Base 1 of a trapezoid is 10 ft base is 8 feet the height is 5 ft the area is?

Mathematics

1 answer:

andriy [413]3 years ago

3 0

Answer: The area of the trapezoid is 45 square feet .

Step-by-step explanation:

Formula : Area of trapezoid : $A=\dfrac{1}{2}(a+b)h$ , where a and b are the parallel opposite sides (bases) and h= height.

We are given that : A trapezoid has Base 1 = 10 ft

Base 2 = 8 feet

height = 5 ft

Then, the area of the trapezoid would be (Put a=10 , b= 8 and h=5)

$A=\dfrac{1}{2}(10+8)(5)$

$A=\dfrac{1}{2}(18)(5)$

$A=(9)(5)=45\ \text{square feet}$

Therefore , the area of the trapezoid = 45 square feet

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nikitadnepr [17]

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Step-by-step explanation:

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Step 2:

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3 years ago

Read 2 more answers

Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$