Question

Need help understanding how to work these math problems

Answer 1

Answer:

1. B. x = 11

2. E. $m\angle BEF=140^{\circ}$

3. D. 4.3 in

4. D. 40.8 ft

Step-by-step explanation:

1. By Angle Addition Postulate,

$m\angle KLM=m\angle KLV+m\angle VLM$

Since

$m\angle KLV=34^{\circ}\\ \\m\angle KLM=14x+19\\ \\m\angle VLM=12x+7,$

then

$14x+19=34+12x+7\\ \\14x-12x=34+7-19\\ \\2x=22\\ \\x=11$

2. By Angle Addition Postulate,

$m\angle FED=m\angle DEB+m\angle BEF$

Since

$m\angle FED=14x+8\\ \\m\angle DEB=22^{\circ}\\ \\m\angle BEF=13x-3,$

then

$14x+8=22+13x-3\\ \\14x-13x=22-3-8\\ \\x=11$

Therefore,

$m\angle BEF=(13\cdot 11-3)^{\circ}=(143-3)^{\circ}=140^{\circ}$

3. The area of trapezoid is

$A_{trapezoid}=\text{Midsegment}\times \text{Height}$

From the diagram,

$\text{Smaller base}=1.2\ in\\ \\\text{Bigger base}=4.2\ in,$

then

$\text{Midsegment}=\dfrac{1.2+4.2}{2}=2.7\ in$

Since the area of trapezoid is $11.6\ in^2,$ then

$11.6\ in^2 =2.7\ in\times \text{Height}\\ \\\text{Height}=\dfrac{11.6\ in^2}{2.7\ in}\approx 4.3\ in$

4. Use formula for the area of the circle to find the radius of the circle:

$A_{circle}=\pi r^2$

So,

$132.7=\pi r^2\\ \\r^2=\dfrac{132.7}{\pi}\\ \\r=\sqrt{\dfrac{132.7}{\pi}}\ ft$

Now, find the circumference of the circle:

$C=2\pi r\\ \\C=2\pi \cdot \sqrt{\dfrac{132.7}{\pi}}=2\sqrt{132.7\pi}\approx 40.8\ ft$