1. Let x x x be the length of the side of the first square and y y y be the length of the side of the second square. Then
4 x + 4 y = 50 x 2 + y 2 = 93.25 \begin{alignedat}{2}
4x+ 4y = 50 \\
x^2+y^2 = 93.25
\end{alignedat} 4 x + 4 y = 50 x 2 + y 2 = 93.25
y = 12.5 − x x 2 + ( 12.5 − x ) 2 = 93.25 \begin{alignedat}{2}
y = 12.5-x \\
x^2+(12.5-x)^2 = 93.25
\end{alignedat} y = 12.5 − x x 2 + ( 12.5 − x ) 2 = 93.25
2 x 2 − 25 x + 63 = 0 , 0 < x < 12.5 2x^2-25x+63=0, 0<x<12.5 2 x 2 − 25 x + 63 = 0 , 0 < x < 12.5
x = 25 ± 2 5 2 − 4 ( 2 ) ( 63 ) 2 ( 2 ) = 25 ± 11 4 x=\dfrac{25\pm\sqrt{25^2-4(2)(63)}}{2(2)}=\dfrac{25\pm11}{4} x = 2 ( 2 ) 25 ± 2 5 2 − 4 ( 2 ) ( 63 ) = 4 25 ± 11 x 1 = 25 − 11 4 = 3.5 , y 1 = 12.5 − 3.5 = 9 x_1=\dfrac{25-11}{4}=3.5, y_1=12.5-3.5=9 x 1 = 4 25 − 11 = 3.5 , y 1 = 12.5 − 3.5 = 9
x 2 = 25 + 11 4 = 9 , y 2 = 12.5 − 9 = 3.5 x_2=\dfrac{25+11}{4}=9, y_2=12.5-9=3.5 x 2 = 4 25 + 11 = 9 , y 2 = 12.5 − 9 = 3.5
3.5 cm and 9 cm
2. Let x x x be the radius of the first circle and y y y be the radius of the second circle. Then
2 π x + 2 π y = 36 π π x 2 + π y 2 = 170 π \begin{alignedat}{2}
2\pi x+ 2\pi y = 36 \pi \\
\pi x^2+\pi y^2 = 170\pi
\end{alignedat} 2 π x + 2 π y = 36 π π x 2 + π y 2 = 170 π
y = 18 − x x 2 + ( 18 − x ) 2 = 170 \begin{alignedat}{2}
y = 18-x \\
x^2+(18-x)^2 = 170
\end{alignedat} y = 18 − x x 2 + ( 18 − x ) 2 = 170
2 x 2 − 36 x + 154 = 0 , 0 < x < 18 2x^2-36x+154=0, 0<x<18 2 x 2 − 36 x + 154 = 0 , 0 < x < 18
x 2 − 18 x + 77 = 0 x^2-18x+77=0 x 2 − 18 x + 77 = 0
x = 18 ± 1 8 2 − 4 ( 1 ) ( 77 ) 2 ( 1 ) = 9 ± 2 x=\dfrac{18\pm\sqrt{18^2-4(1)(77)}}{2(1)}=9\pm2 x = 2 ( 1 ) 18 ± 1 8 2 − 4 ( 1 ) ( 77 ) = 9 ± 2 x 1 = 9 − 2 = 7 , y 1 = 18 − 7 = 11 x_1=9-2=7, y_1=18-7=11 x 1 = 9 − 2 = 7 , y 1 = 18 − 7 = 11
x 2 = 9 + 2 = 11 , y 2 = 18 − 11 = 7 x_2=9+2=11, y_2=18-11=7 x 2 = 9 + 2 = 11 , y 2 = 18 − 11 = 7
7 cm and 11 cm
3. The volume V V V of a cuboid is
V = 5 × x × y V=5\times x\times y V = 5 × x × y
Then
x + y = 20.5 5 x y = 360 \begin{alignedat}{2}
x+ y = 20.5 \\
5xy = 360
\end{alignedat} x + y = 20.5 5 x y = 360
y = 20.5 − x x ( 20.5 − x ) = 72 \begin{alignedat}{2}
y = 20.5-x \\
x(20.5-x) =72
\end{alignedat} y = 20.5 − x x ( 20.5 − x ) = 72
x 2 − 20.5 x + 72 = 0 , 0 < x < 20.5 x^2-20.5x+72=0, 0<x<20.5 x 2 − 20.5 x + 72 = 0 , 0 < x < 20.5
x = 20.5 ± 20. 5 2 − 4 ( 1 ) ( 72 ) 2 ( 1 ) = 10.25 ± 5.75 x=\dfrac{20.5\pm\sqrt{20.5^2-4(1)(72)}}{2(1)}=10.25\pm5.75 x = 2 ( 1 ) 20.5 ± 20. 5 2 − 4 ( 1 ) ( 72 ) = 10.25 ± 5.75 x 1 = 10.25 − 5.75 = 4.5 , y 1 = 20.5 − 4.5 = 16 x_1=10.25-5.75=4.5, y_1=20.5-4.5=16 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 x_2=10.25+5.75=16, y_2=20.5-16=4.5 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
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