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

Given, {tex}\triangle ABC{/tex} in which {tex}AB = BC = CA{/tex} and D is a point on BC such that BD ={tex}\frac { 1 } { 3 }{/tex}BC.
Prove: 9AD² = 7AB².
Construction :Draw {tex}A L \perp B C{/tex}
Proof: In right triangles ALB and ALC, we have
{tex}AB = AC {/tex}(given)
{tex}AL = AL {/tex}(common)
{tex}\therefore \triangle A L B \cong \triangle A L C{/tex} [by RHS axiom]
So, {tex}BL = CL.{/tex}
Thus, BD ={tex}\frac { 1 } { 3 }{/tex}BC and BL ={tex}\frac { 1 } { 2 }{/tex}BC.
In {tex}\triangle ALB{/tex}, {tex}\angle A L B = 90 ^ { \circ }{/tex}
By using pythagoras theorem, we get
{tex}\therefore{/tex} AB² = AL² + BL² ...(i)
In {tex}\triangle ALD{/tex}, {tex}\angle A L D = 90 ^ { \circ }{/tex}
By using pythagoras theorem, we get
{tex}\therefore{/tex} AD² = AL² + DL²
= AL² + (BL - BD)²
= AL² + BL² + BD² - {tex}2 B L \cdot B D{/tex}
= (AL² + BL²) + BD² - {tex}2 B L \cdot B D{/tex}
= AB² + BD² - {tex}2 B L \cdot B D{/tex} [using (i)]
{tex}= B C ^ { 2 } + \left( \frac { 1 } { 3 } B C \right) ^ { 2 } - 2 \left( \frac { 1 } { 2 } B C \right) \cdot \frac { 1 } { 3 } B C{/tex}
{tex}= B C ^ { 2 } + \frac { 1 } { 9 } B C ^ { 2 } - \frac { 1 } { 3 } B C ^ { 2 }{/tex}
{tex}= \frac { 7 } { 9 } B C ^ { 2 } = \frac { 7 } { 9 } A B ^ { 2 }{/tex} [{tex}\because{/tex} BC = AB]
Therefore, 9AD² = 7AB².

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Sin theta is equal to cos theta then find the value of 2 tan theta plus Cos theta

Posted by Sandeep Chaudhary 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Sandeep Chaudhary 6 years, 11 months ago

Thanks for answering my question

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Guddu Guddu 6 years, 11 months ago

Sin square theta=cos square theta (squaring on both sides) 1_ cos square theta= cos square theta (sin square theta =1_ cos square theta) 1= cos square theta +cos square theta 1=2cos square theta 1/2=cos square theta 1/underroot2=cos theta. 1/2=cos square theta 1/2=1_sin square theta. Sin square theta =1_1/2=1/2. Sin theta=1/underroot2. 2 tan theta+cos theta = 2sin theta /cos theta +cos theta =2*1/underroot2 /1/underroot2 +1/underroot2 =2*1+1/underroot2 =2+1/underroot2=2*underroot2 +1/underroot2

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Gautãm . 6 years, 11 months ago

2+cos(theta) or 2+sin(theta)

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solve x and y x+y=3 4x_3y=26

Posted by Shahid Hassan 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Yogita Ingle 6 years, 11 months ago

x + y = 3
x = 3 - y........ (i)

4x - 3y = 26...... (ii)
Put (i) in (ii)
4(3 - y) - 3y = 26
12 - 4y - 3y = 26
-7y = 26 - 12
-7y = 14
y = -2
Y = -2 in (i)
x = 3 - (-2)
x = 3 + 2
x = 5

1Thank You

Guddu Guddu 6 years, 11 months ago

X+y=3 .......... (1) 4x_3y=26..........(2). Multiply equation (1) by 3& equation (2) by1 to make coefficient of y equal 3x+3y=9...........(3). 4x_3y=26..........(4). From equation (3)&(4). X=5&y=_2

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Gautãm . 6 years, 11 months ago

x=5, y=-2

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I've revised NCERT 4 to 5 time but haven't even touched R. D. Sharma What should i do? Do i need to study R. D. Sharma?

Posted by Janhvi Singh 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Abc . 6 years, 11 months ago

No i don't think so that you should study RD Sharma.. Because the question are asked from ncert and ncert exempler only

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Avinash Kumar 6 years, 11 months ago

Halka phulka ☺️☺️

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QR is a tangent at Q, PR parallel PQ where AQ is a chord prove BR is tangent at B

Posted by Parag Dubey 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find roots of quadratic equation: x - 1/x=3 , x is not equal to 0

Posted by Nonu Kansal 6 years, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Prashant Chaudhary 6 years, 11 months ago

Hlo mehak Aapne mujhe pehchana ki nahi

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Ram Kushwah 6 years, 11 months ago

x2-1=3x

x2-3x-1=0

{tex}x = {3\pm \sqrt{9+4} \over 2}={3\pm\sqrt{13}\over 2}{/tex}

1Thank You

Prabhjeet Singh 6 years, 11 months ago

X-1/3=3 x(x)-1/x=3 x2-1/x=3 x2-1=3x x2-3x-1=0 comparing equation with ax2+bx+c=0 a=1,b=-3,c=-1 answer

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Nonu Kansal 5 years, 8 months ago

Please answer me

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Find 25th term of the A.P , -5 , -5/2 , 0 ,5/2.

