Write the formula for finding the area of a trapezoid.

\(\displaystyle A=\frac{h}{2}(b_1+b_2)\)

Substitute the givens and solve for either base.

\(\displaystyle 30=\frac{10}{2}(5+b_2)\)

\(\displaystyle 30=5(5+b_2)\)

\(\displaystyle 6=5+b_2\)

\(\displaystyle b_2=1\)

Write the formula for the area of a trapezoid.

\(\displaystyle A=\frac{h}{2}(b_1+b_2)\)

Substitute all the given values and solve for the base.

\(\displaystyle 15=\frac{3}{2}(8+b_2)\)

\(\displaystyle 10=8+b_2\)

\(\displaystyle b_2=2\)

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. Since this problem provides the length for both of the bases as well as the total perimeter, the missing sides can be found using the following formula: Perimeter= Base one \(\displaystyle +\) Base two \(\displaystyle +\) \(\displaystyle 2\)(leg), where the length of "leg" is one of the two equivalent nonparallel sides. 

Thus, the solution is:

\(\displaystyle 35=19+6+2(leg)\)

\(\displaystyle 35=25+2(leg)\)

\(\displaystyle 2(leg)=35-25\) 

\(\displaystyle 2(leg)=10\)
\(\displaystyle leg=\frac{10}{2}=5\)

Check the solution by plugging in the answer:

\(\displaystyle 35=19+6+2(5)\)

\(\displaystyle 35=25+10\)

\(\displaystyle 35=35\)

In order to solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. 

This problem provides the lengths for each of the bases as well as the height of the isosceles trapezoid. In order to find the length for one of the two equivalent nonparallel legs of the trapezoid, first use the height of the trapezoid to form right triangles on the interior of the trapezoid that each have a base length of \(\displaystyle 2\). See image below:

 
Note: the base length of \(\displaystyle 2\) can be found by subtracting the lengths of the two bases, then dividing that difference in half: 

\(\displaystyle 11-7=4\)

\(\displaystyle 4\div2=2\)

Now, apply the formula \(\displaystyle a^2+b^2=c^2\), where \(\displaystyle c=\) the length for one of the two equivalent nonparallel legs of the trapezoid. 

Thus, the solution is:

\(\displaystyle 2^2+4^2=c^2\)

\(\displaystyle 4+16=c^2\)

\(\displaystyle 20=c^2\)

\(\displaystyle c=\sqrt{20}\)

In this problem the lengths for each of the bases and the height of the isosceles trapezoid is provided in the question prompt. In order to find the length for one of the two equivalent nonparallel legs of the trapezoid (side \(\displaystyle x\)), first use the height of the trapezoid to form right triangles on the interior of the trapezoid that each have a base length of \(\displaystyle 3\).

The base of the interior triangles is equal to \(\displaystyle 3\) because the difference between the two bases is equal to \(\displaystyle 6\). And, this difference must be divided evenly in half because the isosceles trapezoid is symmetric--due to the two equivalent nonparallel sides and the two nonequivalent parallel bases.  

Now, apply the pythagorean theorem: \(\displaystyle a^2+b^2=c^2\), where \(\displaystyle c=x\).

\(\displaystyle 3^2+15^2=c^2\)

\(\displaystyle 9+225=c^2\)

\(\displaystyle c^2=234\)

\(\displaystyle c=\sqrt{234}=\sqrt{9\times 26}=\sqrt{9}\sqrt{26}=3\sqrt{26}\)

Thus, \(\displaystyle x=3\sqrt{26}\)

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. Since this problem provides the length for both of the bases as well as the total perimeter, the missing sides can be found using the following formula: Perimeter= Base one \(\displaystyle +\) Base two \(\displaystyle +\) \(\displaystyle 2\)(leg), where the length of "leg" is one of the two equivalent nonparallel sides. 

Thus, the solution is:

\(\displaystyle 39=17+8+2(leg)\)

\(\displaystyle 39=25+2(leg)\)

\(\displaystyle 2(leg)=39-25\)

\(\displaystyle 2(leg)=14\)

\(\displaystyle leg=\frac{14}{2}=7\)

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. 

Therefore, use the given information to apply the formula: 

Perimeter= Base one \(\displaystyle +\) Base two \(\displaystyle +\) \(\displaystyle 2\)(leg), where the length of "leg" is one of the two equivalent nonparallel sides.

Thus, the solution is:

\(\displaystyle 119=45+2(24) +Base\)

\(\displaystyle 119=48+45+Base\)

\(\displaystyle 119=93+Base\)

\(\displaystyle Base=119-93=26\) 

In order to solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. 

This problem provides the lengths for each of the bases as well as informataion regarding the height of the isosceles trapezoid. In order to find the length for one of the two equivalent nonparallel legs of the trapezoid, use the height of the trapezoid to form right triangles on the interior of the trapezoid that each have a base length of \(\displaystyle 1\). The interior triangle base length of \(\displaystyle 1\) can be found by subtracting the lengths of the two bases, then dividing that difference in half: 

\(\displaystyle 15-13=2\)

\(\displaystyle \frac{2}{2}=1\)

In order to calculate the exact height of the isosceles trapezoid (as well as the interior triangle), find \(\displaystyle \frac{1}{3}\) of the larger base. Since the largest base of the trapezoid is \(\displaystyle 15\), the height of the trapezoid is: \(\displaystyle \frac{1}{3}\times15=\frac{15}{3}=5\)


Now you have enough information to apply the formula \(\displaystyle a^2+b^2=c^2\), where \(\displaystyle c=\) one of the missing sides. 

The final solution is:

\(\displaystyle 1^2+5^2=c^2\)

\(\displaystyle 1+25=c^2\)

\(\displaystyle c^2=26\)

\(\displaystyle c=\sqrt{26}\) 

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. Since this problem provides the length for both of the bases as well as the total perimeter, the missing sides can be found using the following formula: Perimeter= Base one \(\displaystyle +\) Base two \(\displaystyle +\) \(\displaystyle 2\)(leg), where the length of "leg" is one of the two equivalent nonparallel sides. 

Thus, the solution is: 

\(\displaystyle 372=163+115+2(leg)\)

\(\displaystyle 372=278+2(leg)\)

\(\displaystyle 2(leg)=372-278=94\)

\(\displaystyle leg=\frac{94}{2}=47\)


Check the solution by plugging in the answer:

\(\displaystyle 372=163+115+2(47)\)
\(\displaystyle 372=372\)

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. 

Therefore, use the given information to apply the formula: 

Perimeter= Base one \(\displaystyle +\) Base two \(\displaystyle +\) \(\displaystyle 2\)(leg), where the length of "leg" is one of the two equivalent nonparallel sides.

Thus, the solution is:

\(\displaystyle 301=100+2(55) +Base\)

\(\displaystyle 301=100+110+Base\)

\(\displaystyle 301=210+Base\)

\(\displaystyle Base=301-210=91\)