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Ask QuestionPosted by C Abbi 5 years, 2 months ago
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Sabnam Patri Sabnam Patri 5 years, 2 months ago
Seema Cma 5 years, 2 months ago
Shimpi Saha 5 years, 2 months ago
Gaurav Seth 5 years, 2 months ago
Supplementary angles and complementary angles are defined with respect to the addition of two angles. If the sum of two angles is 180 degrees then they are said to be supplementary angles, which forms a linear angle together. Whereas if the sum of two angles is 90 degrees, then they are said to be complementary angles, and they form a right angle together.
Posted by Muskan Mishra 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
In the given figure, O is the centre of a circle and ∠BOD =150°. Find the values of x and y.
Answer:
It is given that O is the centre of a circle and ∠BOD =150o
We know that
Reflex ∠BOD = (360o – ∠BOD)
By substituting the values
Reflex ∠BOD = (360o – 150o)
By subtraction
Reflex ∠BOD = 210o
Consider x = ½ (reflex ∠BOD)
By substituting the value
x = 210/2
So we get
x = 105o
We know that
x + y = 180o
By substituting the values
105o + y = 180o
On further calculation
y = 180o – 105o
By subtraction
y = 75o
Therefore, the value of x is 105o and y is 75o.
Posted by Sunil Hadapad 5 years, 2 months ago
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Posted by Amisha Kumari 5 years, 2 months ago
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Nitin Tripathi 5 years, 2 months ago
Gaurav Seth 5 years, 2 months ago
All the no.s from 1 to 100 are integers. Integers are no.s other than decimals, for eg., 5 is an integer, 5.1 is non-integer i.e., decimal. Integers are both positive and negative.
Posted by Sharanya Rumandla 5 years, 2 months ago
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Anushka Agarwal 5 years, 2 months ago
Posted by Archit Gourha 5 years, 2 months ago
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Yogita Ingle 5 years, 2 months ago
X5 - x
= x ( x4 - 1)
=x (x2 - 1) (x2 + 1)
= x ( x - 1) ( x+ 1) (x2 + 1)
Posted by Sameer Kumar 5 years, 2 months ago
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Ashirwad Chauhan 5 years, 2 months ago
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Gaurav Seth 5 years, 1 month ago
10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.
The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)

Solution:
Say h = height of the frame of lampshade, looks like cylindrical shape
r = radius
Total height is h = (2.5+30+2.5) cm = 35cm and
r = (20/2) cm = 10cm
Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh
= (2×(22/7)×10×35) cm2
= 2200 cm2
Hence, 2200 cm2 cloth is required for covering the lampshade.
11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)
Solution:
Radius of the circular end of cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh+πr2
= 2×(22/7)×3×10.5+(22/7)×32= 1584/7
Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm2
So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be needed for the competition.
Posted by Sujal Jaiswal 5 years, 2 months ago
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Anushka Agarwal 5 years, 2 months ago
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Sabnam Patri Sabnam Patri 5 years, 2 months ago
Posted by ꧁╾━╤デ╦︻ ℛ??????࿐ 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a).
Solution to (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a)
Posted by Yamini Kandpal 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
The steps of construction to follow:
Step 1: Draw a ray OP.
Then, take O as the centre and any radius draw an arc cutting OP at Q.
Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.
Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.
Hence, .
Justification:
We need to justify,
So, join OR and OS and RQ. we obtain
By construction OQ = OS = QR.
So, is an equilateral triangle. Similarly
is an equilateral triangle.
So,
Now, that means
.
Then, join AS and AR:
Now, in triangles OSA and ORA:
(common)
(Radii of same arcs)
(radii of the same arcs)
So,
Therefore,
and
Hence, justified.
Posted by Aniket Yadav 5 years, 2 months ago
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Rahul Kumar 5 years, 2 months ago
Posted by Parvathy Roshan 5 years, 2 months ago
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Vikramdeep Singh Sekhon 5 years, 2 months ago
Posted by Bhumika Mittal 5 years, 2 months ago
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Kiran Barthwal 5 years, 2 months ago
Yogita Ingle 5 years, 2 months ago
1) y = 4x
Put x = -3 and y = a
a= 4 ( -3)
a = - 12
2) y = 4x
Put x = b and y = 4
4 = 4 b
b= 4/4 = 1
Posted by Saloni Dixit 5 years, 2 months ago
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Posted by Ayush Kumar Sinha 5 years, 2 months ago
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Anushka Agarwal 5 years, 2 months ago

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Anushka Agarwal 5 years, 2 months ago
1Thank You