Question

1.The perimeter of a RT angled triangle is 120cm and it's area is 480cm^2 calculate the sides.

Answer 1

Answer:

Explanation:

The area of a triangle is given by

A=hb2  

where  h  is the height and  b  is the base. In the case of a right-angle triangle,  h=rsinθ ,  b=rcosθ  where  r  is the hypotenuse and  θ  is the angle between the hypotenuse and the base side. So

A=rsinθrcosθ2=r2sinθcosθ2 .

Similarly, the perimeter  P  is given by

P=r+h+b=r(1+sinθ+cosθ).  

If we square the perimeter, we get

P2=r2(1+(sinθ+cosθ))2=  

r2(1+2(sinθ+cosθ)+cos2θ+2cosθsinθ+sin2θ=  

r2(1+2(sinθ+cosθ)+1+2cosθsinθ=  

2r2(1+sinθ+cosθ+sinθcosθ)  

Noting

P=r(1+sinθ+cosθ)  

2A=r2sinθcosθ  

We can substitute into the equation for  P2  to get

P2=2r2(1+sinθ+cosθ+sinθcosθ)=  

2r2(1+sinθ+cosθ)+2r2sinθcosθ=2rP+4A  

So  P2=2rP+4A , and we can solve for  r :

r=P2−4A2P  

You have  P=60 ,  A=120 , so  A=2P  

r=P2−8P2P=P−82=26  

Given the radius and area, we can find the angle \theta fairly easily, from the double-angle identity:

sin2θ=2sinθcosθ  

Noting

A=r2sinθcosθ2  

A=r2sin2θ4  

sin2θ=4Ar2  

I tried fiddling with various identities to see if I could come up with a nice expression for \theta, but haven’t been able to. However:

θ=arcsin4Ar22  

4A=480,r2=(4)(169) , so  4Ar2=120169=.71 ,  θ=.39  

So  sinθ=0.384 ,  cosθ=0.923 . and the sides are 26, 24, 10

hope this helps u

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