Question

the area of a triangle inscribed in a circle is 60cm^2 and the radius of the circumscribed circle is 8cm. If the 2 sides of the inscribed triangle measure 10cm and 12cm, find the length of the third size.

Answer 1

area of triangle = 60cm^2
Radius of circle = 8cm
Triangle side A = 10cm
Triangle side B = 12cm
Triangle side C = ???

Answer 2

I dont know how to get the answer pls help

Answer 3

now ive never tried this before, but couldn't we use side-side-side to work out the angles where AB meet, then Side-angle-side to work out side C?

so angle would be found from Cosine rule
z^2 = x^2 + y^2 - 2xy Cos (Z)
8^2 = 12^2 + 8^2 - 2*12*8 Cos Z
64 = 144 + 64 - 192 * Coz Z
-144 = - 192 * Coz Z
144 = 192 * Coz Z
144/192 = Coz Z
angle Z = cos^-1 (144/192) = 41.41 degrees

do the same for triangle made up from 8,8,10 (side A)
z^2 = x^2 + y^2 - 2xy Cos (Z)
8^2 = 10^2 + 8^2 - 2*10*8 Cos Z
64 = 100 + 64 - 160 * Coz Z
-100 = - 160 * Coz Z
100 = 160 * Coz Z
100/160 = Coz Z
angle Z = cos^-1 (100/160) = 51.32 degrees

sub of 2 angles = 51.32 + 41.41 = 92.73 degrees


now do the same for the big triangle

c^2 = a^2 + b^2 - 2ab cos C
c^2 = 10^2 + 12^2 - 2*10*12 * cos 92.73
c^2 = 100 + 144 - 240 * cos 92.73
c^2 = 244 + 11.43
c^2 = 232.56
c = 15.25 ish

sorry man, never dealt with circum-triangles before... but this makes logical sense to me... maybe too longwinded tho?

Answer 4

@BeefJerkyLover ?

Answer 5

think i spot a mistake here, sorry:
now do the same for the big triangle

c^2 = a^2 + b^2 - 2ab cos C
c^2 = 10^2 + 12^2 - 2*10*12 * cos 92.73
c^2 = 100 + 144 - 240 * cos 92.73
c^2 = 244 + 11.43
c^2 = 255.43
c = 15.98 ish

Answer 6

Try this
Area of triangle=\[\sqrt{s(s-a)(s-b)(s-c)}=60\].......Heron's formulae
Squaring both sides
s(s-a)(s-b)(s-c)=3600
multiply throughout by 16
16s(s-a)(s-b)(s-c)=57600
2s(2s-2a)(2s-2b)(2s-2c)
(a+b+c)(b+c-a)(a+c-b)(a+b-c)=57600
now put a=10 and b=12 then solve for c....

Answer 7

ok... cool, never seen that before, but what's "s" in the above eqn? @Salmon ?

Answer 8

s is semi-perimeter i.e
s=(a+b+c)/2

Answer 9

ah, ok... thanks...
why did u multiply by 16?

Answer 10

so that it become easy to substitute
2s=a+b+c
rather than bringing fractions into the equation and increase the steps by substituting s=(a+b+c)/2

Answer 11

gotcha, cheers man

Answer 12

Let the two given sides be a and b and they intersect at an angle Q
Area, A=1/2(axbxsinQ)
1/2(12x10sinQ)=60
sinQ=1
Q=90 degrees.
using cosine rule:
c^2=a^2 +b^2 -2xaxbxcosQ
c^2=12^2 + 10^2-2x2x10xcos90
c^2=144+100-0
c^2=244
c=15.62cm

Answer 13

perfect @UhuruThomas , very elegant solution dude

Answer 14

Thanks. you can even use the Heros formula to prove that
a=10cm b=12cm and c=15.62cm
s=1/2(10+12+15.62)=18.81cm
Area=sqrt of s(s-a)(s-b)(s-c)
Area= sqrt of {18.81(18.81-10)(18.81-12)(18.81-15.62)}
Area=59.99999987cm^2 which is appxmtly 60cm^2

Answer 15

what is the use of the 8 radius of the circle then? no use?