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Express 0.2353535... in p/q form

Posted by Sanjana Venkatesh 5 years, 1 month ago

2 answers

Nandini 18 5 years, 1 month ago

Let 0.2353535...be x x=0.2353535... 100 * x =0.2353535...*100 100x =23.53535... 100x=23.3 + .2353535...(23.3+.23535..=23.53535) 100x=23.3 + x (0.23535..=x) 100x - x =23.3 99x = 23.3 x =23.3/99 x=23.3 * 10/99 * 10 x = 233/990

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Yogita Ingle 5 years, 1 month ago

Given a number 0.2353535…….

We need to prove 0.2353535… = 0.235‾ can be expressed in the form of p/q, where p and q are integers and q ≠zero

Proof:

Let us assume that

x = 0.2353535…

x = 0.235 ——————(i)

On Multiplying both sides by 100 of equation (i) we get,

100x = 100 × 0.2353535…

100x = 23.53535————–(ii)

Subtracting equation (i) from equation (ii) we get,

100x – x = 23.53535 – 0.2353535…

99x = 23.2999965

x = 23.2999965/99

x = 233/990

x = 0.2353535

Hence, x = 0.2353535…= 0.235‾ can be expressed in the form of p/q as 233/ 990 and here q=990 (q≠zero)

Hence proved.

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ANSWER

in the given fig , ∆PQC and ∆PRC are such that QC = PR and PQ=CR prove that angle PCQ = angle CPR

Posted by Hitesh Chandora 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Nandini 18 5 years, 1 month ago

in triangle PQC and triangle PRC, Given - QC= PR PQ=CR To prove:- angle PCQ= angle CPR Proof:- In triangle PQC and triangle PRC:- QC=PR [GIVEN] PQ=CR [GIVEN] andPC=PC [COMMON] by SAS similarly criterion, triangle PQC is congruent to triangle PRC. therefore by C.P.C.T angle PCQ =angle CPR prooved.

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Yogita Ingle 5 years, 1 month ago

In triangle PQC and triangle PRC,
Given - QC= PR
PQ=CR
To prove:- angle PCQ= angle CPR

Proof:- In triangle PQCand triangle PRC:-

QC=PR [GIVEN]
PQ=CR [GIVEN]
andPC=PC [COMMON]
by SAS similarly criterion, triangle PQC is congruent to triangle PRC.
therefore by C.P.C.T
angle PCQ =angle CPR

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ANSWER

What is non-euclidian geometry?

Posted by Gaurav Kumar 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Is Topper Is Topper 5 years, 1 month ago

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Yogita Ingle 5 years, 1 month ago

Non Euclidean Geometry:

All the attempts to prove Euclid’s fifth postulate using the first 4 postulates failed. But they led to the discovery of several other geometries, called non-Euclidean geometries. Now the geometry of the universe we live in has been shown to be a non-Euclidean geometry. In fact, it is called spherical geometry. In spherical geometry, lines are not straight. They are parts of great circles (i.e., circles obtained by the intersection of a sphere and planes passing through the centre of the sphere). As per fist postulate of Euclid, lines AN & BN should not meet, but in the image below, these lines meet at point N. It has been proved that Euclidean geometry is valid only for the figures in the plane. On the curved surfaces, it fails.

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ANSWER

Q-13. Construct an equilateral triangle with 5cm and justify the construction. Write steps of construction. ~~~~~~~~Only this much~~~~

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Kumar 5 years, 1 month ago

Step 1- measure a line segment of 5cm (let A and B ) Step 2- now draw an arc of 5cm from point A. Step 3- and draw an arc of 5 cm from point B. Step 4- let the point of intersection of two arcs be C. Step 5- join points A,B and C. To justify your answer: Measure all sides and angles. You will observe that each side is of 5cm and each angle is of 60⁰.

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ANSWER

Q-12. Prove that angles opposite to equal sides of an isosceles triangle are equal.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 1 month ago

Given: In the isosceles ∆XYZ, XY = XZ.

To prove ∠XYZ = ∠XZY.

Construction: Draw a line XM such that it bisects ∠YXZ and meets the side YZ at M.

Proof:

Statement

1. In ∆XYM and ∆XZM,

(i) XY = XZ

(ii) XM = XM

(iii) ∠YXM = ∠ZXM

2. ∆XYM ≅ ∆XZM

3. ∠XYZ = ∠XZY. (Proved)

Reason

(i) Given.

