146,548 views
35 votes
35 votes
Can you please help with problem 23

Can you please help with problem 23-example-1
User Carebear
by
2.8k points

2 Answers

28 votes
28 votes

#22

  • Find the height using the trigonometric ratio.
  • Trigonometric ratio used here is:


\boxed{\mathfrak{\sin( \theta)=(perpendicular)/(hypotenuse)}}

  • θ = 45°
  • Perpendicular = ?
  • Hypotenuse = 22


\tt \: \sin(45 \degree) = (p)/(22)


\tt \: ( √(2) )/(2) = (p)/(22)


\tt \: 22 √(2) = 2p


\tt11 √(2) = p \: or \: p = 11 √(2)

Therefore th value of perpendicular is 15.5 when rounded off...

#23

  • Now, We will find the area with the help of height that we just found using the trigonometric ratio.
  • The formula to find area of parallelogram is:


\boxed{ \mathfrak{area = base * height}}

  • Base = 26 in. [Since opposite sides of parallelogram are equal and parallel]
  • Height = 15.5 in.


\sf \: area = 26 * 15.5 \\ \sf \: area = 403 \: {in}^(2)

Thus, The area of parallelogram is 403 inch²...~

User Malko
by
3.1k points
23 votes
23 votes


\\ \rm\longmapsto sin45=(h)/(22)


\\ \rm\longmapsto h=22sin45=22(0.71)=15.6in

  • Parallel sides of a parallelogram are equal.


\\ \rm\longmapsto Area


\\ \rm\longmapsto Base(Height)


\\ \rm\longmapsto 26(15.6)=405.6in^2

User JrBenito
by
2.6k points
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