Given AB and CD are the chords of a circle with centre at O such that AB = CD

To Prove: \(\angle\)AOB = \(\angle\)COD
Proof: In \(\Delta\)AOB and \(\Delta\)COD
AO = CO        (radii of same circle)
AB = CD         (given)
BO = DO        (radii of same circle)
\(\Delta\)AOB\(\cong \) \(\Delta\)COD   (SSS)
\(\angle\)AOB = \(\angle\)COD  (c.p.c.t.) 2 Hence Proved.

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

So, when \(y=\frac{7x-35}{5}\)

Rs. 23760

\(18\sqrt { 2 } \)cm²

c=\(126^{ 0 }\)

\(\frac{1}{7}=0.142857142857 \ldots=0 . \overline{142857}\) and \(\frac{2}{7}=0.28571428571428 \ldots=0 . \overline{285714}\)
Here, the two decimal expansions are non-terminating recurring.
Hence, 1/7​ and 2/7​ are two rational numbers.
We know, between any two rational numbers, there are infinitely many irrational numbers.
An irrational number has non-terminating non-recurring decimal expansions.
Then an irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\) is 0.15015001500015.
Similarly, 0.21020020002... is another irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\)

Required probability=\(\frac { 600 }{ 1000 } =\frac { 3 }{ 5 } \)

Theorem

Theorem

X=\(180^{ 0 }\)-x\(\Rightarrow \)x\(90^{ 0 }\)

2x+0y-5=0

Degree of \(x^3+3x^2-4x+3\)​​​​​​​ is 3

Steps of construction:
i) Draw a line segment QR = 6cm.At point Q construct an angle = 60^o i.e ㄥXQR = 60^o 

ii) Cut a line segment QS = 2 cm from the line segment QT extended on opposite side of line segment XQ.
(As PR > PQ and PR - PQ = 2cm) join SR
iii) Draw perpendicular bisector AB of line segment SR,which intersects QX at point P.Join PQ,PR. ΔPQR is the required triangle.

We know that \(\sqrt{5}=\sqrt{4+1}=\sqrt{2^{2}+1^{2}}\)

Draw a right angled ∆OBA, such that
OB = 2 units, AB = 1unit and ∠OBA = 90°
Now, by using Pythagoras theorem, we have
OA² = OB² + AB² = 2² +1²
⇒ OA = \(\sqrt{4+1}=\sqrt{5}\)
Now, take O as centre, OA =√5 as radius, draw an arc which intersects the line at point C.
Hence, the point C represents √5.

h = 77 cm
2r = 4 cm
r = 2 cm
2R = 4.4 cm
R = 2.2 cm
(i) Inner curved surface area \(=2\pi rh\)
 \(=2\times \frac { 22 }{ 7 } \times 2\times 77=968{ cm }^{ 2 }\)
(ii) Outer curved surface area =
 \(=2\times \frac { 22 }{ 7 } \times 2.2\times 77=1064.8{ cm }^{ 2 }\)
(iii) Total surface area
 \(=2\pi Rh+2\pi rh+2\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) \)
\(\\ =1064.8+2\times \frac { 22 }{ 7 } \times 2\times 77+2\times \frac { 22 }{ 7 } \times \left\{ { \left( 2.2 \right) }^{ 2 }-{ \left( 2 \right) }^{ 2 } \right\} \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times \left( 4.84-4 \right) \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times 0.84\)
\(\\ =1064.8+968+5.28=2038.08{ cm }^{ 2 }.\)

36000 cm²

k=7

We know that
\(x^3+y^3+z^3=3xyz\)
\(=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\) Using Identify VII
\(=(0)(x^2+y^2+z^2-xy-yz-zx)\) \(\because x+y+z=0\)
= 0
\(\Rightarrow x^3+y^3+z^3=3xyz\) 

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) AB and CD intersect atO0
\(\therefore\) \(\angle\)AOD = \(\angle\)BOC
(Vertically opp. angles) ...(i)
In \(\triangle\)AOD and \(\triangle\)BOC, we have 
\(\angle\)AOD = \(\angle\)BOC ...(ii)
\(\angle\)DAO = \(\angle\)CBO = 90° (Given)
and AD = BC (Given)
\(\triangle AOD\cong \triangle BOC\)
(By AAS congruence criterion)
\(\Rightarrow\) OA = OB (By c.p.c.t.)
i.e., O is the mid-point of AB
Hence, CD bisects AB.
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

(i) Required number of students = 8 + 32 = 40
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =\(\frac { Frequency\ of\ Class }{ Width\ of\ class } \times Minimum\ class-size\)

Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as heights.
Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork and Dilligence.