Question

The length of the sides of a triangle are 4cm,6cm and 8cm.The length of perpendicular from the opposite vertex to the side whose length is 8cm,is equal to

Answer 1

Given :

The length of the sides of a triangle are 4cm,6cm and 8cm

To Find :

The length of perpendicular from the opposite vertex to the side whose length is 8cm.

Solution :

The perpendicular drawn from the opposite vertex to the side whose length is 8 cm divides the side into 4 cm each.

Now, let the corners of the triangle be A, B and C such that AB = 4 cm, BC = 6 cm, CA = 8 cm.

So the perpendicular is drawn from B to the side C. Let the point on AC where the perpendicular bisects be D.

We need to find the length of BD.

Using Pythagoras theorem in right angled triangle BDC,

           $BC^{2} = DC^{2} + BD^{2}$

           BD   =   $\sqrt{BC^{2} - DC^{2} }$

                   =   $\sqrt{6^{2} - 4^{2} }$

                   = $\sqrt{36 - 16}$

                   = $\sqrt{20}$

                   = 4.47 cm.

∴ The length of perpendicular from the opposite vertex to the side whose length is 8cm, is equal to 4.47 cm.

           

Answer 2

Given:

Triangle is a scalene triangle with sides: a = 4 cm, b = 6 cm, c = 8 cm

To Find: Length of perpendicular on side with length 8 cm

Calculation:

• Semi - perimeter of triangle (s) = $\frac{a+b+c}{2}$ = $\frac{4+6+8}{2} = 9$

Area of a triangle is given as:

• Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

• Area = $\sqrt{9*5*3*1} =\sqrt{135} =11.61$

Now, assuming 8 cm as base, area of triangle  = $\frac{1}{2} *b*h=\frac{1}{2}*8*h$

• Area = $\frac{1}{2} *8*h$

• 11.61 =  $\frac{1}{2} *8*h$

• h = 2.90

Answer:

Height of perpendicular is 2.90.