The largest area of the rectangle is 2 square units.
We have a right angled isosceles triangle with hypotenuse 4 units.
so its base will be equal to the perpendicular , let it be B
So we know ,
B² + P² = H²
2B² = H²
B² = 16/2
B = 2√2
So base is equal to perpendicular = 2√2 units
Now let consider a rectangle inside the triangle with length 2x and the width y , and the area will become
A = 2xy ....(1)
we see from the figure the length of the perpendicular AD drawn on the hypotenuse is
BD² + AD² = AB²
AD² = AB² - BD²
= 8 - 4
AD² = 4
AD = 2 units
Now we see triangle ABD and AEG are similar to each other , So
AD/BD = AG/GE
2/2 = 2-y/x
therefore
x = 2-y
put this value in eqn (1)
A = 2(2-y)y
A = 4y - 2y²
Now to maximize the area we take the derivative of the equation and then put it equal to zero
∴ 4 - 4y = 0
y = 1
∴ x = 2-y = 2-1 = 1 units
So area will be maximum when length and width are 1 units each.
therefore , largest area (A) of the rectangle is = 2xy = 2 square units.
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