CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Draw a line segment {tex}AB = 6 \ cm.{/tex}
At A, construct {tex}\angle{/tex}{tex}BAZ{/tex} = {tex}30 ^ { \circ }{/tex} and at B, construct {tex}\angle{/tex}{tex}ABY {/tex}= {tex}60 ^ { \circ }{/tex}
Suppose AZ and BY intersect at C.
{tex}\triangle{/tex}{tex}ABC {/tex} so obtained is the given triangle.
Below AB, make an acute {tex}\angle{/tex}BAX.
Along AX, mark off 8 points {tex}B_1, B_2, B_3, B_4, B_5, B_6, B_7, B_8{/tex}
such that {tex}AB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7 = B_7B_8{/tex}
Join B₆B.
From B₈, draw {tex}B_8B'||B_6B {/tex} meeting AB produced at B'.
From B', draw B'C'{tex}\|{/tex}BC, meeting AZ at C'.
Thus, {tex}\triangle{/tex}AB'C' is the required similar triangle.

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Find the centre of a circle passing through the points (6,-6) (3,-7) and(3,3)

Posted by Khushi Agrawal 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let the given points are A(6,-6), B(3,-7), C(3,3)
And P(x,y) be the centre of the circle. So, AP = BP = CP (radii of the circle)

Taking AP = BP and Squaring both sides, we get,
{tex}\Rightarrow A P ^ { 2 } = B P ^ { 2 }{/tex}
{tex}\Rightarrow ( x - 6 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 }{/tex}(by using distance formula)
{tex}\Rightarrow x ^ { 2 } - 12 x + 36 + y ^ { 2 } + 12 y + 36{/tex} ={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex}
{tex}\Rightarrow - 12 x + 6 x + 12 y - 14 y + 72 - 58 = 0{/tex}
{tex}\Rightarrow - 6 x - 2 y + 14 = 0{/tex}
{tex}\Rightarrow 3 x + y - 7 = 0{/tex} ……….(i)

Again, taking BP = CP and squaring both sides, we get,
{tex}\Rightarrow B P ^ { 2 } = C P ^ { 2 }{/tex}
{tex}\Rightarrow ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y - 3 ) ^ { 2 }{/tex}
{tex}\Rightarrow x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex} ={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } - 6 y + 9{/tex}
{tex}\Rightarrow - 6 x + 6 x + 14 y + 6 y + 58 - 18 = 0{/tex}
{tex}\Rightarrow 20 y + 40 = 0{/tex}
{tex}\Rightarrow y = - 2{/tex}

Putting the value of y in eq. (i),
{tex}3x + y - 7 = 0{/tex}
{tex}\Rightarrow 3 x = 9{/tex}
{tex}\Rightarrow x = 3{/tex}
Hence the coordinates of P i.e. centre of circle are (3,-2).

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The ten's digit of a two digit number is 3 times the unit digit. The sum of the numbers and the unit digit is 32. Find the numbers

Posted by Aryan Yadav 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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Tik tok is the best app for us if agree then write yes and like

Posted by Jayjagdish Rout 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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If SecQ + TanQ=p Find the value of CosecQ

Posted by Ishant Kumawat 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Ishant Kumawat 6 years, 9 months ago

This is wrong

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Aryan Yadav 6 years, 9 months ago

Sec + tan=p 1/cos+sin/cos=p Sin=pcos-1 We know that cosec =1/sin Therefore,cosec=1/pcos-1

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Khy last 5 years ke board question paper mil sakate h

Posted by Swati Garhewal 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Usha ❤❤❤ 6 years, 9 months ago

I think Google pr mil jayenge..bt pichle saal hi to board pattern chnge hua h to nxt year k sample paper hi bst choice h

1Thank You

Rajputana Rajputana 6 years, 9 months ago

Yes

1Thank You

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How to guess that whether in median table frequency is given or CF (if it is not particularly mentioned)

Posted by Xyz Abc 6 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Vaibhav Verma 6 years, 9 months ago

If the series is in ascending order then it is cf &if not it is frequency

2Thank You

Shashi Prabha 6 years, 9 months ago

if table is less than type then given frequency is cf. otherwise it is f. in morethan type it is not f.

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Alok Kumar 6 years, 9 months ago

Commulative frequency

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If the roots of the equation (b-c)x²+(c-a)x+(a-b)=0 are equal prove that 2b=a+c

Posted by Pragya Bharti 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Alok Kumar 6 years, 9 months ago

Here; quadratic equation is in the form of ax sqaure +bx +c =0..then..a=(b-c),b=(c-a),c=(a-b)

1Thank You

Diya Namdev 6 years, 9 months ago

(b-c)+(a-b) b+b=a+c 2b=a+c

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If the quadratic px2-2underroot5+15 =0 has two equal roots then find the value of p

Posted by Mohd Moazzam 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Bahubali Gaming 6 years, 9 months ago

Underroot 5x hai ki nahi

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The sum of 4 concucutive intergers is 32 and the ratio of the 1st and last with two middle terms is 7:15 find 4 numbets

Posted by Kaustubh Pathade 6 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Kaustubh Pathade 6 years, 9 months ago

How

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Satyanarayan Sharma In 6 years, 9 months ago

Let the consecutive interger is x

1Thank You

Honey ??? 6 years, 9 months ago

2,6,10,14

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A ladder rests against a wall at angle alpha to the horizontal. Its foot is pulled away from the wall through a distance 'a', so that it slides a distance 'b' down the wall making an angle beta with the horizontal. Show that a/b=(cos alpha -cos beta) /sin beta - sin alpha

Posted by Anushka Tiwari 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let PQ be the ladder such that is top Q is on the wall OQ.
The ladder is pulled away from the wall through a distance a, so Q slides and takes position Q'.
Clearly, {tex}PQ = P'Q'.{/tex}
In {tex}\Delta 's{/tex} {tex}POQ \ and \ P'OQ', {/tex}we have
{tex}\sin \alpha = \frac{{OQ}}{{PQ}},\cos \alpha = \frac{{OP}}{{PQ}},\sin \beta = \frac{{OQ'}}{{P'Q'}},\cos \beta = \frac{{OP'}}{{P'Q'}}{/tex}
{tex} \Rightarrow \sin \alpha = \frac{{b + y}}{{PQ}},\cos \alpha \frac{x}{{PQ}},\sin \beta = \frac{y}{{PQ}},\cos \beta = \frac{{a + x}}{{PQ}}{/tex}
{tex} \Rightarrow \sin \alpha - \sin \beta = \frac{{b + y}}{{PQ}} - \frac{y}{{PQ}}{/tex} and
{tex}\cos \beta - \cos \alpha = \frac{{a + x}}{{PQ}} - \frac{x}{{PQ}}{/tex}
{tex} \Rightarrow \sin \alpha - \sin \beta = \frac{b}{{PQ}}{/tex} and
{tex}\cos \beta - \cos \alpha = \frac{a}{{PQ}}{/tex}
{tex} \Rightarrow \frac{a}{b} = \frac{{\cos \alpha - \cos \beta }}{{\sin \beta - \sin \alpha }}{/tex}