Question

ABC is an traingle,angleABC=90 and M is a midpoint of AC prove AM=BM=CM​

Answer 1

Answer:

Step-by-step explanation:

Midpoint of hypotenuse of right - angle triangle is also  

the circumference  

Also the angle subscribed by triangle in  

a semicircle is right -angle  

∴ In this circle,  

AM=MC=MB= radius ...(1)  

Given, BM=  

177

​  

cmAB+BC=30cm

BY Pythagoras theorem, In ΔABC

AC  

2

=AB  

2

+BC  

2

 

(AM+MC)  

2

=(AB+BC)  

2

−2AB.BC   (∵AC=AM+MC=MB+MB)

(2MB)  

2

=(30)  

2

−2AB.BC

(2  

117

​  

)  

2

−(30)  

2

=−2AB.BC

468−900=−2AB.BC

AB.BC=  

2

432

​  

=216

In ΔABC, BC is Base & AB is height  

∴A(ΔABC)=  

2

1

​  

BC×AB

=  

2

1

​  

×216

A(ΔABC)=108cm  

2

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