Q7:

$\triangle ABC$ is right angled at $A$ (Fig 9.16). $AD$ is perpendicular to $BC$. If $AB = 5$ cm, $BC = 13$ cm and $AC = 12$ cm, Find the area of $\triangle ABC$. Also find the length of $AD$.

Given Variables & Initial Setup

We are given a right-angled triangle, $\triangle ABC$, with the right angle located at vertex $A$. The dimensions of the sides are provided as follows:

• Length of leg $AB = 5 \text{ cm}$
• Length of leg $AC = 12 \text{ cm}$
• Length of hypotenuse $BC = 13 \text{ cm}$
• $AD$ is the altitude dropped from the right angle $A$ to the hypotenuse $BC$, meaning $AD \perp BC$.

Step 1: Calculating the Area of $\triangle ABC$

The area of any triangle is given by the standard formula:

$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $

[Because $\triangle ABC$ is right-angled at $A$, the two legs forming the right angle ($AB$ and $AC$) are mutually perpendicular. Therefore, one leg can act as the base while the other acts as the corresponding height.]

Let the base be $AC = 12 \text{ cm}$ and the height be $AB = 5 \text{ cm}$. Substituting these values into the area formula yields:

$ \text{Area of } \triangle ABC = \frac{1}{2} \times AC \times AB $

$ \text{Area of } \triangle ABC = \frac{1}{2} \times 12 \text{ cm} \times 5 \text{ cm} $

$ \text{Area of } \triangle ABC = 6 \times 5 = 30 \text{ cm}^2 $

Step 2: Formulating the Equation for Altitude $AD$

[The area of a geometric figure is invariant; it remains constant regardless of which side is chosen as the base for the calculation.]

We can recalculate the area of $\triangle ABC$ by choosing the hypotenuse $BC$ as the new base. The corresponding height for this base is the perpendicular altitude $AD$.

$ \text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} $

$ \text{Area of } \triangle ABC = \frac{1}{2} \times BC \times AD $

Step 3: Solving for the Length of $AD$

We already established from Step 1 that the area of the triangle is $30 \text{ cm}^2$. We also know the length of the hypotenuse $BC = 13 \text{ cm}$. Substituting these knowns into our new area equation:

$ 30 = \frac{1}{2} \times 13 \times AD $

To isolate $AD$, multiply both sides of the equation by 2:

$ 60 = 13 \times AD $

Divide both sides by 13:

$ AD = \frac{60}{13} \text{ cm} $

Converting this fraction to a decimal provides the approximate length:

$ AD \approx 4.615 \text{ cm} $

Final Solution: The area of $\triangle ABC$ is $30 \text{ cm}^2$, and the length of the altitude $AD$ is $\frac{60}{13} \text{ cm}$ (approximately $4.62 \text{ cm}$).