+0  
 
+1
752
3
avatar+153 

a.Determine the missing side length:

 

 [asy] size(4cm);void mra(pair u, pair v, pair w, real r){ draw(v+(u-v)/length(u-v)*r--v+(u-v)/length(u-v)*r+(w-v)/length(w-v)*r--v+(w-v)/length(w-v)*r);} pair a=(0,0); pair b=(28,0); pair c=(0,-21);draw(a--b--c--a);mra(c,a,b,2.5);label(

 

b. Determine the missing side length

 

 

[asy] size(4cm);void mra(pair u, pair v, pair w, real r){ draw(v+(u-v)/length(u-v)*r--v+(u-v)/length(u-v)*r+(w-v)/length(w-v)*r--v+(w-v)/length(w-v)*r);} pair a=(0,42/sqrt(85)); pair b=(-36/sqrt(85),0); pair c=(49/sqrt(85),0);draw(a--b--c--a);mra(c,a,b,0.75);label(

 

 

c. Determine the missing side length

[asy] size(4cm);void mra(pair u, pair v, pair w, real r){ draw(v+(u-v)/length(u-v)*r--v+(u-v)/length(u-v)*r+(w-v)/length(w-v)*r--v+(w-v)/length(w-v)*r);} pair k=(0,0); pair l=(3,sqrt(3)); pair n=(3,0);label(rotate(30)*

 Mar 28, 2019
 #1
avatar+4586 
+4

1. 28^2+21^2=c^2,\(c=35.\)

2. 6^2+7^2=c^2, c^2=85, \(c=\sqrt{85}.\)

3. \(c^2+(\sqrt{3})^2=(2\sqrt{3})^2, c^2=9, c=3.\)

.
 Mar 28, 2019
 #2
avatar+1354 
+3

Ok, all of these are relativaly simple.

The formula you will be working with for all of these is a^2+b^2=c^2

Im not going to give you the answers only explain how to do it.

#1. 28 and 21 are a and b. Insert them into the formula to get 28^2+21^2=c^2

Solve for 28^2+21^2 and get 1225

now you are at 1225=c^2

Your answer will be the square root of 1225.

#2. In this problem 6 and 7 are a and b.

Put them in the equation above to get 36+49=c^2

Your answer for this will be the square root of 85

#3. In this problem 2 times the square root of 3 is c and the square root of 3 is a.

Put them into the equation above to get the square root of 3 squared (3) + b^2=2times the square root of 3 squared (2 times the square root of 3.

This gets us to the Square root of 3+ b^2=2 times the square root of 3

Minus Square root of 3 from right side to get b^2 equals the square root of three. so b equals the square root of the square root of three

Have A nice day!

 Mar 28, 2019
 #3
avatar+153 
0

Thanks!

SoulSlayer615  Mar 30, 2019

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