Someone please help me in this question it is kind of confusing it is a question under pre calculus in first year calculus i wil
l 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) <span>but we don't know the y-coordinate of B </span>
<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span> now look at the right triangle BOA <span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span> 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 <span>so its area A(x)=2x*(-x+1)= -2x²+2x
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