Question

(c) Express the diagonal of a rectangle, D, in terms of its width, w, if the area of the rectangle is 13 square feet.
D(w) =

Answer 1

we are asked to express the diagonal of a rectangle in terms of the width in which the area of the polygon is equal to 13 square feet. The area of teh rectangle is equal to length times width. That is 13 = lw; diagonal using pythagorean theorem is sqrt ( l2 + w2). Thus d =  sqrt ( (13/w)2 + w2) = sqrt ( (169/w2 + w2) = sqrt ( (169 + w4)/w2) = 1/w  sqrt (169 + w4)

Answer 2

Answer:

Hence, the diagonal of rectangle in terms of width(w) is:

$D(w)=\dfrac{\sqrt{169+w^4}}{w}$

Step-by-step explanation:

Let the length of the rectangle=l

and width of the rectangle is:w.

We know that the diagonal D of the rectangle is given by with the help of the Pythagorean theorem as:

$D=\sqrt{l^2+w^2}-------(1)$

Also the area of rectangle(A) is given as:

$A=l\times w$

Area of rectangle=13=l×w.

$l=\dfrac{13}{w}$

Hence putting the value of l in equation (1) we get:

$D=\sqrt{(\dfrac{13}{w})^2+w^2}=\sqrt{\dfrac{13^2}{w^2}+w^2}\\\\D=\sqrt{\dfrac{169}{w^2}+w^2}=\sqrt{\dfrac{169+w^4}{w^2}}=\dfrac{\sqrt{169+w^4}}{w}$

Hence, the diagonal of rectangle in terms of width(w) is:

$D(w)=\dfrac{\sqrt{169+w^4}}{w}$