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What is the perimeter of the trapezoid shown?

 

[asy] unitsize(1.5cm);pair a=(0,0); pair b=(1,sqrt(3)); pair c=(3,sqrt(3)); pair d=(4,0);draw(a--b--c--d--a);label(

 

 

b. What is the area of the trapezoid shown? Express your answer in simplest exact form.

 Apr 11, 2019
edited by SoulSlayer615  Apr 11, 2019
 #1
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+3

 

Drop a perpendicular from the top left corner down to the base.  You've formed a 30-60-90 triangle so you can use the sine of 30o to determine that the base of that triangle is 1.  Same on the right hand side.  Add up all the sides and segments of sides and get

 

sides of trapezoid and top total 2 + 2 + 2

 

bases of triangles and center segment of trapezoid base total 1 + 2 + 1

 

So the trapezoid perimeter is 6 + 4 = 10

 

.

 Apr 11, 2019
 #3
avatar+100571 
+4

a ) If we draw a perpendicular to the base from the top right vertex....we have a 30-60-90 right triangle

 

The distance, D, from the bottom right vertex to the inersection of the perpendicular and the base = 

 

cos 60” =  D /2

 

2(1/2)  = 1

 

So.....by symmetry.....the length of the bottom base = 1 + 2 + 1  = 4

And the perimeter of the trapezoid is  bottom base + top base + sides  =

 

4 + 2 + 2 + 2  =         10 units

 

b) The height of the right triangle we created  = √3

 

The area =  (1/2)(sum of bases) (height)  =  

 

(1/2)(4 + 2)√3  =

 

(1/2) (6)√3 =

 

3√3  units^2 

 

 

cool cool  cool

 Apr 11, 2019
edited by CPhill  Apr 12, 2019
 #4
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+2

SOLUTION:

 

\((A)\) Since we know the bottom left and bottom right angles of this trapezoid, we can construct two different triangles with angles \(60 ^{\circ}\) and  \(90 ^{\circ}\). The trianges have angles  \(90​​ ^{\circ}\)and \(60 ^{\circ}\) and since the angles of a triangle sum up to \(180 ^{\circ}\), it is a " \(90 ^{\circ}, 60 ^{\circ}, 30^{\circ}\) triange". " \(90 ^{\circ}, 60 ^{\circ}, 30^{\circ}\) trianges" have the following properties:

 

Hypotenuse = \(2x\)

 

Larger Triange Side = \(x\sqrt{3}\)

 

Smaller Triange Side = \(x\)

 

In the trapezoid, the larger triangle side equals the altitude thus the altidute equals \( \sqrt{3}\) since the hypotenuse equals \(2\) and therefore \((2x = 2), x = 1\). Since the bottom base of the trapezoid merely equals \(x + x + 2 = 4\) and we already have the top base, we can find the perimeter. The perimeter equals the sum of both bases and the two other sides of the trapezoid thus the perimeter equals \(4 + 2 + 2 + 2 = \boxed{10}\) 

 

\((B)\) As you have most likely realized, the altitude isn't utilized whatsoever in the first problem, however, it is needed in order to compute the area. The formula of the area of a trapezoid is \((\frac{a+b}{2})h\)(\(a\) and \(b\) are the bases of a trapezoid and \(h\) is the altitude). Since we already know the bases and altitude of this specific trapezoid, we can quite feasibly compute the area. \((\frac{a+b}{2})h = (\frac{4 + 2}{2})\sqrt{3} = (\frac{6}{2})\sqrt{3} = \boxed{3\sqrt{3}}\)

 

RB - ∃

 Apr 12, 2019

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