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Ayush Raj 6 years, 2 months ago
Posted by Tejas Agrawal 6 years, 2 months ago
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Gaurav Seth 6 years, 2 months ago
S=30+30+48/2
s=54
area=√54(54-30)(54-30)(54-48)
area=432m²
area×2/18
432×2/18
864/18
48m²
Therefore,every cow will get 48m² to graze
Posted by Dilip Kumar 6 years, 2 months ago
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Gaurav Seth 6 years, 2 months ago
let x,y,z be the angle of a triangle
now, if x+y+z=180
if x=90,theny+z=90,x=y+z
then it is right angle triangle
if x>y+z
then x must be greater than90
otherwise either x is 90 or less than90
so it is a abtuse angle triangle
Posted by Kamal Jyoti 6 years, 2 months ago
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Posted by The Mesmerizers 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Given √2 is irrational number.
Let √2 = a / b wher a,b are integers b ≠ 0
we also suppose that a / b is written in the simplest form
Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2
∴ 2b2 is divisible by 2
⇒ a2 is divisible by 2
⇒ a is divisible by 2
∴ let a = 2c
a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2
∴ 2c2 is divisible by 2
∴ b2 is divisible by 2
∴ b is divisible by 2
∴a are b are divisible by 2 .
this contradicts our supposition that a/b is written in the simplest form
Hence our supposition is wrong
∴ √2 is irrational number.
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Sia ? 6 years, 2 months ago
Given : ABCD is a parallelogram.
BC=CQ.
To prove :
ar(ΔBCP)=ar(ΔDPQ)
Construction :
Join AC
Proof :
ar(ΔBPC) = ar (ΔAPC) [Triangles on the same base and between same parallels ]
Similarly ,
ar(ΔADC) = ar(ΔADQ) [Triangles on the same base and between same parallels ]
ar(ΔADC) = ar(ΔADP) +ar(ΔAPC)
ar(ΔADQ) = ar(ΔADP) + ar(ΔDPQ)
ar(ΔADC) = ar(ΔADQ) and ,
ar(ΔADP) is common.
{tex}\therefore{/tex} ar(ΔAPC) = ar(ΔDPQ)
But ,
ar(ΔAPC) = ar(ΔBPC)
∴ ar(ΔBPC) = ar(ΔDPQ)
Hence ,proved.
Posted by Gulafshan Nayeem 6 years, 2 months ago
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Ankit Raj 6 years, 2 months ago
Posted by Gulafshan Nayeem 6 years, 2 months ago
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Anku Singla 6 years, 2 months ago
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