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Gaurav Seth 4 years, 3 months ago
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Let ABCD and EFCD be two parallelograms lying on the same base CD and between the same parallels AF and CD.
We have to prove that area of the parallelograms ABCD and EFCD are equal.
In 
ADE and BCF,
DAE = CBF (Corresponding angles)
AED = BFC (Corresponding angles)
AD = BC (Opposite sides of the parallelogram ABCD)
ADE 
BCF (By ASA congruence rule)
We know that congruent figures have the same area.
area (ADE) = area (BCF) … (1)
Now, we have:
area (ABCD) = area(ADE) + area (EDCB)
= area (BCF) + area (EDCB) [Using (1)]
= area (EDCF)
Thus, the areas of the two parallelograms ABCD and EFCD are the same.
Posted by Anchal Mishra 4 years, 3 months ago
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Posted by Rushil Pandya 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Curved Surface Area / Lateral Surface Area of Cube : 4a²
where a is the side of cube
Total Surface area of Cuboid :
2 ( lb + bh + hl ) where l is length , b is breadth and h is height .
Posted by M. Pranesh 4 years, 3 months ago
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Yogita Ingle 4 years, 3 months ago
Area of a ΔABC, given sides a, b, c by Heron’s formula (Also known as Hero’s Formula) :
Find semi perimeter (s ) = (a+b+c)/2
Area=√[s(s-a)(s-b)(s-c)]
Posted by Vijay Laxmi Sharma 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Question : If x=2-√3, find the value of (x-1/x) of power 3
Answer :
x = 2 - √3--( 1 )
1/x = 1/( 2 - √3 )
= ( 2 + √3 ) / [ ( 2 - √3 )(2 + √3 ) ]
= ( 2 + √3 ) / [ 2² - ( √3 )² ]
= ( 2 + √3 ) / ( 4 - 3 )
= 2 + √3 ----( 2 )
x - 1/x
= 2 - √3 - ( 2 + √3 )
= 2 - √3 - 2 - √3
= - 2√3 ---( 3 )
Therefore ,
( x - 1/x )³ = ( - 2√3 )³
= - 24√3
Posted by Smita Tripathi 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Given, (x+y)3 - (x-y)3 - 6y (x+y) (x-y)
Answer : 8y3
Explanation:
(x+y)3 - (x-y)3 - 6y (x+y) (x-y)
= (x + y)3 - (x - y)3 - 6y(x2 - y2)
Now, Let x + y = a, x - y = b
Then, ab = (x + y) (x - y) = x2 - y2
and a - b = (x + y) - (x - y) = 2y
Now, (x + y)3 - (x - y)3 - 6y (x2 - y2)
= (x + y)3 - (x - y)3 - 3(2y) (x2 - y2)
= a3 - b3 - 3(a - b) ab
= (a - b)3 | Using Identity VII
= {(x + y) - (x - y)}3
= (2y)3 = 8y3
Posted by Vijay Laxmi Sharma 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Answer : 1 / 9
Explanation :
Given : Expression
To find : The value of the expression .
Solution :
Expression
We can write 81 as ,
Therefore, the value of expression is
Posted by Rohan Mahto 4 years, 3 months ago
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Posted by Afra Muhammed Asafali 4 years, 3 months ago
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Posted by Femi Binod 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
A reflex angle is an angle which is more than 180 degrees and less than 360 degrees.
A reflex angle is any angle that is more than 180 degrees (half circle) and less than 360 degrees (full circle). A reflex angle will always have either an obtuse or an acute angle on the other side of it.
Posted by Krishnanandha Ar 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Long division
x^2 - x + 1
____________
x+1 [ x^3 + 1
[ x^3 + x^2
----------------
-x^2 + 1
-x^2 - x
------------
x + 1
x + 1
----------
0
so, the answer is:
x² - x + 1
Using factor theorm:
a³ + b³ = (a + b)(a² – ab + b²)
So, x³ + 1 = x³ + 1³ = (x+1)(x² - 2x + 1)
and (x³ + 1) / (x+ 1) = (x+1)(x² - 2x + 1) / (x + 1)
The (x + 1) brackets cancel out and we are left with
x² - 2x + 1
Posted by Shalini Maurya 4 years, 3 months ago
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Posted by Sanyam Bhardwaj 4 years, 3 months ago
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Yogita Ingle 4 years, 3 months ago
2x2-7x-15
= 2x2-10x+ 3x -15
= 2x( x -5) + 3 (x -5)
= (2x +3)(x-5)
Posted by Alisha Mariam 4 years, 3 months ago
- 1 answers
Gaurav Seth 4 years, 3 months ago
Simplify (a+2b+3c)^2- (a-2b-3c)^2-6b^2-9bc
(a+2b+3c+a-2b-3c) (a+2b+3c-a+2b+6c) - 3b(2b+3c){a^2-b^2 = (a+b)(a-b)}
= (2a)(2b+6c)-3b(2b+3c)
= 4ab+12ac-3b(2b+3c)
= 4a(b+3c)-3b(2b+3c)
Posted by Swati Tiwari 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
root 3 on a number line
OR
Step 1. Take 1 unit from “0” and draw a perpendicular at that point.
Step 2. Join line ie., hypotenuse
Step 3. Keep compass point on “0” and pencil point on top of triangle and draw an arc.
Step 4. The point obtained is √2
Step 5. Now draw 1 unit perpendicular at √2 and similarly draw hypotenuse.
Step 6. Similarly draw an arc as we did in √2 case and point obtained on number line is √3.
Posted by ꧁ ༒ShꭵveຮH༒ ꧂ 4 years, 3 months ago
- 1 answers
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