CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Let ABCD be a square When diagonals are constructed we get two triangle that is ∆ ADC and ∆ BCD We have to prove that these triangles are congruent (same) to justify that diagonals are equal. In ∆ADC and ∆ BCD AD =BC (sides) DC =DC (common) Angle ADC = angle BCD (90) So ∆ADC is congruent to ∆BCD AC =BD (CPCT :- coresponding part of congruent triangle ) AC and BD are diagonals of square ABCD In same square let us take the ∆AOD and ∆BOC Note:- here O is the intersection point of diagonals of square In ∆AOD and ∆BOC AD = BC Angle ADO = Angle OBC Angle DAO =Angle OCB (Alternative interior angle) ∆AOD and ∆BOC are congruent AO = OC (CPCT) DO = OB (CPCT) Hence diagonals bisect each other Same way when we prove ∆AOD is congruent to ∆DOC using SSS congruence rule we will get Angle AOD = Angle DOC (CPCT)---------- note 1 let angle AOD = angle 1 let angle DOC = angle 2 When we add angle 1 and 2 we will get 180° Angle 1 + Angle 2 = 180° From note 1 Angle 1= angle 2 So, angle 1+ angle 1 = 180° 2(angle 1) = 180° Angle 1 = 90° Angle 1, angle 2 = 90° Hence diagonals of a square are equal and bisect each other at right angle Hope you got it!!! ☺️☺️

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(i) What is the probability of selecting a male participant to inaugurate the meeting? (ii) If a team leader is to be selected from the participants. What is the probability of selecting a female team leader of age less than 30 years? OR (iii) If 4 more female participants of age less than 30 years and 5 more male participants of age more than 30 years joins the meeting, what is the probability of selecting a male participant of age less than 30 years to give vote of thanks? (iv) What is the probability of selecting a female participant of age more than 30 years to conclude the meeting?

Posted by [email protected] Morning 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

[email protected] Morning 2 years, 11 months ago

Answer

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0.345bar

Posted by Zainab Bushra 2 years, 11 months ago

CBSE > Class 09 > Mathematics

0 answers

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Prove that if 5=5a then what is the 4=1a ? If 4=1 is 0.25

Posted by Twing Gaming Gamerz 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Ravi Shankar Kumar 2 years, 11 months ago

Hlo

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Prove 5is an irrational

Posted by Ujjwal Pandit 2 years, 11 months ago

CBSE > Class 09 > Mathematics

3 answers

Shamitha Shami 2 years, 11 months ago

Actually 5 is a rational number because it can be represented in the form of p/q Ex:- 5/1

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Rashi Kaushik 2 years, 11 months ago

Sample paper solve set2

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Mohit Yadav 2 years, 11 months ago

5 is not a irrational number

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How we can proof that a given no. Is rational or irrational

Posted by Ali Talib 2 years, 11 months ago

CBSE > Class 09 > Mathematics

3 answers

Shamitha Shami 2 years, 11 months ago

* Hey 5/3 is rational number

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Shamitha Shami 2 years, 11 months ago

If we can represent a number in the form of p/q then it is a rational number if cannot then it is a irrational number Ex:- 5/3 is irrational √2 is irrational

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Ujjwal Pandit 2 years, 11 months ago

We let any no. In p/q form and squaring both side then q divide q2 then they divide q also

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If p(x) =5x - 4x^2 + 3 then find p (-1)

Posted by Oas&Ojas Kedia 2 years, 11 months ago

CBSE > Class 09 > Mathematics

3 answers

Dena Joshi 2 years, 11 months ago

-6

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Shamitha Shami 2 years, 11 months ago

Here substitute -1 in place of x in the equation 5x-4x^2+3 = 5(-1) -4(-1)^2 + 3 = -5 -4(1) + 3 = -5-4+3 = -9+3 =-6 hope you got it!!! ☺️☺️☺️

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Vedshree Patil 2 years, 11 months ago

It 2

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We observe that the cuboid is made of six rectanglar faces sum of area of all the six rectangles= _________ square units

Posted by Sameer John Santhi 2 years, 11 months ago

CBSE > Class 09 > Mathematics

2 answers

Sushamita Singh 2 years, 11 months ago

2(lb+bh+hl) , l = length h = height b = breadth

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Harshit Agarwal 2 years, 11 months ago

2(lb+bh+hl)

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t²+4t+3

Posted by Sumit Lohani 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Viswajeet Reddy Bheemireddy 2 years, 11 months ago

We know that factors of 3 are 1,3 Hence, 4t=1t+3t So, t²+4t+3 t²+1t+3t+3 t(t+1)+3(t+1) ((taking t and 3 common)) (t+3)(t+1) Hence this is your answer.

