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As base of triangle inscribed in a semi circle is 2rand altitude is r then area of two right angled triangle with base and height with r is 2r.... Thank you???

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Find the root of equation X+ROOT X-2

Posted by Deva Sam 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Avinash Saigal 6 years, 11 months ago

There can't be any root of this equation as there should not any other vaue of variable( i.e. x) except whole number. But here x has the value1/2. So, not possible thank you??

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Can anyone tell me the all areas and perimeter of common shapes which are used in chapter (areas related to circles)

Posted by Anshika Pal???? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Puja Sahoo? 6 years, 11 months ago

(1) For a circle of radius r , we have : (a) Circumference of circle = 2πr (b ) Area of circle = πr² (2) For a semicircle of radius r , we have : (a ) Perimeter = (πr + 2r ) (b ) Area = 1/2 × πr² (3 ) For a ring having outer radius = R and ineer radius = r , we have : (a ) Area of the ring = π( R² - r² ) (4 ) Let an arc ACB make an angle ¢° at the centre of a circle of radius r . Then , we have : (a ) Length of minor arc = 2πr ¢ / 360° (b ) Length of major arc = ( 2πr - 2πr theta/360 ) (c ) Area of minor sector = πr² ¢ / 360° Or 1/2 × Radius × arc length ). ( d ) Area of major sector = ( π²r - πr² theta/360) (e ) Area of minor segment = ( πr² theta /360 - 1/2 r² sin theta ) (f ) Area of major segment = [ πr² - ( area of minor segment ) ] (g) Perimeter of sector = ( 2r + 2πr theta /360)

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Which questions are important from chapter surface area and volume ??

Posted by Anshika Pal???? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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A hemisphical tank full of water is emptied by a pipe at the rate of 25/7 litrr per sec how much Tim e Will it take to empty half the tank if it is 3m in diameter

Posted by Aanchal Sharma 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have, Radius of hemispherical tank = {tex} \frac { 3 } { 2 } m{/tex}
{tex} \therefore{/tex} Volume of the tank ={tex} \frac { 2 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 3 } { 2 } \right) ^ { 3 } \mathrm { m } ^ { 3 } = \frac { 99 } { 14 } \mathrm { m } ^ { 3 }{/tex}
Volume of the water to be emptied = {tex} \frac { 1 } { 2 } \times \frac { 99 } { 14 } \mathrm { m } ^ { 3 } = \frac { 99 } { 28 } \mathrm { m } ^ { 3 } = \frac { 99 } { 28 } \times 1000 \text { litres } = \frac { 99000 } { 28 }{/tex}litres
Since {tex} \frac { 25 } { 7 }{/tex} litres of water is emptied in one second. Therefore,
Total time taken to empty half the tank i.e {tex} \frac { 99000 } { 28 }{/tex} litres of water = {tex} = \frac { 99000 } { 28 } \div \frac { 25 } { 7 }{/tex}seconds
{tex} = \frac { 99000 } { 28 } \times \frac { 7 } { 25 }{/tex}seconds
{tex} = \frac { 99000 } { 28 } \times \frac { 7 } { 25 } \times \frac { 1 } { 60 }{/tex}minutes
= 16.5 minutes

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Or and and concept from probability

Posted by Muskan Gupta 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Prove that : [ {cosA - sinA + 1} ÷ {cosA + sinA - 1} ] = cosecA + cotA...???

Posted by Saurabh Singh Lodhi 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Rahul Raj 6 years, 11 months ago

It's ncert example and already solved

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What must be subtracted or added to PX equal to 8 x 4 + 14 x cube minus 2 X square + 8 x minus 12 so that 4 x square + 3 x minus 2 is factor of P i ¤》□♡●□♧□♡●♤▪》¤♤▪■♤●fjX

Posted by Ritick Mishra 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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If tan if tan theta

Posted by Ritick Mishra 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Ankita ?? Arpita☺️ 6 years, 11 months ago

What?????

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D.J Alok 6 years, 11 months ago

Please complete your question first...

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How to find n term on ap in question how to recognize from the question

Posted by Durga Appa 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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o is the centre of circle such that diameter AB= 13 cm & AC= 12cm. BC is joined . find the area of the shaded region

Posted by Ritu Singh 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Kiran Yadav 6 years, 11 months ago

Which part is shaded

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What is the mark scheme in board 10th

Posted by Ramneek Virk 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Indian #Proud To Be Indian# 6 years, 11 months ago

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AD is median of ∆ABC and E is the mid point of AD IF BE is produced to meet AC at F, then AF = one third of AC

Posted by Kiran Yadav 6 years, 4 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 4 months ago

Given AD is the median of ΔABC and E is the midpoint of AD.
Through D, draw DG || BF.
In ΔADG, E is the midpoint of AD and EF || DG.
By converse of midpoint theorem, we have
F is midpoint of AG and AF = FG ................. (1)
Similarly, in ΔBCF
D is the midpoint of BC and DG || BF
G is midpoint of CF and FG = GC .................(2)
From equations (1) and (2), we get
AF = FG = GC ...............................................(3)
From the figure we have, AF + FG + GC = AC
AF + AF + AF = AC [from (3)]
3 AF = AC
Hence, AF= {tex}\frac { 1 } { 3 }{/tex}AC.

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Prove that tangent drawn at any point of a circle perpendicular to the radius through the point of contact.

Posted by Abhishek Pandey 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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| | | {R J } .....! ! !: ; 6 years, 11 months ago

It's given in the book ........

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Rahul Raj 6 years, 11 months ago

It requires figure bro how one can solve it here

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the decorative block shown in figure is made of two solids -a cube and a hemisphere. The base of the block is a cube with edge 5cm ,and the hemisphere fixed on the top has a diameter of 4.2cm. find the total surface area of the block

Posted by Varun Hs 6 years, 4 months ago

CBSE > Class 10 > Mathematics

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Sia ? 6 years, 4 months ago

Let S be the total surface area of the decorative block. Then,
S = Total surface area of the cube - Base area of hemisphere + Curved surface area of hemisphere

{tex}\Rightarrow S = \left( 6 \times 5 \times 5 - \pi r ^ { 2 } + 2 \pi r ^ { 2 } \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \left( 150 + \pi r ^ { 2 } \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \left\{ 150 + \frac { 22 } { 7 } \times ( 2.1 ) ^ { 2 } \right\} \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \{ 150 + 22 \times 0.3 \times 2.1 \} \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow{/tex} S = (150 + 13.86) cm² = 163.86 cm²