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# What is the perimeter of the trapezoid below?(a) 52(b) 72(c) 75(d) 80

Last updated date: 23rd Jul 2024
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Hint: First of all find the length of the side CD. To find the length of the side CE consider ABCE as a rectangle and use the fact that the opposite sides of a rectangle are equal and use the relations CE = AB and BC = AE. To find the length of the side DE consider AD as hypotenuse (H), DE as the base (B) and AE as the perpendicular (P) of the right angle triangle AED. Use the Pythagoras theorem given as ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$ to find the value of DE. Add CE and DE to find the length of CD and finally take the sum of the length of all the sides of the trapezium ABCD to get the answer.

Complete step by step solution:
Here we have been provided with a trapezium with the following measurements shown below. We are asked to calculate the perimeter of the trapezium. First we need to find the length of the fourth side that is CD.

Now, clearly we can consider ABCE as a rectangle and we know that the opposite sides of a rectangle are equal so we have the following relations: -
$\Rightarrow$ AE = BC = 15 ……. (1)
$\Rightarrow$ AB = CE = 20 ……. (2)
Now, in the right angle triangle AED we have AD as hypotenuse (H), DE as the base (B) and AE as the perpendicular (P). Therefore, using the Pythagoras theorem given as ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$ we get,
$\Rightarrow {{\left( AD \right)}^{2}}={{\left( AE \right)}^{2}}+{{\left( DE \right)}^{2}}$
Substituting the values AD = 17 and AE = 15 {relation (1)} we get,
\begin{align} & \Rightarrow {{\left( 17 \right)}^{2}}={{\left( 15 \right)}^{2}}+{{\left( DE \right)}^{2}} \\ & \Rightarrow {{\left( DE \right)}^{2}}=289-225 \\ & \Rightarrow {{\left( DE \right)}^{2}}=64 \\ \end{align}
Taking square root both the sides we get,
$\Rightarrow$ DE = 8
Therefore the total length of the side CD is sum of the length of the sides CE and DE, so we get,
$\Rightarrow$ CD = CE + DE
Substituting the obtained value of DE and using relation (2) for the value of CE we get,
$\Rightarrow$ CD = 20 + 8
$\Rightarrow$ CD = 28
Now, the perimeter of the trapezium ABCD is the sum of the length of all the sides of the trapezium, so we have,
$\Rightarrow$ Perimeter = AB + BC + CD + DA
$\Rightarrow$ Perimeter = 20 + 15 + 28 + 17
$\therefore$ Perimeter = 80
So, the correct answer is “Option d”.

Note: You must remember the properties of certain special quadrilaterals like parallelogram, square, rectangle, rhombus etc. as any of their properties may be used in the questions where we need to find the area or the perimeter of a figure. Also remember the formula for the area of a trapezium given as $\dfrac{1}{2}\times$ sum of parallel sides $\times$ perpendicular distance between the parallel sides.