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Ask QuestionPosted by Naman Sharma 7 years, 1 month ago
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Posted by Shivam Tiwari 7 years, 1 month ago
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Sujal Rai 7 years, 1 month ago
Yogita Ingle 7 years, 1 month ago
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
Posted by Sonu Kumar Madheshya 7 years, 1 month ago
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Sujal Rai 7 years, 1 month ago
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Yogita Ingle 7 years, 1 month ago
2x - 3y + a = 0
2(2) - 3(3) + a = 0
4 - 9 + a = 0
a = 5
2x + 3y - b + 2 = 0
2(2) + 3(3) - b + 2 = 0
4 + 9 - b + 2 = 0
15 - b = 0
b = 15
Posted by Tushar Singh Rajput 5 years, 8 months ago
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Posted by Lav Agraawal 7 years, 1 month ago
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Yogita Ingle 7 years, 1 month ago
Tangent: A tangent is a line that touches a circle at only one point. A tangent is perpendicular to the radius at the point of contact. The point of tangency is where a tangent line touches the circle.
Secant: When the line touches the circle at two points: It is called a secant. A secant line intersects the circle in two points.
Anurag Ojha 7 years, 1 month ago
Posted by Ankit Rajput 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
In {tex}\Delta{/tex}BMC and {tex}\Delta{/tex} EMD, we have
MC = MD [{tex}\because{/tex} M is the mid-point of CD]
{tex}\angle \mathrm { CMB } = \angle E M D{/tex} [Vertically opposite angles]
and, {tex}\angle M B C = \angle M E D{/tex} [Alternate angles]
So, by AAS-criterion of congruence, we have
{tex}\therefore \quad \Delta B M C \cong \Delta E M D{/tex}
{tex}\Rightarrow{/tex} BC = DE .......(i)
Also, AD = BC [{tex}\because{/tex} ABCD is a parallelogram] ...... (ii)
Adding (i) and (ii),we get,
AD + DE = BC + BC
{tex}\Rightarrow{/tex} AE = 2 BC ...(iii)
Now, in {tex}\Delta{/tex}AEL and {tex}\Delta{/tex}CBL, we have
{tex}\angle A L E = \angle C L B{/tex} [Vertically opposite angles]
{tex}\angle E A L = \angle B C L{/tex} [Alternate angles]
So, by AA-criterion of similarity of triangles, we have
{tex}\Delta A E L \sim \Delta C B L{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { A E } { C B }{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { 2 B C } { B C }{/tex} [Using equations (iii)]
{tex}\Rightarrow \quad \frac { E L } { B L } = 2{/tex}
{tex}\Rightarrow{/tex} EL = 2 BL
Posted by Manish Kumar 7 years, 1 month ago
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Sia ? 6 years, 6 months ago
Check sample papers here : <a href="https://mycbseguide.com/cbse-sample-papers.html">https://mycbseguide.com/cbse-sample-papers.html</a>
Posted by Arpan Ghule 7 years, 1 month ago
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Sia ? 6 years, 6 months ago
Get RD Sharma solutions with notes : <a href="https://mycbseguide.com/course/cbse-class-10-mathematics/1202/">https://mycbseguide.com/course/cbse-class-10-mathematics/1202/</a>
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