The length of a rectangle is 5 greater than 3 times the width. The area of the rectangle is 22 cm². Find the length and the width.
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Answer (2)
0\\b-the\ width\ of\ a\ rectangle,\ \ \ b>0\\\\a=3b+5\ \ \ \wedge\ \ \ A=22\ cm^2\\ \\A=a\cdot b\ \ \ \Leftrightarrow\ \ \ (3b+5)\cdot b=22\ \ \Leftrightarrow \ \ 3b^2+5b-22=0\\ \\\Delta=5^2-4\cdot3\cdot(-22)=25+264=289\ \ \Rightarrow\ \ \sqrt{\Delta} =17\\ \\b_1= \frac{-5-17}{2\cdot3} = \frac{-22}{6}=-3 \frac{2}{3} <0,\ \ \ \ \ \ \ \ \ b_2= \frac{-5+17}{2\cdot3} = \frac{12}{6}=2\\ \\b=2\ \ \ \Rightarrow\ \ \ \ a=3 b+5=3\cdot2+5=6+5=11"> a − t h e l e n g t h o f a rec t an g l e , a > 0 b − t h e w i d t h o f a rec t an g l e , b > 0 a = 3 b + 5 ∧ A = 22 c m 2 A = a ⋅ b ⇔ ( 3 b + 5 ) ⋅ b = 22 ⇔ 3 b 2 + 5 b − 22 = 0 Δ = 5 2 − 4 ⋅ 3 ⋅ ( − 22 ) = 25 + 264 = 289 ⇒ Δ = 17 b 1 = 2 ⋅ 3 − 5 − 17 = 6 − 22 = − 3 3 2 < 0 , b 2 = 2 ⋅ 3 − 5 + 17 = 6 12 = 2 b = 2 ⇒ a = 3 b + 5 = 3 ⋅ 2 + 5 = 6 + 5 = 11
A n s . t h e l e n g t h i s 11 c m , t h e w i d t h i s 2 c m .
The length of the rectangle is 11 cm and the width is 2 cm, calculated by forming a quadratic equation based on the relationships between width, length, and area. The appropriate formulas allowed us to find the values systematically. The quadratic formula helped us identify viable dimensions for the rectangle.
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