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Ask QuestionPosted by Rohini Girase 3 years, 10 months ago
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Posted by Pinky Jain 3 years, 10 months ago
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Indhu Indhu 3 years, 10 months ago
Gaurav Seth 3 years, 10 months ago
the formula to find total surface area of cone is
πr(r+l)
to find l use l²=r²+h²
∴l²=3²+4²
=9+16
=25
∴l=5cm
now using π r ( r + l )
3.14 *3(3+5)
3.14 *3(8)
9.42*8
75.36 cm²
Posted by Naman Dubey 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
A circle’s perimeter is called the circumference. The symbol of the circumference is C.
The Circumference (or) perimeter of a circle = 2πR
where,
R is the radius of the circle
π is the mathematical constant with an approximate (up to two decimal points) value of 3.14
Again,
Pi (π) is a special mathematical constant; it is the ratio of circumference to diameter of any circle.
where C = π D
C is the circumference of the circle
D is the diameter of the circle
For example: If the radius of the circle is 4cm then find its circumference.
Given: Radius = 4cm
Circumference = 2πr
= 2 x 3.14 x 4
= 25.12 cm
Posted by Riya Sahu 3 years, 10 months ago
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Posted by Amatullah Lightwala 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
1. State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Solution:
True
Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.
i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….
Real numbers – The collection of both rational and irrational numbers are known as real numbers.
i.e., Real numbers = √2, √5, 0.102…
Every irrational number is a real number, however, every real numbers are not irrational numbers.
(ii) Every point on the number line is of the form √m where m is a natural number.
Solution:
False
The statement is false since as per the rule, a negative number cannot be expressed as square roots.
E.g., √9 =3 is a natural number.
But √2 = 1.414 is not a natural number.
Similarly, we know that there are negative numbers on the number line but when we take the root of a negative number it becomes a complex number and not a natural number.
E.g., √-7 = 7i, where i = √-1
The statement that every point on the number line is of the form √m, where m is a natural number is false.
For more click on the given link:
<a data-ved="2ahUKEwj58_izssrtAhUs4zgGHbGcCaUQFjALegQICRAC" href="https://mycbseguide.com/blog/ncert-solutions-class-9-maths-exercise-1-2/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-class-9-maths-exercise-1-2/&ved=2ahUKEwj58_izssrtAhUs4zgGHbGcCaUQFjALegQICRAC" rel="noopener" target="_blank">NCERT Solutions for Class 9 Maths Exercise 1.2 ...</a>
Posted by Hãrjåß Käûr 3 years, 10 months ago
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Posted by Saranya B 3 years, 10 months ago
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Posted by Teju Kn 3 years, 10 months ago
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Shubham Singh 3 years, 10 months ago
Yogita Ingle 3 years, 10 months ago
Since we want five numbers, we write 3/5 and 4/5 So multiply in numerator and denominator by 5+1 =6 we get
Posted by Anuj Yadav 3 years, 10 months ago
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Posted by Anuj Yadav 3 years, 10 months ago
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Posted by Avneesh Kumar 3 years, 10 months ago
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Yogita Ingle 3 years, 10 months ago
base = 10 cm and height = 3 cm
Area of parallelogram = base × height sq. units
= 10 × 3
= 30 sq. cm
Posted by Vanshika Dik 3 years, 10 months ago
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Indhu Indhu 3 years, 10 months ago
Posted by Ishita Budhwar 3 years, 10 months ago
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Yogita Ingle 3 years, 10 months ago
Step's of construction
(1) Draw BC = 6 cm.
(2) Construct ∠CBX = 60°.
(3) Along BX , set off BP = 9 cm.
(4) Join CP
(5) Draw the perpendicular bisector of CP tp intersect BP at A .
(6) Join AC. Then, ∆ABC is the required triangle.
Posted by Asha Rani 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
From the figure, we know that,
AB and CD intersect each other at point O.
Let the two pairs of vertically opposite angles be,
1st pair – ∠AOC and ∠BOD
2nd pair – ∠AOD and ∠BOC
To prove:
Vertically opposite angles are equal,
i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC
From the figure,
The ray AO stands on the line CD.
We know that,
If a ray lies on a line then the sum of the adjacent angles is equal to 180°.
⇒ ∠AOC + ∠AOD = 180° (By linear pair axiom) … (i)
Similarly, the ray DO lies on the line AOB.
