Question

3. The area of an isosceles trapezoid is 77 square inches, the height measures 4 inches and the legs are
5 inches long. Find the lengths of the bases.
​

Answer 1

Answer:

In an isosceles trapezoid, its height and congruent side form a right triangle on both ends.

The congruent side (5 inches) is the hypotenuse and its height (4 inches) is one of the legs of the triangle. To determine the other leg, apply the Pythagorean formula which is:

c^2=a^2+b^2

Plug-in c=5 and a=4.

5^2=4^2+b^2

25=16+b^2

25-16=16-16+b^2

9=b^2

sqrt9=sqrt(b^2)

3=b

So, the length of the other leg of the right triangle is 3 inches.

Next, let the shorter base of the isosceles trapezoid be x.

Since two right triangles are formed on both ends of an isosceles trapezoid, then the longer base is:

longer base= x+3+3=x+6

From here, apply the formula of area of trapezoid which is:

A=1/2(b_1+b_2)h

Plug-in A=77 in^2, h= 4 in, b_1=x and b_2=x+6 .

77=1/2(x+x+6)*4

Then, simplify the equation.

77=1/2(2x+6)*4

77=2(2x+6)

77=4x+12

And solve for x.

77-12=4x+12-12

65=4x

65/4=(4x)/4

16.25=x

Hence, the length of the shorter base is 16.25 inches.

And, plug-in x=16.25 to x+6 to get the length of the longer base.

longer base=x+6=16.25+6=22.25

Thus, the lengths of the bases of the isosceles trapezoid are 16.25 inches and 22.25 inches.

Answer 2

Answer:

$\omega$ be a complex number such that $2 \omega + 1 = z$ , where $z = \sqrt{-3}. If \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ [tex] \left | \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - {\omega}^{2} -1 & {\omega}^{2} \\ 1 & {\omega}^{2} & {\omega}^{7} \end{array} \right |$