Recall how to find the area of a parallelogram:

$\displaystyle \text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

$\displaystyle x(x+5)=150$

Expand this equation.

$\displaystyle x^2+5x=150$

$\displaystyle x^2+5x-150=0$

Now factor this equation.

$\displaystyle (x-10)(x+15)=0$

Solve for $\displaystyle x$.

$\displaystyle x=10, x=-5$

Since lengths of bases and heights can only be positive, $\displaystyle x=10$.

Notice that the length of the base is given by the expression $\displaystyle x+5$. Substitute in the value of $\displaystyle x$ to find the length of the base.

$\displaystyle \text{Length of Base}=x+5$

$\displaystyle \text{Length of Base}=10+5$

$\displaystyle \text{Length of Base}=15$

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

$\displaystyle x(x+5)=500$

Expand this equation.

$\displaystyle x^2+5x=500$

$\displaystyle x^2+5x-500=0$

Now factor this equation.

$\displaystyle (x-20)(x+25)=0$

Solve for $\displaystyle x$.

$\displaystyle x=20, x=-25$

Since lengths of bases and heights can only be positive, $\displaystyle x=20$.

Notice that the length of the base is given by the expression $\displaystyle x+5$. Substitute in the value of $\displaystyle x$ to find the length of the base.

$\displaystyle \text{Length of Base}=x+5$

$\displaystyle \text{Length of Base}=20+5$

$\displaystyle \text{Length of Base}=25$

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

$\displaystyle x(x+37)=164$

Expand this equation.

$\displaystyle x^2+37x=164$

$\displaystyle x^2+37x-164=0$

Now factor this equation.

$\displaystyle (x-4)(x+41)=0$

Solve for $\displaystyle x$.

$\displaystyle x=4, x=-41$

Since lengths of bases and heights can only be positive, $\displaystyle x=4$.

Notice that the length of the base is given by the expression $\displaystyle x+37$. Substitute in the value of $\displaystyle x$ to find the length of the base.

$\displaystyle \text{Length of Base}=x+37$

$\displaystyle \text{Length of Base}=4+37$

$\displaystyle \text{Length of Base}=41$

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

$\displaystyle x(x+24)=256$

Expand this equation.

$\displaystyle x^2+24x=256$

$\displaystyle x^2+24x-256=0$

Now factor this equation.

$\displaystyle (x-8)(x+32)=0$

Solve for $\displaystyle x$.

$\displaystyle x=8, x=-32$

Since lengths of bases and heights can only be positive, $\displaystyle x=8$.

Notice that the length of the base is given by the expression $\displaystyle x+24$. Substitute in the value of $\displaystyle x$ to find the length of the base.

$\displaystyle \text{Length of Base}=x+24$

$\displaystyle \text{Length of Base}=8+24$

$\displaystyle \text{Length of Base}=32$

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

$\displaystyle x(x-8)=128$

Expand this equation.

$\displaystyle x^2-8x=128$

$\displaystyle x^2-8x-128=0$

Now factor this equation.

$\displaystyle (x-16)(x+8)=0$

Solve for $\displaystyle x$.

$\displaystyle x=16, x=-8$

Since lengths of bases and heights can only be positive, $\displaystyle x=16$.

Notice that the length of the base is given by the expression $\displaystyle x-8$. Substitute in the value of $\displaystyle x$ to find the length of the base.

$\displaystyle \text{Length of Base}=x-8$

$\displaystyle \text{Length of Base}=16-8$

$\displaystyle \text{Length of Base}=8$

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\displaystyle \text{height}=\sqrt{13^2-7^2}=\sqrt{169-49}=\sqrt{120}$

Now, use the height to find the area of the parallelogram.

$\displaystyle \text{Area}=17\times\sqrt{120}=186.23$

Remember to round to $\displaystyle 2$ places after the decimal.

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\displaystyle \text{height}=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}$

Now, use the height to find the area of the parallelogram.

$\displaystyle \text{Area}=12\times\sqrt{21}=54.99$

Remember to round to $\displaystyle 2$ places after the decimal.

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\displaystyle \text{height}=\sqrt{15^2-9^2}=\sqrt{225-81}=12$

Now, use the height to find the area of the parallelogram.

$\displaystyle \text{Area}=12\times20=240$

Remember to round to $\displaystyle 2$ places after the decimal.

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\displaystyle \text{height}=\sqrt{6^2-4^2}=\sqrt{36-16}=\sqrt{20}$

Now, use the height to find the area of the parallelogram.

$\displaystyle \text{Area}=16\times\sqrt{20}=62.61$

Remember to round to $\displaystyle 2$ places after the decimal.

Recall how to find the area of a parallelogram:

$\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\displaystyle \text{height}=\sqrt{8^2-5^2}=\sqrt{64-25}=\sqrt{39}$

Now, use the height to find the area of the parallelogram.

$\displaystyle \text{Area}=15\times\sqrt{39}=93.67$

Remember to round to $\displaystyle 2$ places after the decimal.