Question

the sides of a triangle are 50 cm, 78 cm and 112 cm . find the altiude to the smallest side

Answer 1

Let
a = 50cm
b = 78cm
c = 112

S = a+b+c/2 

S = 50+78+112/2

S = 240/2 = 120

Area = $\sqrt{s(s-a)(s-b)(s-c)}$

  = $\sqrt{120(120-50)(120-78)(120-112)}$

  =$\sqrt{2822400}$

  = 1680 $cm^{2}$

Longest side = 120cm
corresponding to this side as the base the altitude will be smallest.

area =1/2 x base x altitude

=1680 = 1/2 x 112 x altitude

=altitude = 1680 x 2/112  

= 30cm 



HOPE IT HELPS YOU :-)

Answer 2

Answer:

30 cm

Step-by-step explanation:

here

a = 50cm

b = 78cm

c = 112

S = a+b+c/2  

S = 50+78+112/2

S = 240/2 = 120

Area =  \sqrt{s(s-a)(s-b)(s-c)}  

 =  \sqrt{120(120-50)(120-78)(120-112)}  

 = \sqrt{2822400}  

 = 1680  cm^{2}  

Longest side = 120cm

so the maximum length of altitude possible will be:

area =1/2 x base x altitude

=1680 = 1/2 x 112 x altitude

=altitude = 1680 x 2/112  

= 30cm