Calculate Trapezoid Area: Isosceles Triangle with Height 8 and Base 17

Question

Given that the triangle ABC is isosceles,
and inside it we draw EF parallel to CB:

171717888AAABBBCCCDDDEEEFFFGGG53 AF=5 AB=17
AG=3 AD=8
AD the height in the triangle

What is the area of the trapezoid EFBC?

Video Solution

Solution Steps

00:00 What is the area of trapezoid EFBC?
00:03 EF is parallel to CB
00:06 According to corresponding angles, AG is also perpendicular to EF
00:14 Let's use the Pythagorean theorem on triangle AFG
00:18 Let's use the theorem to find GF
00:21 Let's substitute appropriate values according to the given data
00:28 Let's isolate GF
00:34 Let's solve the powers and then take the square root
00:37 And we got the length of GF
00:44 Now let's use the Pythagorean theorem in triangle ABD
00:50 Let's use the theorem to find DB
00:55 Let's substitute values according to the given data
01:02 Let's isolate DB and take the square root
01:07 This is the size of DB
01:16 In an isosceles triangle, the height is also the median
01:21 Therefore GF equals EG
01:27 The same thing happens in the larger triangle ABC
01:30 Therefore DB equals CD
01:35 Let's substitute the values
01:40 Let's use the formula for calculating trapezoid area
01:44 (Sum of bases(EF,CB) multiplied by height(GD)) divided by 2
01:50 Base EF equals EG plus GF
01:54 Base CB equals CD plus DB
01:58 Height GD equals AD minus AG
02:06 Let's substitute appropriate values in our equation

Step-by-Step Solution

To find the area of the trapezoid, you must remember its formula:(base+base)2+altura \frac{(base+base)}{2}+\text{altura} We will focus on finding the bases.

To find GF we use the Pythagorean theorem: A2+B2=C2 A^2+B^2=C^2  In triangle AFG

We replace:

32+GF2=52 3^2+GF^2=5^2

We isolate GF and solve:

9+GF2=25 9+GF^2=25

GF2=259=16 GF^2=25-9=16

GF=4 GF=4

We will do the same process with side DB in triangle ABD:

82+DB2=172 8^2+DB^2=17^2

64+DB2=289 64+DB^2=289

DB2=28964=225 DB^2=289-64=225

DB=15 DB=15

From here there are two ways to finish the exercise:

  1. Calculate the area of the trapezoid GFBD, prove that it is equal to the trapezoid EGDC and add them up.

  2. Use the data we have revealed so far to find the parts of the trapezoid EFBC and solve.

Let's start by finding the height of GD:

GD=ADAG=83=5 GD=AD-AG=8-3=5

Now we reveal that EF and CB:

GF=GE=4 GF=GE=4

DB=DC=15 DB=DC=15

This is because in an isosceles triangle, the height divides the base into two equal parts then:

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

CB=DB×2=15×2=30 CB=DB\times2=15\times2=30

We replace the data in the trapezoid formula:

8+302×5=382×5=19×5=95 \frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95

Answer

95


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