Question

4. one of the longest sides of the triangle is 20 m. the other side is 10 m. area of the triangle is 80 m 2 . what is the another side of the triangle'

Answer 1

Refer the image attached:

We will start with finding the height of the triangle:

Area of triangle is given to us that is $A=80 m^2$

Area of triangle $=\frac{h\times b}{2}$

We have assumed that the base length of the triangle is 20 m, plugging the values of area and the height we get:

$80=\frac{20 \times h}{2} =\frac{20h}{2}$

$h=\frac{80 \times 2}{20} =8$

So the height of the triangle is 8 meters.

Now, we have split the base of the triangle into two pieces, lets say the length of one part is 'a' units, so the rest of the base length will be (20-a) units.

Now, using the pythagoras theorem to find the length of the third side:

$x^2=h^2+(20-a)^2$

Now from the smaller triangle we can have:

$10^2=a^2+h^2$

Solving for 'a' we get:

$a= \sqrt{10^2-h^2}=\sqrt{100-64}=\sqrt {36}=6$

The base length of the other triangle is $20-a=20-6=14$

Now using the pyhtagoras theorem for the other triangle we get,

$x= \sqrt{h^2+14^2}=\sqrt{64+196}=\sqrt{264}=16.12$

So the length of the third side is 16.12 meters.