in a triangle find the length of the other two sides if the perimetre is 19cm

Question

this is a geometric sequence. the length of the shortest side is 4cm and the perimetre is 19cm. find the lengths of the other two sides

Answer 1

Use the cosine rule: a^2=b^2+c^2-2bccosA. If c=4, then a+b=19-4=15, so b=15-a.

a^2=225-30a+a^2+16-8(15-a)cosA; 0=241-30a-(120-8a)cosA

cosA=(241-30a)/(120-8a). Since -1<cosA<1, -1<(241-30a)/(120-8a)<1.

Take the case 120-8a>0 first (a<15cm):

241-30a<120-8a; 121<22a; 11/2<a so 5.5<a<15; AND

8a-120<241-30a; 38a<361; a<19/2; a<9.5cm, so 5.5<a<9.5 fits both inequalities.

Now the case 120-8a<0 (a>15cm):

241-30a>120-8a; 4<a<5.5; AND

8a-120>241-30a; a>121/38. Both inequalities fail to meet the requirement a>15.
 

So now we have the other two sides between 5.5cm and 9.5cm (sum of 15cm). The angle between them is given by:

cosA=(241-30a)/(120-8a)=76/76=1 or -44/44=-1, the range 0 to 180. So as long as the sides are between these limits we can have any angle between them. Just as an example pick 7.5cm, making an isosceles triangle:

cosA=4/15, A=74.53°. The 3 sides are 4cm, 7.5cm and 7.5cm. Perimeter=19cm.

When the angle between the sides is 90 degrees, a=241/30=8.033c, b=6.967cm.