Question

What is the perimeter of the isosceles trapezoid that has vertices of $A\left( -3,5 \right),B\left( 3,5 \right),C\left( 5,-3 \right)$ and $D\left( -5,-3 \right)?$

Answer 1

We know that the perimeter of a trapezoid is the sum of the lengths of the sides of the trapezoid. Generally, it is the total length obtained by adding the lengths of the sides of the polygons. To find out the lengths of the sides, we will use the formula $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.$

Complete step-by-step answer:
Let us consider the given isosceles trapezoid $ABCD$ whose vertices are given by $A\left( -3,5 \right),B\left( 3,5 \right),C\left( 5,-3 \right)$ and $D\left( -5,-3 \right)$

Certainly, as we know, a trapezoid has four sides.
We can say that the sides of the trapezoid are $AB,BC,CD$ and $DA.$
As we know, we can find the lengths of the sides using the formula $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.$
Now let us consider the side $AB.$
So, we know that the side $AB$ is the line joining the points $A\left( -3,5 \right),B\left( 3,5 \right)$
Now, we will get the side $AB=\sqrt{{{\left( 3-\left( -3 \right) \right)}^{2}}+{{\left( 5-5 \right)}^{2}}}=\sqrt{{{6}^{2}}+0}=6.$
And now we will find the side $BC$ using the same formula.
And since $BC$ is the line joining the points $B\left( 3,5 \right),C\left( 5,-3 \right),$ we will get $BC=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( -3-5 \right)}^{2}}}=\sqrt{4+64}=68.$
Similarly, now we will find the side $CD$ using the formula written above.
As we know, $CD$ is the line joining the points $C\left( 5,-3 \right),D\left( -5,-3 \right).$
Now, we will get $CD=\sqrt{{{\left( 5-\left( -5 \right) \right)}^{2}}+{{\left( -3-\left( -3 \right) \right)}^{2}}}=\sqrt{{{10}^{2}}+0}=10.$
Now, finally, we will find the fourth side $DA.$
Now, we know that $DA$ is the line joining the points $D\left( -5,-3 \right),A\left( -3,5 \right).$
Now, we will get the side as $DA=\sqrt{{{\left( -5-\left( -3 \right) \right)}^{2}}+{{\left( -3-5 \right)}^{2}}}=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -8 \right)}^{2}}}=\sqrt{68}.$
Now, we can find the perimeter of the given isosceles trapezoid by adding the sides we have obtained.
Now, we will get the perimeter as $6+10+\sqrt{68}+\sqrt{68}.$
Hence the perimeter is $16+2\sqrt{68}\approx 32.5.$

Note: As we know, the isosceles trapezoid has two of its sides equal. And so, we can find a vertical symmetry of an isosceles triangle that shows the two equal parts of the trapezoid. We should remember that there will not be any horizontal symmetry for a trapezoid.