### Theory:

Now, we shall find the area of the right-angled triangle.

Let us consider a rectangle. When we cut a rectangle through the diagonal, we make $$2$$ right-angled triangles.

In this right-angled triangle, the base of the angle containing $$90^{\circ}$$ is considered as base ($$b$$) $$units$$, and the other side is considered as height ($$h$$) $$units$$.

Then, the area of the right-angled triangle $$ABC$$ is given by:

$$2\times$$ Area of the right-angled triangle $$=$$ Area of the rectangle

$$2\times$$ Area of the right-angled triangle $$= l \times b$$

Area of the right-angled triangle $$=\frac{1}{2} (l \times b)$$

Here, the base and height of the right-angled triangle are the length and breadth of the rectangle, respectively.

Therefore, the area of the right-angled triangle is $$\frac{1}{2} (b \times h)$$$$sq. \ units$$.
Example:
1. Let the base and height of the right-angled triangle be $$6 \ cm$$ and $$10 \ cm$$, respectively. Find the area of the right-angled triangle.

Solution:

Let $$b$$ denote the base and $$h$$ denote the height. Then, $$b=6 \ cm$$ and $$h=10 \ cm$$.

Substituting the values in the area of the right-angled triangle formula, we have:

Area, $$A=\frac{1}{2}(b \times h)$$

$$A=\frac{1}{2}(6 \times 10)$$ $$sq. cm$$

$$A=\frac{1}{2}\times 60$$ $$sq. cm$$

$$A=30 \ sq. cm$$

Therefore, the area of the right-angled triangle is $$30$$  $$sq. \ cm$$.