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Samridhi ?? 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
Altitude of equilateral triangle = h cm
Let the side of triangle = x cm
As we all know the altitude bisects the side of triangle .
A.T.Q
Area of angled triangle ADC,
area = 1/2 × base × height
= 1/2 × x/2 ×h
= 2xh/2
= xh cm²
Area of triangle ABC = 2 × area of triangle ADC
= 2xh cm²
Posted by Rinku Jain Dilip Jain 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
1. First, we draw a line OA of length 1
2. Now, we draw a perpendicular of length 1 on point A as AB
3. From, Pythagoras Theorem, OB = √2
4. Now, take an arc of length OB, and draw it on the number line which meets as E.
So, at E, we can represent √2 as shown in the figure.
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Posted by Prateek Kumar 5 years, 5 months ago
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Sia ? 5 years, 5 months ago
Whole numbers are positive numbers, including zero, without any decimal or fractional parts. They are numbers that represent whole things without pieces. The set of whole numbers is represented mathematically by the set: {0, 1, 2, 3, 4, 5...}.
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Sia ? 5 years, 5 months ago
Given, {tex}2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }{/tex}
We know that {tex}a ^ { m } \cdot a ^ { n } = a ^ { ( m +n) }{/tex}.
So, {tex}2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } } = ( 2 ) ^ { \frac { 2 } { 3 } + \frac { 1 } { 3 } }{/tex}
{tex}= ( 2 ) ^ { \frac { 2+1} { 3} }\\ = ( 2 ) ^ { \frac { 3 } { 3 } }\\=(2)^1\\=2{/tex}
Therefore, the value of {tex}2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }{/tex} will be {tex}2{/tex}.
Posted by Mahi Vats 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
The distance of a point from x-axis is equal to its y coordinate.
So, the distance of the point (-3, -2) from x-axis is 2 units.
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Abhsiehk Kumar 5 years, 5 months ago
1Thank You