Q8:

$\triangle ABC$ is isosceles with $AB = AC = 7.5$ cm and $BC = 9$ cm (Fig 9.17). The height $AD$ from $A$ to $BC$, is $6$ cm. Find the area of $\triangle ABC$. What will be the height from $C$ to $AB$ i.e., $CE$?

Given Variables & Initial Setup

We are given an isosceles triangle, $\triangle ABC$, with the following dimensional properties:

• Length of equal sides: $AB = AC = 7.5 \text{ cm}$
• Length of the base: $BC = 9 \text{ cm}$
• Altitude (height) corresponding to base $BC$: $AD = 6 \text{ cm}$

We are tasked with determining two values:

1. The total area of $\triangle ABC$.
2. The length of the altitude $CE$, which is the perpendicular dropped from vertex $C$ to the side $AB$.

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

The fundamental theorem of Euclidean geometry states that the area of any triangle can be calculated using the length of any chosen base and its corresponding perpendicular height (altitude). The formula is given by:

$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} $

Taking $BC$ as the base, the corresponding height is the altitude dropped from vertex $A$, which is $AD$. Substituting the given values:

• $\text{Base } (b_1) = BC = 9 \text{ cm}$
• $\text{Height } (h_1) = AD = 6 \text{ cm}$

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

$ \text{Area of } \triangle ABC = \frac{1}{2} \times 9 \text{ cm} \times 6 \text{ cm} $

$ \text{Area of } \triangle ABC = 9 \times 3 \text{ cm}^2 = 27 \text{ cm}^2 $

Step 2: Establishing the Invariance of Area to Find $CE$

[Per the invariant property of the area of a polygon], the area of a rigid geometric figure remains constant regardless of the perspective or the specific base-height pair used for calculation. Therefore, if we reorient our perspective and treat side $AB$ as the base, the corresponding height is the altitude $CE$.

We set up the area equation using the new base-height pair:

• $\text{Base } (b_2) = AB = 7.5 \text{ cm}$
• $\text{Height } (h_2) = CE \text{ (Unknown)}$

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

Step 3: Algebraic Isolation and Calculation of $CE$

Since the area is strictly invariant, we equate the expression from Step 2 to the calculated area from Step 1 ($27 \text{ cm}^2$):

$ 27 = \frac{1}{2} \times 7.5 \times CE $

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

$ 54 = 7.5 \times CE $

Next, divide both sides by $7.5$:

$ CE = \frac{54}{7.5} $

To simplify the division, multiply the numerator and the denominator by $10$ to eliminate the decimal:

$ CE = \frac{540}{75} $

Divide both the numerator and the denominator by their greatest common divisor, which is $15$:

$ CE = \frac{540 \div 15}{75 \div 15} = \frac{36}{5} $

Converting the improper fraction to a decimal:

$ CE = 7.2 \text{ cm} $

Final Solution: The area of $\triangle ABC$ is exactly $27 \text{ cm}^2$, and the height from $C$ to $AB$ (altitude $CE$) is $7.2 \text{ cm}$.