Question

15. In trapezoid ABCD, AB - 13.BC -15, and CD-13. Altitudes BF and CE are both drawn to side AD.

B
15
с
(a) If altitude BF has a length of 12, then find the
perimeter of trapezoid ABCD.
13 points)
13
13
A
F
E
D
(b) Find mZBCD to the nearest tenth of a degree,
13 points)

Answer 1

Check the picture below, so the perimeter of the trapezoid is really 13 + 15 + 13 + x + 15 + x, or just 56 + 2x.

`\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies √(c^2-a^2)=b \qquad \begin{cases} c=\stackrel{hypotenuse}{13}\\ a=\stackrel{adjacent}{12}\\ b=\stackrel{opposite}{x}\\ \end{cases}`

`√(13^2 - 12^2)=b\implies √(169 - 144)=b\implies √(25)=b\implies 5=b \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{perimeter}{56+2(5)\implies 66}~\hfill ~ \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ cos(\theta)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\implies \theta =cos^(-1)\left( \cfrac{12}{13} \right)\implies \theta \approx 22.62^o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\measuredangle BCD}{\theta +90^o~~\approx 112.62^o}~\hfill`

Answer 2

(a) The perimeter of trapezoid ABCD is 56.

(b) The measure of angle BCD is 44.4 degrees.

In trapezoid ABCD, (AB=13. BC=15) , and (CD=13) Altitudes are both drawn to side
`(\overline{AD})`

Part (a)

Since trapezoid ABCD has parallel bases AB and CD, we know that ∠B and ∠C are supplementary angles. Therefore, m∠B+m∠C=180

Since altitudes BF and CE are drawn to AD, we know that △ABC and △CDA are right triangles.

Using the Pythagorean Theorem in △ABC, we get:

`AC^2 = BC^2 - AB^2 = 15^2 - 13^2 = 225 - 169 = 56`

Taking the square root of both sides, we get:

`AC = √(56) = 2√(14)`

Using the Pythagorean Theorem in △CDA, we get:

`AD^2 = CD^2 + AC^2 = 13^2 + (2√(14))^2 = 169 + 56 = 225`

Taking the square root of both sides, we get:

`AD = √(225) = 15`

Now that we know the lengths of all four sides of the trapezoid, we can find the perimeter:

Perimeter
`= AB + BC + CD + AD = 13 + 15 + 13 + 15 = \boxed{56}`

Part (b)

To find m∠BCD, we can use the tangent function.

`tan(m\angle BCD) = (BF)/(CD) = (12)/(13)`

Using the inverse tangent function (also known as arctangent or tan^-1), we get:

`m\angle BCD = arctan((12)/(13)) \approx \boxed{44.4^\circ}`