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Yogita Ingle 6 years, 10 months ago
Let √2 + √3 = (a/b) is a rational no.
On squaring both sides , we get
2 + 3 + 2√6 = (a2/b2)
So,5 + 2√6 = (a2/b2) a rational no.
So, 2√6 = (a2/b2) – 5
Since, 2√6 is an irrational no. and (a2/b2) – 5 is a rational no.
So, my contradiction is wrong.
So, (√2 + √3) is an irrational no.
Posted by Arsh Raj 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Construction : Join A to B
we have,
OP = diameter
{tex}\Rightarrow {/tex} OQ + QP = diameter
{tex}\Rightarrow {/tex} Radius + QP = diameter
{tex}\Rightarrow {/tex} OQ - PQ = radius
Thus OP is the hypotenuse of right angled {tex}\triangle {/tex}AOP
So, In {tex}\angle A O P , \sin \theta = \frac { A O } { O P } = \frac { 1 } { 2 }{/tex}
{tex}\theta = 30 ^ { \circ }{/tex}
Hence, {tex}\angle A P B = 60 ^ { \circ }{/tex}
Now, In {tex}\triangle A O P{/tex}
AP = AB
So, {tex}\angle P A B = \angle P B A = 60 ^ { \circ }{/tex}
{tex}\therefore \triangle A P B{/tex} is an equilateral triangle.
Posted by #Aditi~ Angel???? 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Volume of the bowl {tex}= \frac { 2 } { 3 } \pi r ^ { 3 }{/tex}
Volume of liquid in the bowl {tex}= \frac { 2 } { 3 } \pi \times ( 18 ) ^ { 3 } \mathrm { cm } ^ { 3 }{/tex}
Volume of liquid left after wastage {tex}= \frac { 2 } { 3 } \pi \times ( 18 ) ^ { 3 } \times \frac { 90 } { 100 } \mathrm { cm } ^ { 3 }{/tex}
Volume of one bottle {tex}= \pi r ^ { 2 } h{/tex}
Now ,According to question
Volume of 72 bottles = volume of liquid after wastage
{tex}\Rightarrow{/tex}{tex}\pi \times ( 3 ) ^ { 2 } \times h \times 72 = \frac { 2 } { 3 } \pi \times ( 18 ) ^ { 3 } \times \frac { 90 } { 100 }{/tex}
{tex}\Rightarrow{/tex}{tex}\pi \times ( 9) \times h \times 72 = \frac { 2 } { 3 } \pi \times ( 18 ) ^ { 3 } \times \frac { 90 } { 100 }{/tex}
{tex}\Rightarrow{/tex}{tex}h = \frac { \frac { 2 } { 3 } \pi \times ( 18 ) ^ { 3 } \times \frac { 90 } { 100 } } { \pi \times ( 9 ) \times 72 }=5.4cm{/tex}
Hence, height of each bottles is 5.4 cm
Posted by Priya Mehta 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
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Given: In ∆ ABC, AD is the median
Construction: Draw AE ⊥BC
Now since AD is the median
∴ BD = CD =
BC ....... (1)
In ∆ AED
AD2 = AE2 + DE2 (Pythagoras theorem)
⇒ AE2 = AD2 – DE2 ......... (2)
In ∆ AEB
AB2 = AE2 + BE2
= AD2 – DE2 + BE2 (from (2))
= (BD + DE)2 + AD2 – DE2 (∵ BE = BD + DE)
= BD2 + DE2 + 2BD·DE + AD2 – DE2
= BD2 + AD2 + 2·BD·DE
In ∆ AED
AC2 = AE2 + EC2
= AD2 – DE2 + EC2 (from (5))
= AD2 – DE2 + (DC – DE)2
= AD2 – DE2 + DC2 + DE2 – 2DC·DE
= AD2 + DC2 – 2DC·DE
Adding (3) and (4) we get
AB2 + AC2 =
BC2 + AD2 + BC·DE + AD2 +
BC2 – BC·DE
⇒ 2 (AB2 + AC2) = BC2 + 4AD2
⇒ 4AB2 + BC2 = 2AB2 + BC2
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Guddan_ ?? 6 years, 10 months ago
2Thank You