Question

nidhi has lawn in the form of isosceles ∆ of side 13 and 10 m a.she want to divide into 2∆ parts of equal area how she can do it. justify
b.what is the area of ∆ thus each form

Answer 1

From the described triangle, we can draw a perpendicular bisector of the 10m side which will pass through the opposite vertex (because it is an isosceles triangle). This perpendicular bisector will clearly divide the area of the triangle into half.
Considering one of the 2 congruent triangles hence formed, 
It is a right angled triangle with base length = 5m, and hypotenuse = 13m.
using Pythagoras theorem,
$5^{2} + X^{2} = 13^{2}$
$=> X = \sqrt{169 - 25}$
$=> X = \sqrt{144}$
=> X = 12m
Hence, we got the altitude of the triangle = 12m
Now,
total area of the isosceles triangle = $\frac{(base)x(height)}{2}$
=> Area = 60 $m^{2}$

Therefore, after dividing the area into two equal parts, area of each part $= 30 m^{2}$