(ii) Common side.

(iii) XM bisects ∠YXZ.

2. By SAS criterion.

3. CPCTC.

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ANSWER

Q-11. ABC is a triangle in which altitudes BE and CF to side AC and AB respectively. Show that ∆ABE = ∆ACF.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 5 years, 1 month ago

In ΔABE and ΔACF,

∠BAE=∠CAF (Common angle)

∠AEB=∠AFC ....(∵BE⊥AC and CF⊥AB)

BE=CF (Given that altitudes are equal)

By AAS criterion of congruence,

ΔABE≅ΔACF

Hence,

AB=AC (by CPCT)

1Thank You

Harsh Grover 5 years, 1 month ago

Angle A common angle E and angle F=90 AC =AB

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ANSWER

Q-10. Construct an angle of 105°.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 1 month ago

<i>Step of Construction:</i>

(i) After making 90° angle take L and N as centre and draw two arcs cutting each other at S.

(ii) Join SO.

Then, ∠SOA = 105°.

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ANSWER

Q-9. AB is a line segment and line I is its perpendicular bisector. If a point P line on I . Show that P is equidistant from A and B.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 1 month ago

since , AB is a line segment,
L is drawn perpendicular to AB,
A point 'p' lies on line L.
( as shown in the diagram in the attachment )
To prove : P is equidistant from A and B.
Prove :- In ∆AOP and ∆BOP ,
OP = OP ( common side )
POA = POB
AO = OB
∆AOP ∆BOP (By S.A.S.)
=> AP = BP ( By C.P.C.T. )
Hence , "P" is equidistant from A and B .

1Thank You

ANSWER

Q-8. Line-segment AB is parallel to another line-segment CD. O is the mid-point of AD. Show that (I) ∆AOB = ∆DOC (II) O is also the mid-point of BC.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 5 years, 1 month ago

In △AOB and △DOC

∠BAO=∠CDO (alternate angles as AB∥CD and BC is the transversal)

∠AOB=∠DOC(vertically opposite angles)

OA=OD(given)

∴△AOB≅△DOC by ASA rule.

1Thank You

Meenkashi Singh 5 years, 1 month ago

Given: AB is parallel to another line segment CD. O is the midpoint OF AD In ΔAOB and ΔDOC ∠AOB=∠COD ...(Vertically opposite angle ) ∠BAO=∠CDO ...(Given AB parallel to DC and AD meet both lines so alternate angles are equal) AO=OD ....(O is the midpoint of AD ) ΔAOB≅ΔDOC ...ASA test So, BO=CO Then, O is the midpoint of BC.

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ANSWER

Q-7. In the given figure, ∆PQC and ∆PRC are such that QC = PR and PQ = CR, Prove that /PCQ=/CPR.

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 5 years, 1 month ago

In triangle PQC and triangle PRC,
Given - QC= PR
PQ=CR
To prove:- angle PCQ= angle CPR

Proof:- In triangle PQCand triangle PRC:-
QC=PR [GIVEN]
PQ=CR [GIVEN]
andPC=PC [COMMON]
by SAS similarly criterion, triangle PQC is congruent to triangle PRC.
therefore by C.P.C.T
angle PCQ =angle CPR

1Thank You

Meenkashi Singh 5 years, 1 month ago

PQ=PR (Given) ∴∠PQR=PRQ (Angles opposite to equal sides) QC=RC (Given) △PQC≅△CPR By SAS criteria

1Thank You

ANSWER

• Solve the following . Q-6. In the given figure, AC=AE , AB=AD and /BAD = /EAC. Show that BC=DE Please friends help me

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 1 month ago

It is given that ∠BAD=∠EAC

∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]

∴∠BAC=∠DAE

In △BAC and △DAE

AB=AD (Given)

∠BAC=∠DAE (Proved above)

AC=AE (Given)

∴△BAC≅△DAE (By SAS congruence rule)

∴BC=DE (By CPCT)

2Thank You

ANSWER

https://drive.google.com/file/d/1WIsflJsMBhYO_P5X-0XzCjp6xPntL1av/view?usp=drivesdk. Friend this is the link.. Please help me please ökay..

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

rini suchandra

Posted by Vandana Kapoor 5 years, 1 month ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

ABCD is a parallelogram and EFCD is a rectangle. Also,AL perpendicular DC. Prove that (1) ar (ABCD) = ar (EFCD) (2) ar (ABCD) = DC × AL

Posted by Riya Kumari 5 years, 1 month ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 1 month ago

Given that:- ABCD is a parallelogram and EFCD is a rectangle.

