Find the area and perimeter of an isosceles right triangle, each of whose equal sides measures 10 cm. [Given: $\sqrt{2}=1.41$]

Equal sides of measure $=10cm$
Here, base and perpendicular are equal sides.
Area of a triangle $=\frac{1}{2}\times Base\times Height$ 
                              $=\frac{1}{2}\times 10\times 10$
                              $=50$

From Pythagorean Theorem:
Hypotenuse${}^2=$ Base${}^2+$ Perpendicular${}^2$
                        $=10^2+10^2$
                        $=200$
or Hypotenuse $=10\sqrt{2}cm=10\times 1.41cm=14.1cm$
Perimeter $=$ Sum of all the sides $=10+10+14.1=34.1$
Therefore, Area of required triangle is $50c{m}^2$ and perimeter $=34.1cm$