Someone me in this question it is kind of confusing it is a question under pre calculus in first year calculus i will post the picture after this.. so the question is . the accompanying figure shows a rectangle inscribed in a isoceles right triangle whose hypotenuse is 2 units long.. a) express the y-coordinate of p in terms of x (you might start by writing and equation for the line ab). b) express the area of the rectangle in terms of x.

Neither P, nor A are on the sketch 
I guess P is the upper right corner of the rectangle 
and A=(0,1) 

P belongs to the line going through (1,0) and B(0,y) 
but we don't know the y-coordinate of B 

the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 
now look at the right triangle BOA 
his hypotenuse is AB=sqrt2 and the the kathete OA is 1 
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1 
so the coordinates of B are (0,1) 

the line going through (1,0) and (0,1) is L(x)=-x+1 

P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1) 

b) so if that's P, the height of the rectangle is -x+1 and the width=2x 
so its area A(x)=2x*(-x+1)= -2x²+2x 

I hope my answer has come to your help. Thank you for posting your question here in .

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