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Two chord ab and cd are 6 cm & 12 cm respectively and distance between both is 3 cm find the radius of circle

Posted by Arnav Jain 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 11 months ago

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. Let the radius of the circle be r cm. Draw OP {tex}\perp{/tex} AB and OQ {tex}\perp{/tex} CD. Since AB || CD and OP {tex}\perp{/tex} AB, OQ {tex}\perp{/tex} CD. Therefore, points O, Q, and P are collinear.

Clearly, PQ = 3 cm
Let OQ = x cm. Then, OP = (x + 3) cm
Applying Pythagoras theorem in right triangles OAP and OCQ, we obtain
OA² = OP² + AP² and OC² = OQ² + CQ²
{tex}\Rightarrow{/tex} r² = (x + 3)² + 3² and r² = x² + 6² [{tex}\because{/tex} AP = {tex}\frac{1}{2}{/tex} AB = 3 cm and CQ = {tex}\frac{1}{2}{/tex}CD = 6 cm]
{tex}\Rightarrow{/tex} (x + 3)² + 3² = x² + 6² [On equating the values of r²]
{tex}\Rightarrow{/tex}x² + 6x + 9 + 9 = x² + 36 {tex}\Rightarrow{/tex} 6x = 18 {tex}\Rightarrow{/tex} x = 3
Putting the value of x in r² = x² + 6², we get
r² = 3² + 6² = 45 {tex}\Rightarrow{/tex} r = {tex}\sqrt{45}{/tex} cm = 6.7 cm
Hence, the radius of the circle is 6.7 cm.

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In the given figure, ÐPQR = 120° where P, Q and R are points on the circle with centre O. Find ÐOPR

Posted by Khushee Bhogayta 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 11 months ago

Here, ∠PQR=100^∘

Take a point S in the major arc. Join PS and RS.

{tex}\because{/tex} PQRS is a cyclic quadrilateral.
{tex}\therefore \angle \mathrm { PQR } + \angle \mathrm { PSQ } = 180 ^ { \circ }{/tex}
|The sum of either pair of opposite angles of a cyclic quadrilateral is 180^o
{tex}\Rightarrow 100 ^ { \circ } + \angle P S R = 180 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 180 ^ { \circ } - 100 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 80 ^ { \circ }{/tex} ......... (1)
Now, {tex}\angle \mathrm { POR } = 2 \angle \mathrm { PSR }{/tex}
|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
{tex}= 2 \times 80 ^ { \circ } = 160 ^ { \circ }{/tex} ........ (2) |Using (1)
In {tex}\triangle O P R{/tex}
{tex}\because O P = O R{/tex} |Radii of a circle
{tex}\therefore \angle \mathrm { OPR } = \angle \mathrm { ORP }{/tex} ....... (3)
|Angles opposite to equal sides of a triangle are equal
In {tex}\triangle O P R{/tex}
{tex}\angle O P R + \angle O R P + \angle P O R = 180 ^ { \circ }{/tex} | Sum of all the angles of a triangle is 180^o
{tex}\Rightarrow \angle \mathrm { OPR } + \angle \mathrm { OPR } + 160 ^ { \circ } = 180 ^ { \circ }{/tex} |Using (2) and (1)
{tex}\Rightarrow 2 \angle O P R + 160 ^ { \circ } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow 2 \angle O P R = 180 ^ { \circ } - 160 ^ { \circ } = 20 ^ { \circ }{/tex}
{tex}\Rightarrow \angle \mathrm { OPR } = 10 ^ { \circ }{/tex}