Solid Geometry Solutions
Topics || Problems
Find the area of the rectilinear figure shown, if it is the difference between two isosceles trapezoids whose corresponding sides are parallel.
Solution:
Extend lines then we can create 3 similar triangles
- Triangle A: Triangle with altitude = 6 and base = z
- Triangle B:Triangle with altitude = 8 and base = 3
- Triangle C: Triangle with altitude = 2 and hypotenuse = x
By AAA theorem we can prove that these triangles are similar. Thus if these triangles are similar we can use ratio and proportion
\( \frac{6}{z} = \frac{8}{3}\)
\( z = \frac{9}{4}\)
Calculate the hypotenuse of the triangle B: \( 8^2 + 3^2 = c^2 \)
\( c = \sqrt{73}\)
Ratio and Proportion between triangles B and C.
\( \frac{2}{x} = \frac{8}{\sqrt{73}} \)
\( x = \frac{\sqrt{73}}{4}\)
Calculate the length of \(y\): \(y = ~18 ~- ~2x ~- ~2z\)
\( y = 18 - (2) \frac{9}{4} - (2) \frac{\sqrt{73}}{4} \)
\( y = \frac{27-\sqrt{73}}{2}\)
Find the value of \(q\): \(q = 18 - 2x\)
\( q = 18 - (2) \frac{\sqrt{73}}{4} \)
\( q = \frac{36-\sqrt{73}}{2}\)
\( A_{Trap(outside)} = \frac{1}{2}(8)(12+18) = \: 120 \: in^2\)
\( A_{Trap(outside)} = 120 \:in ^2\)
\( A_{Trap(inside)} = \frac{1}{2}(6)(\frac{27-\sqrt{73}}{2}+\frac{36-\sqrt{73}}{2})\)
\( A_{Trap(inside)} = 68.86798876 \:in ^2\)
\( A =120 - 68.86798876 \:in ^2\)
\( A =51.13 \: in ^2\). Answer