Question

Consider triangle ABC in the diagram below:

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

$80=n(10)$

$n=8\ sides$