A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $7$ $7$ , what is the largest possible area of the triangle?

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A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $7$ $7$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-10-08T04:03:00

Area of the largest triangle possible $= 87.4551$ $=87.4551$

Explanation:

The angles are $\frac{π}{2}, \frac{3 π}{8}, \frac{π}{8}$ $pi/2,(3pi)/8,pi/8$
It is a right angle triangle.

Smallest side $= 7$ $=7$ It is also the height.
$:.a/sin(pi/12)=b/sin((3pi)/8)=c/sin(pi/2)$
$7/sin(pi/12)=b/sin((3pi)/8)=c/1$
$c=7/sin(pi/12)=27.0459=$ hypotenuse of the triangle.
$b=((7*sin((3pi)/8))/sin(pi/12))$
$=24.9872$ . It is also the base.

Area of the triangle $=(1/2)*b*h=(1/2)*7*24.9872=87.4551$

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A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $7$ $7$ , what is the largest possible area of the triangle?

1 Answer

Explanation:

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Explanation:

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A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $7$ $7$ , what is the largest possible area of the triangle?