Question

HELP PLZ
21. Find the perimeter of the triangle:
2
A
4715

Answer 1

16
`√(5)`+4
`√(15)` or 51.26902

Explanation:

Consider one of the smaller, right-angled triangles within the main triangle, and use sine to find the hypotenuse of the small triangles:

sinA=opposite/hypotenuse

sin60=4
`√(15)`/hypotenuse

hypotenuse=
`(4√(15) )/((√(3) )/(2) )`

hypotenuse=8(
`(√(15) )/(√(3) )`)

hypotenuse=8
`√(5)`

So the left and right hand sides of the larger triangle are both 8
`√(5)` respectively.

To find the base, once again consider the smaller triangle, and this time, use pythagorus:

a²=b²+c²

(8
`√(5)`)²=(4
`√(15)`)²+c²

320=240+c²

80=c²

c=4
`√(15)`

So the perimeter is 8
`√(5)`+8
`√(5)`+4
`√(15)`=16
`√(5)`+4
`√(15)`=51.26902