Question

The length of a rectangle is 5 yd less than three times the width, and the area of the rectangle is 50 yd. Find the dimensions of the rectangle.

Answer 1

$\large\huge\green{\sf{Answer:-}}$

$Answer:The width is 5m and the length is 10m.Step-by-step explanation:Rectangle:Has two dimensions: Width(w) and length(l).It's area is:A = w*lA=w∗lThe length of a rectangle is 5m less than three times the widthThis means that l = 3w - 5l=3w−5The area of the rectangle is 50m^(2)This means that A = 50A=50 . SoA = w*lA=w∗l50 = w*(3w - 5)50=w∗(3w−5)3w^{2} - 5w - 50 = 03w2−5w−50=0Solving a quadratic equation:Given a second order polynomial expressed by the following equation:ax^{2} + bx + c, a\neq0ax2+bx+c,a=0 .This polynomial has roots x_{1}, x_{2}x1,x2 such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})ax2+bx+c=a(x−x1)∗(x−x2) , given by the following formulas:x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}x1=2∗a−b+△x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}x2=2∗a−b−△\bigtriangleup = b^{2} - 4ac△=b2−4acIn this question:3w^{2} - 5w - 50 = 03w2−5w−50=0Soa = 3, b = -5, c = -50a=3,b=−5,c=−50\bigtriangleup = (-5)^{2} - 4*3*(-50) = 625△=(−5)2−4∗3∗(−50)=625w_{1} = \frac{-(-5) + \sqrt{625}}{2*3} = 5w1=2∗3−(−5)+625=5w_{2} = \frac{-(-5) - \sqrt{625}}{2*3} = -3.33w2=2∗3−(−5)−625=−3.33Dimension must be positive result, soThe width is 5m(in meters because the area is in square meters).Length:l = 3w - 5 = 3*5 - 5 = 10l=3w−5=3∗5−5=10The length is 10 meters$