⇒ ∠AOD + ∠BOD = 180° (By linear pair axiom) … (ii)
From equations (i) and (ii),
We have,
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD – – – – (iii)
Similarly, the ray BO lies on the line COD.
⇒ ∠DOB + ∠COB = 180° (By linear pair axiom) – – – – (iv)
Also, the ray CO lies on the line AOB.
⇒ ∠COB + ∠AOC = 180° (By linear pair axiom) – – – – (v)
From equations (iv) and (v),
We have,
∠DOB + ∠COB = ∠COB + ∠AOC
⇒ ∠DOB = ∠AOC – – – – (vi)
Thus, from equation (iii) and equation (vi),
We have,
∠AOC = ∠BOD, and ∠DOB = ∠AOC
Therefore, we get, vertically opposite angles are equal.
Hence Proved.
Posted by Aarya Gupta 3 years, 10 months ago
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Posted by Diya Jogi 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Posted by P R 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Gaurav Seth 3 years, 10 months ago
- Draw a line AB by measuring 9.3cm
- From the point, B add 1cm and mark it as C
- Mark the point of bisection by a compass and say it as ‘O’
- Measure AO which is the radius and draws a semi-circle.
- From B draw a perpendicular AB touching the semi-circle and mark as D
- Draw an arc on the number line by taking compass pointer on B and pencil on D
- The point which intersects the number line is the square root of 9.3
Posted by Jasnoor Kaur Sandha 3 years, 10 months ago
- 2 answers
Posted by Gargi Rajput 3 years, 10 months ago
- 2 answers
Gaurav Seth 3 years, 10 months ago
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
Let AB and CD be two equal cords (i.e. AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.
It is now to be proven that the line segments AE = DE and CE = BE
Construction Steps:
Step 1: From the center of the circle, draw a perpendicular to AB i.e. OM ⊥ AB
Step 2: Similarly, draw ON ⊥ CD.
Step 3: Join OE.
Now, the diagram is as follows-
Proof:
From the diagram, it is seen that OM bisects AB and so, OM ⊥ AB
Similarly, ON bisects CD and so, ON ⊥ CD
It is known that AB = CD. So,
AM = ND — (i)
and MB = CN — (ii)
Now, triangles ΔOME and ΔONE are similar by RHS congruency since
OME = ONE (They are perpendiculars)
OE = OE (It is the common side)
OM = ON (AB and CD are equal and so, they are equidistant from the centre)
∴ ΔOME ΔONE
ME = EN (by CPCT) — (iii)
Now, from equations (i) and (ii) we get,
AM+ME = ND+EN
So, AE = ED
Now from equations (ii) and (iii) we get,
MB-ME = CN-EN
So, EB = CE (Hence proved).
Posted by Rajnish Tiwari 3 years, 10 months ago
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Posted by Puram Kowshika Gayathri 3 years, 10 months ago
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Posted by Payal Kumari 3 years, 10 months ago
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K. K. 3 years, 10 months ago
Posted by Nitin Kumar Jain 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
Given: A parallelogram ABCD and AC is its diagonal .
To prove : △ABC ≅ △CDA
Proof : In △ABC and △CDA, we have
∠DAC = ∠BCA [alt. int. angles, since AD | | BC]
AC = AC [common side]
and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC]
∴ By ASA congruence axiom, we have
△ABC ≅ △CDA
Posted by Subhojit Mukherjee 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Posted by Suraj Yadav 3 years, 10 months ago
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Suraj Yadav 3 years, 10 months ago
Suraj Yadav 3 years, 10 months ago
Suraj Yadav 3 years, 10 months ago
Posted by Aniket Kumar 3 years, 10 months ago
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Aniket Kumar 3 years, 10 months ago
Aniket Kumar 3 years, 10 months ago
Posted by Arul S 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Posted by P R 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
Given: ABCD is a trapezium where AB||CD and AD = BC
To prove: ABCD is cyclic.
Construction: Draw DL⊥AB and CM⊥AB.
Proof: In ΔALD and ΔBMC,
AD = BC (given)
DL = CM (distance between parallel sides)
∠ALD = ∠BMC (90°)
ΔALD ≅ ΔBMC (RHS congruence criterion)
⇒ ∠DAL = ∠CBM (C.P.C.T) (1)
Since AB||CD,
∠DAL + ∠ADC = 180° (sum of adjacent interior angles is supplementary)
⇒ ∠CBM + ∠ADC = 180° (from (1))
⇒ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)
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