Answer :
We have the following situation:
• The diagonal of a rectangle is equal to 8√2 inches.
,• The width is 8 inches less than the length
And we need to find the dimensions of the rectangle.
Then we can proceed as follows:
1. We know that all the internal angles of a rectangle are right angles, and the diagonals are congruent. We also know that the width of this rectangle is 8 inches less than the length:
[tex]\begin{gathered} w=l-8 \\ \\ d=8\sqrt{2} \end{gathered}[/tex]2. Then we can draw the situation as follows:
3. Now, we can apply the Pythagorean Theorem as follows:
[tex]\begin{gathered} (8\sqrt{2})^2=l^2+w^2=l^2+(l-8)^2 \\ \\ \text{ Then we have:} \\ \\ l^2+(l-8)^2=(8\sqrt{2})^2 \\ \\ l^2+(l-8)^2=8^2(2) \\ \\ l^2+(l-8)^2=64(2)=128 \\ \\ l^2+(l-8)^2=128 \end{gathered}[/tex]4. We have to expand the binomial expression on the left side of the equation:
[tex]\begin{gathered} l^2+l^2-2(8)(l)+8^2=128 \\ \\ 2l^2-16l+64=128 \\ \\ 2l^2-16t+64-128=0 \\ \\ 2l^2-16l-64=0 \end{gathered}[/tex]5. Now, we need to apply the quadratic formula to find the value of l as follows:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},ax^2+bx+c=0[/tex]6. Then we have that:
[tex]\begin{gathered} 2l^{2}-16l-64=0 \\ \\ a=2,b=-16,c=-64 \\ \\ \text{ Then we have:} \\ \\ x=\frac{-(-16)\pm\sqrt{(-16)^2-4(2)(-64)}}{2(2)} \\ \\ x=\frac{16\pm\sqrt{256+512}}{2(2)}=\frac{16\pm\sqrt{768}}{4} \end{gathered}[/tex]7. Now, to simplify the radicand, we need to find the factors of 768:
[tex]768=2^8*3[/tex]Now, we have:
[tex]\begin{gathered} \sqrt{768}=\sqrt{2^8*3}=(2^8*3)^{\frac{1}{2}}=2^{\frac{8}{2}}*3^{\frac{1}{2}}=2^4*\sqrt{3}=16\sqrt{3} \\ \\ \sqrt{768}=16\sqrt{3} \end{gathered}[/tex]8. Then the values for l are two possible ones:
[tex]\begin{gathered} x=\frac{16\pm\sqrt{768}}{4}=\frac{16\pm16\sqrt{3}}{4} \\ \\ x=\frac{16+16\sqrt{3}}{4},x=\frac{16-16\sqrt{3}}{4} \\ \\ x_1=\frac{16}{4}(1+\sqrt{3}),x_2=\frac{16}{4}(1-\sqrt{3}) \\ \\ x_1=4(1+\sqrt{3}),x_2=4(1-\sqrt{3}) \end{gathered}[/tex]We can see that x2 gives us a negative value, and since we are finding a length, which is a positive value, then the value for l is:
[tex]\begin{gathered} x_1=4(1-\sqrt{3})\approx−2.92820323028 \\ \\ x_2=4(1+\sqrt{3})\approx10.9282032303\text{ }\rightarrow\text{ This is the value we are finding.} \\ \text{ This is positive.} \end{gathered}[/tex]9. Now, we have that:
[tex]\begin{gathered} l=4(1+\sqrt{3}) \\ \\ \text{ Since }w=l-8,\text{ then we have:} \\ \\ w=4(1+\sqrt{3)}-8=4+4\sqrt{3}-8=4-8+4\sqrt{3}=-4+4\sqrt{3} \\ \\ w=4(-1+\sqrt{3})=4(\sqrt{3}-1)\approx2.92820323028 \\ \\ w=4(\sqrt{3}-1) \end{gathered}[/tex]10. Now, we can check both values as follows:
[tex]\begin{gathered} l^2+w^2=d^2 \\ \\ (4(1+\sqrt{3})^)^2+(4(\sqrt{3}-1))^2=(8\sqrt{2})^2 \\ \\ 128=128\text{ }\rightarrow\text{This result is always true.} \end{gathered}[/tex]Therefore, in summary, the dimensions of the rectangle are:
[tex]\begin{gathered} l=4(1+\sqrt{3})=4(\sqrt{3}+1) \\ \\ w=4(\sqrt{3}-1) \end{gathered}[/tex]
