in a square ABCD with side AB=2,two points M and N are at adjacent sides of the square such that MN is parallel to diagonal BD.if x is distance of MN from vertex A and f(x)=area of triangle AMN,then range of f(x) us
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Given: Square ABCD with side AB=2, MN is parallel to diagonal BD, f(x)=area of triangle AMN
To find: Range of f(x)
Solution:
- As we have a square ABCD with side 2, then we can find the diagonal length by using pythagoras theorem.
- So by using Pythagoras theorem:
CD² + BC² = DB²
2² + 2² = 8
DB² = 8
DB = √8
DB = 2√2
- Since, we have given that the length of the vertex A to MN is x, we can also say that range of MN is from 0 to 2√2.
- Also x = 2√2/2
= √2
- So area of triangle AMN can start from 0.
- Now, area of triangle is:
= 1/2 x base x height
= 1/2 x 2√2 x √2
= 2
Answer:
So the range of f(x) is (0 , 2]
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