Question

quadrilaterals skfy and mpwr are shown. mpwr \sim kfys, mp = 4z + 1.9, mr = 2z + 1.6, pw = 7z - 1.5, wr = 3.6, ks = 6z, kf = 10z - 0.35, ys = 5.4, and fy = 9z - 1.05. what is the value of z? round to the nearest tenth, if necessary.

Explanation:

Step1: Use the property of similar quadrilaterals

Since \(MPWR\sim KFYS\), the ratios of corresponding sides are equal. We can use the ratio \(\frac{MR}{YS}=\frac{MP}{KF}\).
Substitute the given expressions: \(\frac{2z + 1.6}{5.4}=\frac{4z+1.9}{10z - 0.35}\).
Cross - multiply: \((2z + 1.6)(10z-0.35)=5.4(4z + 1.9)\).
Expand both sides:

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Move all terms to one side: \(20z^{2}+15.3z-21.6z-0.56 - 10.26 = 0\).
Simplify: \(20z^{2}-6.3z-10.82 = 0\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 20\), \(b=-6.3\), \(c=-10.82\)), the quadratic formula is \(z=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-6.3)^{2}-4\times20\times(-10.82)=39.69+865.6 = 905.29\).
Then \(z=\frac{6.3\pm\sqrt{905.29}}{40}=\frac{6.3\pm30.09}{40}\).
We have two solutions for \(z\):
\(z_{1}=\frac{6.3 + 30.09}{40}=\frac{36.39}{40}=0.90975\) and \(z_{2}=\frac{6.3-30.09}{40}=\frac{-23.79}{40}=-0.59475\).
Since \(z>0\) (lengths of sides are positive, for example, if \(z=-0.59475\), \(MP = 4z + 1.9=4\times(-0.59475)+1.9=-2.379+1.9=-0.479<0\)), we take \(z = 0.90975\approx0.9\).

Answer:

\(z\approx0.9\)