Question

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​

Answer 1

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]