To prove:- ar(ABCD)=DC×AL

Proof:-

∴AB∥CD.....(1)[∵Opposite sides of parallelogram are equal]

As given that EFCD is a rectangle and we know that rectangle is also a parallelogram.

∴EF∥CD.....(2)

From eqn(1)&(2), we hahve

AB∥EF

Now, ABCD and EFCD are two parallelograms with same base CD and between tha same parallels EB and CD.

As we know that parallelogram with same base and in between same parallels are equal in area.

∴ar(ABCD)=ar(EFCD)

As from the fig.

ar(EFCD)=DC×FC

⇒ar(ABCD)=DC×FC.....(3)

AL⊥CD[Given]

∴AFCL is a also a rectangle.

∴AL=FC.....(4)

From eqn(3)&(4), we get

ar(ABCD)=DC×AL

Hence proved that ar(ABCD)=DC×AL.

3Thank You

ANSWER

• Solve the following . Q-6. In the given figure, AC=AE , AB=AD and /BAD = /EAC. Show that BC=DE A __ ________________E / \ \ / / \ \ / / \ \ / / \ \ / / _________\________\/ B. D. C Please friends manage the figure ØK....

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

Evaluate 1. (102)3 2. 140×96 3. (X+4) (x+10)

Posted by Nisha Murali 5 years, 1 month ago

CBSE > Class 09 > Mathematics

3 answers

Nisha Murali 5 years, 1 month ago

Thanks

2Thank You

Abhay Singh 5 years, 1 month ago

Question 1. 102x3=306 2. 140x96=13440 3. (x+4)(x+10) =x+x of 4+10 = 2x of 14 = 28x answer.

3Thank You

Yogita Ingle 5 years, 1 month ago

Let us rewrite (102)3 as (100+2)³

Now using the identity (a+b)³=a³+b³+3ab(a+b), we get

(100+2)³=100³+2³+[(3×100×2)(100+2)]

=1000000+8+(600×102)

=1000000+8+61200

=1061208

Hence, (102)³=1061208

2Thank You

ANSWER

Q-5. In ∆PQR if /R >/Q then (A). QR>PR (B). PQ>PR (C). PQ>PR (D). QR>PR

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 5 years, 1 month ago

Given: In ΔPQR,

∠R >∠Q

⇒ PQ > PR (side opposite to greater angle is greater)

Hence, B is correct.

1Thank You

Ayush Thakur 5 years, 1 month ago

PR>PQ

1Thank You

ANSWER

Q-4. Which of the following can be construed using ruler and pair of compasses only (A). 35° (B). 40° (C). 37.5° (D). 47.5°

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

3 answers

Satyam Kumar 5 years, 1 month ago

(C). 37.5° and (D). 47.5° both can be constructed.

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Yogita Ingle 5 years, 1 month ago

Which of the following can be construed using ruler and pair of compasses only (A). 35° (B). 40° (C). 37.5° (D). 47.5°

Ans c

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Ayush Thakur 5 years, 1 month ago

(C). 37.5°

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ANSWER

Q-3. E and F are respectively the mid-points of equal sides AB and AC of ∆ABC Than_______. (A). BF = AC (B). BE = AF (C). CE = AB (D). BF = AC

Posted by Arun Kushwah 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Yogita Ingle 5 years, 1 month ago

Given E and F are mid point of equal sides AB and AC of triangle ABC

In ΔABF and ΔACE

AB=AC (given )

∠A=∠A (common angle )

AF=AE (halves of equal sides)

∴ΔABF≅ΔACE (SAS rule)

∴BF=CF (CPCT)

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Ayush Thakur 5 years, 1 month ago

(B). BE=AF

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ANSWER

How we can easily solve the problem if don't know

Posted by Aryan Maran 5 years, 1 month ago

CBSE > Class 09 > Mathematics

2 answers

Chandana M Patil 5 years, 1 month ago

We can first we should know the formule and the second we have to understand question and third is we should know that what is given and what we have to find

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Piyush Patel 5 years, 1 month ago

Yes we can

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ANSWER

Make a double bar graph of urban and rural population of Meghalaya according to age groups

Posted by Pakhi Saxena 5 years, 1 month ago

CBSE > Class 09 > Mathematics