A rectangle is to be inscribed in an isosceles right triangle in such a way that one vertex of the rectangle is the intersection point of the legs of the triangle and the opposite vertex lies on the hypotenuse. Find the largest area (in cm 2 ) of the rectangle and its dimensions (in cm) given that the two equal legs of the triangle have length 4.

x  =  2  cm

y  = 2  cm

A(max) =  4 cm²

Step-by-step explanation: See Annex

The right isosceles triangle has two 45° angles and the right angle.

tan 45°  =  1  =  x / 4 - y        or     x  =  4  -  y     y  =  4  -  x

A(r)  =  x* y

Area of the rectangle as a function of x

A(x)  =  x  *  (  4  -  x )       A(x)  =  4*x  -  x²

Tacking derivatives on both sides of the equation:

A´(x)  =  4 - 2*x             A´(x)  =  0            4   -  2*x  =  0

2*x  =  4

x  =  2  cm

And  y  =  4  - 2  =  2  cm

The rectangle of maximum area result to be a square of side 2 cm

A(max)  = 2*2  =  4 cm²

To find out if A(x) has a maximum in the point  x  =  2

We get the second derivative

A´´(x)  =  -2           A´´(x)  <  0   then A(x) has a maximum at  x = 2

answer: b)

step-by-step explanation:

its a dot plot because since the score is not high numbers.

- r3ktforgood ☕

isocolese triangle

we know that the line connect a and b is the hyptonuse

so first find the hypotonuse length using the distance formula, call the length c

since it is an isocolese triangle, the lengths of the legs are the same

using pythagorean theorem,

in this case, a=b since isocolese triangle so we get

or

we can find c so we can find the distances

we know that if we draw 2 circles, each has center at a and b, and find where they intersect, we can the 2 possible locations for c

if a has center at (xa,ya) and b has center at (xb,yb), then we can solve where

and

intersect to find where they intersect

i'm not going to put an example because i have to go to class. i hope this explanation is enough