We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Help?

0
434
1
+193

Two of the altitudes of an acute triangle divide the sides into segments of lengths 5,3,2 and x units, as shown. What is the value of x?

Apr 26, 2018

### 1+0 Answers

#1
+101322
+1

We can equate the area of the the triangle on both sides of an equation

(1/2)*  (3 + 5) * sqrt [ (x + 2)^2 - 3^2 ]   =  (1/2) (x + 2) * sqrt  [ ( 8^2 - 2^2 ]

8 * sqrt [ x^2 + 4x + 4 - 9]  = (x +2 ) sqrt (60)

8 * sqrt [ x^2 + 4x - 5]  = (x+ 2) sqrt (60)          square both sides

64 (x^2 + 4x  - 5 ]  =  (x^2 + 4x + 4 ) * 60

16(x^2 + 4x - 5)  = 15(x^2 + 4x + 4)

16x^2 + 64x - 80  =  15x^2 + 60x + 60

x^2 + 4x - 140  = 0       factor

(x - 10) ( x + 14)  = 0

Setting both factors to 0 and solving for x gives that x  = -14 (reject)  or

x = 10

Apr 26, 2018