In an isosceles trapezoid ABCD, AB=CD=5. The top base = 8 and the bottom base = 14. What is the area of this trapezoid?

3 Answers
May 27, 2018

Answer:

The area is 44

Explanation:

enter image source here

We know that the area of any trapezoid is
#(AD+BC)/2*h#
To find the area we, therefore, only need to calculate h.

We can see from the figure that #FD=3#
(this follows from #AG=FD#).

Therefore #FC=h=4#
(The triplet 3,4,5 is a well known triplet for right triangles, but you also see it from #3^2+4^2=5^2#)

Therefore the area is
#(14+8)/2*4=44#

May 27, 2018

Answer:

#color(blue)("Area"= 44 " square units")#

Explanation:

enter image source here

The area of a trapezoid is given as:

#"Area"=1/2("base"+"top")xxh#

Where #bbh# is the perpendicular distance between the parallel lines.

From the diagram notice the line of symmetry is at the midpoints of the base and top, also notice that the triangle formed on the left has sides #h# and #3#.

The 3 is arrived at by noticing this is #7-4#

We need to find h:

We know the hypotenuse of the triangle is 5, so by Pythagoras' theorem:

#h+sqrt((5)^2-(3)^2)=4#

We can now calculate the area:

#"Area"=1/2(14+8)*4=44#

Answer:

The area of the trapezoid is #44# #"units"^2#

Explanation:

I went ahead and drew the trapezoid:

enter image source here

The area formula for a trapezoid is:

#A = (t + b)/2(h)#

#"t = top base"#
#"b = bottom base"#
#"h = height"#

The measures of the two bases have been given, but the height has not been given. The height of the trapezoid can be found using the Pythagorean Theorem.

As you can see below, I drew the heights of the trapezoid on either side. I marked one as #h#, and put all of the point labels on the diagram:
enter image source here

Before we can use the Pythagorean Theorem, the length of #AF# must be calculated. There are two things to note:

  1. #BC = FE = 8# [As #BCEF# is a parallelogram]
  2. #AF = ED# [As #DeltaAFB ~= DeltaCED#]

Calculate #AF#:

#color(white)(xxx)AF + FE + ED = 14#

#rArrAF + 8 + ED = 14# (substitution)

#rArrAF + ED = 6#

#rArrAF + AF = 6# (substitution)

#rArr2AF = 6#

#rArrAF = 3#

Now use #AF# and #AB# and the Pythagorean Theorem to calculate #h#.
enter image source here

#color(white)(xxx)AF^2 +h^2 = AB^2#

#rArr 3^2 + h^2 = 5^2#

#rArr9 + h^2 = 25#

#rArrh^2 = 16#

#rArrh = sqrt16# [Length can't be negative]

#rArrh= 4#

Now plug that into the original area formula:

#color(white)(xxx)A = (t + b)/2(h)#

#rArrA = (8 + 14)/2(4)#

#rArrA = 22/2(4)#

#rArrA = 11(4)#

#rArrA = 44#

The area of the trapezoid is #44# #"units"^2#