Waste less time on Facebook — follow Brilliant.
×

Altitude and sides of a triangle

Find the lengths of the sides of a triangle with 20, 28, and 35 as the lengths of altitudes. Please help...

Note by Puneet Pinku
1 year, 3 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Area = 1/2 * a * 20 = 1/2 * b * 28 = 1/2 * c * 35 20a = 28b = 35c

a = 7k b = 5k c = 4k

Since a is longest side, then altitude of 20cm falls onto side a (if angle A is obtuse, then the other altitudes could fall outside of triangle a = a₁ + a₂

a₁² + 20² = b² ----> a₁ = √(b² − 400) = √(25k² − 400) = 5 √(k² − 16)
a₂² + 20² = c² ----> a₂ = √(c² − 400) = √(16k² − 400) = 4 √(k² − 25)

a = a₁ + a₂
7k = 5 √(k² − 16) + 4 √(k² − 25) -----> square both sides
49k² = 25 (k² − 16) + 16 √(k² − 25) + 40 √(k² − 16) √(k² − 25)
49k² = 25k² − 400 + 16k² − 400 + 40 √(k⁴ − 41k² + 400)
49k² = 41k² − 800 + 40 √(k⁴ − 41k² + 400)
8k² + 800 = 40 √(k⁴ − 41k² + 400)
k² + 100 = 5 √(k⁴ − 41k² + 400) -----> square both sides again
k⁴ + 200k² + 10000 = 25 (k⁴ − 41k² + 400)
k⁴ + 200k² + 10000 = 25k⁴ − 1025k² + 10000
24k⁴ − 1225k² = 0
k² (24k² − 1225) = 0
k² = 1225/24

k = 35/(2√6)

a = 7k = 245/(2√6)
b = 5k = 175/(2√6)
c = 4k = 70/√6

Ayush Rai - 1 year, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...