Calculate Trapezoid Perimeter: Inside an Isosceles Triangle with Height 8

Question

ABC is an isosceles triangle.

AD is the height of triangle ABC.

555333171717888AAABBBCCCDDDEEEFFFGGG

AF = 5

AB = 17
AG = 3

AD = 8

What is the perimeter of the trapezoid EFBC?

Video Solution

Solution Steps

00:00 Determine the perimeter of the trapezoid EFBC
00:05 Examine all of the given data
00:23 AD is the height according to the given data
00:31 AG is perpendicular to EF according to corresponding angles
00:35 Apply the Pythagorean theorem to the triangle AGF
00:48 Substitute in the relevant values and proceed to solve in order to determine the value of GF
01:05 Isolate GF
01:17 Take the square root
01:35 This is the length of GF
01:42 In an isosceles triangle, the perpendicular is also a median
01:51 EF equals the sum of its segments (GF+EG)
01:58 Apply the Pythagorean theorem to the triangle ADB
02:12 Substitute in the relevant values and proceed to solve in order to determine DB
02:32 Isolate DB
02:49 Take the square root
03:03 This is the length of DB
03:08 The perpendicular in an isosceles triangle is also a median
03:12 Side CB equals the sum of its parts (DB+CD)
03:29 Segment FB equals the entire side AB minus AF
03:39 Substitute in the relevant values and proceed to solve
03:50 FB equals EC due to the fact that EF intersects the triangle's sides accordingly
04:02 Given that we have all the sides of the trapezoid we can proceed to calculate the perimeter
04:13 The perimeter of the trapezoid equals the sum of its sides
04:16 Substitute in the relevant values and proceed to solve
04:34 This is the solution

Step-by-Step Solution

To find the perimeter of the trapezoid, all its sides must be added:

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 the 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 perform the same process with the side DB of the 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

We start by finding FB:

FB=ABAF=175=12 FB=AB-AF=17-5=12

Now we reveal 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 so:

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

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

All that's left is to calculate:

30+8+12×2=30+8+24=62 30+8+12\times2=30+8+24=62

Answer

62


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