Question

a home decorating company is making similarly shaped end tables in different sizes that can be nested together. the image shows a side view of the tables, similar parallelograms dsgk, tmwr, and hcyf, where dsgk ~ tmwr, tmwr ~ hcyf, sg = 2x inches (in.), sd = 24.6 in., dk = 2x in., mw = (x + 6.3) in., mt = 19.68 in., wr = 19.68 in., and hf = 13.44 in. what is the approximate perimeter of the smallest end table hcyf ?

Explanation:

Step1: Use the property of similar parallelograms

Since \(DSGK\sim TMWB\), the ratios of corresponding sides are equal. For parallelograms, \(\frac{SD}{MT}=\frac{SG}{MW}\). Given \(SG = 2x\), \(SD = 24.6\) in, \(MT=19.68\) in, and \(MW=(x + 6.3)\) in. Then \(\frac{24.6}{19.68}=\frac{2x}{x + 6.3}\).
Cross - multiply: \(24.6(x + 6.3)=19.68\times2x\).
Expand: \(24.6x+24.6\times6.3 = 39.36x\).
\(24.6x+154.98=39.36x\).
Subtract \(24.6x\) from both sides: \(154.98=39.36x - 24.6x\).
\(14.76x=154.98\).
Solve for \(x\): \(x=\frac{154.98}{14.76}=10.5\).

Step2: Find the ratio of similarity

The ratio of similarity between \(TMWB\) and \(HCYF\) (since \(TMWB\sim HCYF\)). Let the ratio \(r\) be \(\frac{WR}{HF}\). Given \(WR = 19.68\) in and \(HF = 13.44\) in, \(r=\frac{13.44}{19.68}=\frac{2}{3}\).
The side \(MW=x + 6.3\), substituting \(x = 10.5\), \(MW=10.5+6.3 = 16.8\) in. The corresponding side in \(HCYF\) is \(CY\), and \(CY=\frac{2}{3}\times16.8 = 11.2\) in.

Step3: Calculate the perimeter of \(HCYF\)

For a parallelogram, the perimeter \(P = 2(a + b)\). Here \(a = 13.44\) in and \(b = 11.2\) in.
\(P=2(13.44+11.2)=2\times24.64 = 49.28\approx49.3\) in.

Answer:

The approximate perimeter of the smallest end - table \(HCYF\) is \(49.3\) inches.