarrow \left( IV \right)\)
\(\\ R\rightarrow \left( x-axis \right)\)
\( \\ S\rightarrow \left( y-axis \right) \)

\((-2x+3y+2z)^2\)
\((-2x+3y+2z)^{ 2 }={ \{ (-2x)+3y+2z\} }^{ 2 }\)
\(={ (-2x) }^{ 2 }+{ (3y) }^{ 2 }+{ (2z) }^{ 2 }+2(-2x)(3y)+(3y)(2z)+2(2z)(-2x)\) |Using Identity V
\(=4{ x }^{ 2 }+9{ y }^{ 2 }+4{ z }^{ 2 }-12xy+12yz-8zx\)

Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm an can be the slant height of the cone, then
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+576 } \)
Now, curved surface area=\(\pi\)rl
\(\Rightarrow \ \frac { 22 }{ 7 } \times r\times \sqrt { { r }^{ 2 }+576 } =550\)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =550\times \frac { 7 }{ 22 } \)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =175\)
Squaring both the sides we get
r² (r² + 576) = 30625
(r²)² + 576r²-30625 = 0
Let r² = x
x² + 576x - 30625 = 0
\(\Rightarrow\) x² + 625x - 49x - 30625 = 0
\(\Rightarrow\) x(x + 625)-  49(x + 625) = 0
\(\Rightarrow\) (x + 625)(x - 49) = 0
\(\Rightarrow\) x + 625=0 or x - 49 = 0
\(\Rightarrow\) x = 625 or x = 49
not possible x = 49
\(\therefore\) r²= 49
\(\Rightarrow\) r = 7 cm
Volume & the cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+24\)
= 1232 cm³.

(i) Given: A circle C(O, r) and chord AB = chord AC. AD is bisector of \(\angle\)CAB.
To Prove: Centre O lies on the bisector of \(\angle\)BAC
Construction Join Be, meeting bisector AD of \(\angle\)BAC, at M.

Proof: In triangles BAM and CAM,
AB=AC    (given)
\(\angle\)BAM= \(\angle\)CAM  (given)
AM=AM   (Common)
\(​​\triangle BAM\cong \triangle CAM\)   (SAS)
\(\Rightarrow\)  BM=CM
and \(\angle\)BMA = \(\angle\)CMA
As \(\angle\)BMA + \(\angle\)CMA = 180⁰ (linear pair)
\(\Rightarrow\) \(\angle\)BMA = \(\angle\)CMA = 90⁰
\(\Rightarrow\) AM is the perpendicular bisector of the chord BC
\(\Rightarrow\) AM passes through the centre O.
[\(\because\) Perpendicular bisector of chord of a circle passes through the centre of the circle]
Hence, the centre of the park lies on the angle bisector of \(\angle\)BAC
(ii) Congruency of triangles by SAS axiom (Geometry)
(iii) Cleanliness and respect for labour.

Proof: In \(\triangle\)ADE, we have
AD = AE
\(\angle\)ADE = \(\angle\)AED
180°- \(\angle\)ADE = 180°- \(\angle\)AED
\(\Rightarrow\) \(\angle\)ADB = \(\angle\)AEC
Consider \(\triangle\)ABD and \(\triangle\)ACE
AD =AE
\(\angle\)ADB = \(\angle\)AEC
BD = EC
By SAS congruence,
\(\triangle ADB\cong \triangle AEC\)
By c.p.c.t., AB = AC
\(\therefore\) \(\triangle\)ABC is an isosceles triangle.

We are given that AB = BC and have to prove that
DE = EF.
Let us join A to E intersecting m at G.
Let trapezium ACFD is divided into two triangles, namely ΔACF and ΔAFD.
In ΔACF, it is given that B is the mid-point of AC(AB = BC) and BG II CF (Since m || n)
So, G is the mid-point of AF (By the converse of midpoint theorem)
Now in ΔAFD, we can apply the sam argument as G is the mid-point of AF, GE IIAD so E is the mid-point of DF
i.e., DE = EF
In otherwords l, m and n cut off equal intercepts on q also.

\(={ \left( { x }^{ { a }^{ 2 }-{ b }^{ 2 } } \right) }^{ \frac { 1 }{ a+b } }.{ \left( { x }^{ { b }^{ 2 }-{ c }^{ 2 } } \right) }^{ \frac { 1 }{ b+c } }.{ \left( { x }^{ { c }^{ 2 }-{ a }^{ 2 } } \right) }^{ \frac { 1 }{ c+a } }\)
\(={ x }^{ \frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b } }.{ x }^{ \frac { { b }^{ 2 }-{ c }^{ 2 } }{ a+b } }.{ x }^{ \frac { { c }^{ 2 }-{ a }^{ 2 } }{ c+a } }\)
\(={ x }^{ a-b }.{ x }^{ b-c }.{ x }^{ c-a }\)
\(={ x }^{ 0 }\)