Posted by Nonu Kansal 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Dhyanu Kumar 6 years, 11 months ago

n=25 a=-5 d=a2-a1 d=-5/2-(-5) d=-5/2+5=-5+10/2=5/2 an=a+(n-1)d a25=-5+(25-1)5/2 =-5+12*5 =-5+60 a25=55

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Nonu Kansal 6 years, 11 months ago

Please answer me?

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If x=a, y=b is the solution of the pair of equation x-y=2 and x +y=4 find the value of a and b

Posted by Mohamed Jabrin 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Dhyanu Kumar 6 years, 11 months ago

x=a=3 y=b=1

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Dhyanu Kumar 6 years, 11 months ago

Given= x=a and y=b The equation are x-y=2. equation-1 x+y=4. equation-2 Adding equation 1 form equation 2 x+y=4 x-y=2 2x=6 x=6/2 x=3 Substitute the value of x in eq 2 x+y=4 3+y=4 y=4-3 y=1

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Yogita Ingle 6 years, 11 months ago

x = a , y = b

x - y = 2
a - b = 2....... (i)
x + y = 4
a + b = 4 ....... (ii)
Add (i) and (ii)
2a = 6
a = 6/2 = 3
But a = 3 in (ii)
3 + b = 4
b= 4 - 3
b = 1
So, a = 3 and b = 1

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Show that for odd positive integer to be a perfect square ,it should be of the form 8k+1

Posted by Zain C 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Since, any odd positive integer n is of the form 4m + 1 or 4m + 3.
if n = 4m + 1
n² = (4m + 1)²
= 16m² + 8m + 1
= 8(2m² + m) + 1
So n² = 8q + 1 ........... (i)) (where q = 2m² + m is a positive integer)
If n = (4m + 3)
n² = (4m + 3)²
= 16m² + 24m + 9
= 8(2m² + 3m + 1) + 1
So n² = 8q + 1 ...... (ii) (where q = 2m² + 3m + 1 is a positive integer)
From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.

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Q. If the points (p, q); (m, n) and (p-m, q-n) are collinear. Show that pn=qm.

Posted by Ram Swaroop 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given points are collinear. Therefore

[p {tex}\times{/tex} n + m(q - n) + (p - m) q] - [m {tex}\times{/tex} q + (p - m) n + p (q - n)] = 0
{tex}\Rightarrow{/tex} (pn + qm - mn + pq - mq) - (mq + pn - mn + pq - pn) = 0
{tex}\Rightarrow{/tex} (pn + p q - mn) - (mq - mn + pq) = 0
{tex}\Rightarrow{/tex} pn - mq = 0
{tex}\Rightarrow{/tex} pn = qm

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The points A(1,-2),B(2,3),C(k,2)andD(-4,-3) are the vertices of a parallelogram. Find the value of k.

Posted by Ayush Patel 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

PlZ yaar this app is use to homework help no personal chat plzzzzzzzzzzzzz

Posted by Aj Raghav 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Cutie Pie ? 6 years, 11 months ago

U r right Aryan. ?

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Find the roots of the equation 5x-6x-2=0 by the method of completing the square

Posted by Student Entertainment 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

5x² - 6x - 2 = 0
Multiplying the above equation by 1/5
{tex} \Rightarrow {x^2} - \frac{6}{5}x - \frac{2}{5} = 0{/tex}
{tex}\Rightarrow x ^ { 2 } - \frac { 6 } { 5 } x + \left( \frac { 3 } { 5 } \right) ^ { 2 } - \left( \frac { 3 } { 5 } \right) ^ { 2 } - \frac { 2 } { 5 } = 0{/tex}
{tex}\Rightarrow \left( x - \frac { 3 } { 5 } \right) ^ { 2 } = \frac { 9 } { 25 } + \frac { 2 } { 5 }{/tex}
{tex}\Rightarrow \left( x - \frac { 3 } { 5 } \right) ^ { 2 } = \frac { 9 + 10 } { 25 }{/tex}
{tex}\Rightarrow \left( x - \frac { 3 } { 5 } \right) ^ { 2 } = \frac { 19 } { 25 }{/tex}
{tex}\Rightarrow x - \frac { 3 } { 5 } = \pm \frac { \sqrt { 19 } } { 5 }{/tex}
{tex}\Rightarrow x = \frac { 3 } { 5 } \pm \frac { \sqrt { 19 } } { 5 }{/tex}
{tex}\Rightarrow x = \frac { 3 + \sqrt { 19 } } { 5 } \text { or } x = \frac { 3 - \sqrt { 19 } } { 5 }{/tex}

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Kx+3y=2k+1 & 2(k+1)x+9=7k+1 find k

Posted by Vivek Chaurasiya 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Vivek Chaurasiya 6 years, 11 months ago

Infinity many solution

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Amresh Shukla 6 years, 11 months ago

Mention hi ni hai ki .....infinitely many hai ya exactly one...ya no solutions....??

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Vivek Chaurasiya 6 years, 11 months ago

Solution fast dena jaldi h

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find the middle term of the AP 6 13 20 ................216

Posted by Vineet Jain 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Tanya Gupta 6 years, 11 months ago

a=6, d=7 an= a+(n-1)×d 216=6+(n-1)x7 216=6+7n-7 216=-1+7n 217 = 7n n= 217/7 n= 31 total no. of terms = 31 middle term=1/2 × (31+1) = 1/2 × 32 = 16th a16 = 6+(16-1)×7 = 6+15×7 = 6+105 111

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Prove that the intercept of a tangent between two parallel tangent to a circle subtends a right angle at centre

Posted by Gauri Suthar 6 years, 11 months ago

CBSE > Class 10 > Mathematics