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Jashith Sharma 5 years, 3 months ago
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Gaurav Seth 5 years, 1 month ago
Given: A quadrilateral ABCD in which AB ║ DC and AB = DC.
To Prove: ABCD is a parallelogram
i.e., AB ║ DC and AD ║ BC.
Construction: Join A and C.
Proof : Since AB is parallel to DC and AC is transversal
∠BAC = ∠DCA [Alternate angles]
AB = DC [Given]
And AC = AC [Common side]
⇒ ∆BAC ≅ ∆DCA [By SAS]
⇒ ∠BCA = ∠DAC [By cpctc]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
⇒ AD ║ BC
Now, AB ║ DC (given) and AD ║ BC [Proved above]
⇒ ABCD is a parallelogram
Posted by Samruddhi Helavi 5 years, 3 months ago
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Shaikh Kumari 5 years, 3 months ago
Posted by Vishesh King 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
let x-2=0
then, x=2
put x in (i)
3(2)(2)+6(2)-24=0
12+12-24=0 {use BODMAS rule for solution}...
24-24=0
0=0
Posted by Anushka Raj 5 years, 3 months ago
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Posted by Vaibhav Gavande 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
x + y = 8
xy = 15.
{tex}( x+y )^2 = x^2 + y^2 + 2xy{/tex}
{tex}x^2 + y^2 = ( x + y )^2 - 2xy{/tex}
{tex}= 8^2 - 2×15{/tex}
= 64 - 30
{tex}x^2 + y^2 = 34.{/tex}
... Kumari 5 years, 3 months ago
Posted by Hisham Ichoo 5 years, 3 months ago
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Posted by Naveen Mk 5 years, 3 months ago
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Abhishek Chaudhary 5 years, 3 months ago
2Thank You