Question

Three sides of a triangular plot are 13 m, 14 m and 15 m. Find the area of the plot. How
much minimum distance has a man to walk if he wants to come to the longest side
from the opposite corner?​

Answer 1

First of all let's find the area of triangle.

semi perimeter (s) =(13+14+15)/2=42/2=21 m,

then

Area of the ∆=√[21(21-13)(21-14)(21-15)],

=√[21×8×7×6],

=√(3×7×2×2×2×7×2×3),

=3×7×2×2,

=84 m²

Now,

We need to find the shortest distance to reach its longest side i.e., 15cm (base) from its opposite corner. ( from point A to D )

Following the figure,

We have, base = 15cm

Shortest distance from point A to D is the height of the triangular plot.

So,

We can say that,

Area of the plot = ½ • base • height

Or

→ 84 = ½ • 15 • h

→ 84*2/15 = h

→ 11.2 = h

Hence, the required shortest distance is 11.2cm.

Hope your concepts are clear.

Refer to the attachment.

Tips:

If you are in exam and you don't know how to solve the question you must draw the drawing of the figure which may helps you in solving the question.

Hope you like my answer.✌