Question

pls tell the complete solution for this please

Answer 1

sosososo sorry for I don't know

Answer 2

Answer:

Range of f(x) is (0,2]

Explanation:

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 × base× height

1/2 × 2✓2 +✓2

= 2

So, the Range of f(x) is (0,2]

$hope \: \: it \: \: helps \: \: you$

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