+0  
 
0
272
5
avatar

 

 What is x in the diagram below

                          

 

A. 10 squareroot 14

B.2squareroot10

C.10squareroot4

D.20squareroot3

 Jul 25, 2019
 #1
avatar+6193 
+1

\(\text{well we can get a few equations using the Pythagorean theorem}\\ \text{Let the left hand unlabeled leg be $a$ and the right hand one be $b$}\\ x^2+4^2=a^2\\ x^2+16=a^2\\~\\ a^2+b^2=(10+4)^2= 14^2=196\\~\\ x^2+10^2=b^2\\ x^2+100=b^2\)

 

\(\text{We can add the first and third equations}\\ x^2+16+x^2+100=a^2+b^2\\ 2x^2 + 116 = a^2+b^2\\~\\ \text{and note this equals the left hand side of the second equation}\\ 2x^2+116 = 196\\~\\ 2x^2=80\\ x^2=40\\ x=\sqrt{40}=2\sqrt{10}\\~\\ \text{choice B}\)

.
 Jul 25, 2019
 #2
avatar
0

Can you help me with another one like this one

 Jul 25, 2019
 #3
avatar+6193 
0

only if you post it

Rom  Jul 25, 2019
 #4
avatar+8965 
+2

 

m∠ABD + m∠CBD  =  90°

m∠ABD + m∠BAD  =  90°

So  m∠CBD =  m∠BAD

 

And  m∠BDA  =  m∠CDB  because they are both right angles.

 

So by the AA similarity theorem,  △ABD  ~ △BCD

 

BD / CD  =  AD / BD

 

x / 10  =  4 / x

 

x2 / 10  =  4

 

x2  =  40

 

x  =  sqrt( 40 )

 

x  =  2 sqrt( 10 )

 Jul 25, 2019
edited by hectictar  Jul 25, 2019
 #5
avatar
0

Tan BCD = x / 10 =Cotan BAD =4 / x
Or: x / 10 = 4 / x    Cross multiply
x^2 = 40
x = 2 Sqrt(10)

 Jul 26, 2019

14 Online Users

avatar
avatar
avatar