Question

Consider triangle ABC in the diagram below:



a. Is this triangle acute, right, or obtuse? (p. 318)

b. Find the measure of angle B. You must show your work. (p. 319)

c. Find the area of the triangle. You must show your work. (p. 328)





A rectangle has an area of 42 cm²; the length is 7 cm. Find its width. You must show your work. (p. 327)





(3) What is the area of the trapezoid on the left side above? You must show your work. (p. 334)

(4) What is the area of the figure on the right side above? You must show your work. (p. 329-330)



Find the area of the parallelogram whose base is 18 feet, height is 11 feet.



Please refer to the figure below in which two parallel lines are cut by a transversal.



6. Name two pairs of corresponding angles. (page 313)

7. Name two angles which are supplementary to angle 5. (page 315)



A certain regular polygon has a side length of 10 cm; its apothem is 12.1 cm. The polygon’s area is 484 cm². How many sides does the polygon have? You must show your work

Answer 1

I think a. would be obtuse. and for b. I think you add 118 and 270 together then divide by 360 for c. I think you either multiply or add (I think add) 118+270+ the answer you got for b and I think that's your answer. and honestly that's all I think I know, i really suck at math lol

Answer 2

Answer:

Part 1)

a) ABC is an acute triangle

b) $B=67.6\°$

c) $A=76\ cm^{2}$

Part 2) The width is $6\ cm$

Part 3) $A=182\ cm^{2}$

Part 4) $119\ cm^{2}$

Part 5) $A=198\ ft^{2}$

Part 6) (m∠2 and m∠6), (m∠4 and m∠8)

Part 7) $8\ sides$

Step-by-step explanation:

Part 1) Consider triangle ABC

see the attached figure with letters to better understand the problem

a)  Is this triangle acute, right, or obtuse?

we know that

m∠ACD+$153\°=180\°$ ------> by supplementary angles

In the right triangle ADC

remember that the sum of the internal angles in a triangle is equal to $180\°$

m∠ACD=$180\°-153\°=27\°$

m∠DAC=$180\°-90\°-27\°=63\°$

Find the measure of side DC

$tan(27\°)=\frac{AD}{DC}$

$DC=AD/tan(27\°)$

substitute

$DC=8/tan(27\°)=15.70\ cm$

In the right triangle ADB

Find the measure of side BD

$BD=BC-DC$

substitute

$BD=19-15.70=3.30\ cm$

Find the measure of angle BAD

$tan(BAD\°)=\frac{BD}{AD}$

$tan(BAD\°)=\frac{3.3}{8}$

$BAD\°=tan^{-1}(3.3/8)=22.4\°$

Find the measure of angle A

m∠A=m∠BAD+m∠DAC  

m∠A=$22.4\°+63\°=85.4\°$

$85.4\° < 90\°$ ------> triangle ABC is an acute triangle

b) Find the measure of angle B

Remember that

the sum of the internal angles in a triangle is equal to $180\°$

A+B+C= $180\°$

we have

$A=85.4\°$

$C=27\°$

$B=180\°-85.4\°-27\°=67.6\°$

c) Find the area of the triangle

the area of a triangle is equal to

$A=\frac{1}{2}bh$

we have

$b=BC=19\ cm$

$h=AD=8\ cm$

substitute

$A=\frac{1}{2}19*8=76\ cm^{2}$

Part 2) A rectangle has an area of 42 cm²; the length is 7 cm. Find its width

Let

x------> the length of rectangle

y------> the width of rectangle

the area of rectangle is equal to

$A=xy$

In this problem we have

$A=42\ cm^{2}$

$x=7\ cm$

substitute and solve for y

$42=7y$

$y=42/7=6\ cm$

Part 3) What is the area of the trapezoid on the left side above?

we know that

the area of trapezoid is equal to

$A=\frac{(B1+B2)h}{2}$

where

B1 and B2 are the parallel sides

h is the height

in this problem we have

$B1=11\ cm, B2=17\ cm, h=13\ cm$

substitute

$A=\frac{(11+17)13}{2}$

$A=182\ cm^{2}$

Part 4) What is the area of the figure on the right side above?

we know that

the area of the figure is equal to the area of a smaller rectangle plus the area of a larger rectangle

area of the smaller rectangle is equal to

$A1=4*5=20\ cm^{2}$

area of the larger rectangle is equal to

$A2=(14-5)*11=99\ cm^{2}$

The total area is

$A1+A2=20+99=119\ cm^{2}$

Part 5) Find the area of the parallelogram whose base is 18 feet, height is 11 feet

we know that

The area of the parallelogram is equal to

$A=bh$

where

b is the base

h is the height

in this problem we have

$b=18\ ft, h=11\ ft$

substitute

$A=18*11=198\ ft^{2}$

Part 6) Name two pairs of corresponding angles

we know that

In the figure

m∠2=m∠6 ----->by corresponding angles

m∠4=m∠8 ----->by corresponding angles

m∠1=m∠5 ----->by corresponding angles

m∠3=m∠7 ----->by corresponding angles

Part 7) A certain regular polygon has a side length of 10 cm; its apothem is 12.1 cm. The polygon’s area is 484 cm². How many sides does the polygon have?

we know that

The area of a regular polygon is equal to

$A=\frac{1}{2}Pa$

where

P is the perimeter of the regular polygon

a is the apothem

in this problem we have

$a=12.1\ cm, b=10\ cm, A=484\ cm^{2}$

substitute and solve for P

$484=\frac{1}{2}P(12.1)$

$968=P(12.1)$

$P=968/12.1$

$P=80\ cm$

Remember that the perimeter of a regular polygon is equal to

$P=nb$

where

n is the number of sides

b is the length side of the polygon

we have

$P=80\ cm$

$b=10\ cm$

substitute and solve for n

$80=n(10)$

$n=8\ sides$