Question 1180870



Triangle ABC is an isosceles right triangle with 

1. If {{{a=5cm}}}, find {{{c}}}.

since triangle ABC is an isosceles right triangle, {{{a=b=5cm}}}, and
{{{c=sqrt((5cm)^2+(5cm)^2)}}}
{{{c=sqrt(25cm^2+25cm^2))}}}
{{{c=sqrt(50cm^2)}}}
{{{c=7.07cm}}}

2.
 If {{{b=8cm}}}, find {{{c}}}.

=>{{{a=8cm}}}
{{{c=sqrt((8cm)^2+(8cm)^2)}}}
{{{c=sqrt(64cm^2+64cm^2))}}}
{{{c=sqrt(128cm^2))}}}
{{{c=11.31cm}}}

3. 
If {{{b=12.5cm}}}, find {{{c}}}.
{{{a=12.5cm}}}
{{{c=sqrt((12.5cm)^2+(12.5cm)^2)}}}
{{{c=sqrt(156.5cm^2+156.5cm^2))}}}
{{{c=sqrt(312.5cm^2))}}}
{{{c=17.68cm}}}



Triangle GMA is a 30°-60°-90° triangle.

<a href="https://ibb.co/nLwKzRb"><img src="https://i.ibb.co/nLwKzRb/triangle.png" alt="triangle" border="0"></a>

if GMA is a 30°-60°-90° triangle, then 
The shorter leg, which is opposite to the 30- degree angle, is labeled as{{{ x}}}.
The hypotenuse, which is opposite to the 90-degree angle, is twice the shorter leg length ({{{2x}}}).
The longer leg, which is opposite to the 60-degree angle, is equal to the shorter leg’s product and the square root of three ({{{xsqrt(3)}}}).

in your case {{{m=2x}}},  shorter leg {{{g=x}}}, and longer leg {{{a=xsqrt(3)}}}

4. 
If {{{m=26cm}}}, find {{{a}}} and {{{g}}}.
if hypotenuse {{{m=26cm}}}=>{{{2x=26cm}}}=>{{{x=13cm}}}
then shorter leg {{{g=13cm}}}
and longer leg {{{a=13cm*sqrt(3)}}}

so, check
{{{m^2=a^2+g^2}}}
{{{(26cm)^2=(13cm*sqrt(3))^2+(13cm)^2}}}
{{{676cm^2=169cm^2*3+169cm^2}}}
{{{676cm^2=507cm^2+169cm^2}}}
{{{676cm^2=676cm^2}}}


5. 
If {{{a=30cm}}}, find {{{m}}} and {{{g}}}.
if {{{a=30cm}}} then {{{xsqrt(13)=30}}}=>{{{x=30/sqrt(3)}}} and {{{2x=2*(30/sqrt(3))=60/sqrt(3)}}}

=>{{{m=60/sqrt(3)cm}}}=>{{{m=(60sqrt(3)/3)cm}}}=>{{{m=20sqrt(3)cm}}}
=>{{{g=(30/sqrt(3))cm}}}=>{{{g=((30sqrt(3))/3)cm}}}=>{{{g=10sqrt(3)cm}}}

check:

{{{m^2=a^2+g^2}}}
{{{(20sqrt(3)cm)^2=(30cm)^2+(10sqrt(3)cm)^2}}}
{{{400*3cm^2=900cm^2+100*3cm^2}}}
{{{1200cm^2=900cm^2+300cm^2}}}
{{{1200cm^2=1200cm^2}}}