Answer to Question #152859 in Geometry for Hardy

Question

1.The sum of the perimeters of two squares is 50 cm and the sum of the areas is 93.25 cm^2. Find the side length of each square.

2. The sum of the circumferences of two circles is 36π cm and the sum of the areas is 170π cm^2. Find the radius of each circle.

3. A cuboid has sides of length 5 cm, x cm and y cm. Given x+y=20.5 and the volume of the cuboid is 360 cm^3, find the value of x and the value of y.

Answer 1

1. Let $x$ be the length of the side of the first square and $y$ be the length of the side of the second square. Then

$\begin{alignedat}{2} 4x+ 4y = 50 \\ x^2+y^2 = 93.25 \end{alignedat}$

$\begin{alignedat}{2} y = 12.5-x \\ x^2+(12.5-x)^2 = 93.25 \end{alignedat}$

$2x^2-25x+63=0, 0<x<12.5$

$x=\dfrac{25\pm\sqrt{25^2-4(2)(63)}}{2(2)}=\dfrac{25\pm11}{4}$

$x_1=\dfrac{25-11}{4}=3.5, y_1=12.5-3.5=9$

$x_2=\dfrac{25+11}{4}=9, y_2=12.5-9=3.5$

3.5 cm and 9 cm

2. Let $x$ be the radius of the first circle and $y$ be the radius of the second circle. Then

$\begin{alignedat}{2} 2\pi x+ 2\pi y = 36 \pi \\ \pi x^2+\pi y^2 = 170\pi \end{alignedat}$

$\begin{alignedat}{2} y = 18-x \\ x^2+(18-x)^2 = 170 \end{alignedat}$

$2x^2-36x+154=0, 0<x<18$

$x^2-18x+77=0$

$x=\dfrac{18\pm\sqrt{18^2-4(1)(77)}}{2(1)}=9\pm2$

$x_1=9-2=7, y_1=18-7=11$

$x_2=9+2=11, y_2=18-11=7$

7 cm and 11 cm

3. The volume $V$ of a cuboid is

$V=5\times x\times y$

Then

$\begin{alignedat}{2} x+ y = 20.5 \\ 5xy = 360 \end{alignedat}$

$\begin{alignedat}{2} y = 20.5-x \\ x(20.5-x) =72 \end{alignedat}$

$x^2-20.5x+72=0, 0<x<20.5$

$x=\dfrac{20.5\pm\sqrt{20.5^2-4(1)(72)}}{2(1)}=10.25\pm5.75$

$x_1=10.25-5.75=4.5, y_1=20.5-4.5=16$

$x_2=10.25+5.75=16, y_2=20.5-16=4.5$

4.5 cm and 16 cm or 16 cm and 4.5 cm