Question

The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?

• A. 48
• B. 60
• C. 70
• D. 80

Hint:

Always draw a diagram for geometry problems. Use algebraic equations for lengths and apply theorems like Pythagoras. Double-check calculations and compare with options. If discrepancies arise, highlight them.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks to find the area of an isosceles triangle, given its altitude to the base and its perimeter.

Step 2: Key Formula or Approach:
1. Properties of an isosceles triangle: The altitude to the base bisects the base and is perpendicular to it. The two non-base sides are equal.
2. Pythagorean theorem: In a right-angled triangle, $a^2 + b^2 = c^2$.
3. Area of a triangle: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Step 3: Detailed Explanation:
Let the isosceles triangle be ABC, with AB = AC.
Let the base be BC. Let D be the midpoint of BC.
The altitude AD = 8 cm. (This forms two right-angled triangles, ADB and ADC).
Let BD = CD = \( x \) cm. So, the base BC = \( 2x \) cm.
Let the equal sides be AB = AC = \( y \) cm.
Perimeter information:
Perimeter = AB + AC + BC = $y + y + 2x = 2y + 2x$.
Given Perimeter = 32 cm.
So, $2y + 2x = 32 \Rightarrow y + x = 16 \Rightarrow y = 16 - x$. (Equation 1)
Pythagorean theorem in triangle ADB:
$AD^2 + BD^2 = AB^2$
$8^2 + x^2 = y^2$
$64 + x^2 = y^2$. (Equation 2)
Substitute Equation 1 into Equation 2:
$64 + x^2 = (16 - x)^2$
$64 + x^2 = 16^2 - 2 \times 16 \times x + x^2$
$64 + x^2 = 256 - 32x + x^2$
Cancel $x^2$ from both sides:
$64 = 256 - 32x$
$32x = 256 - 64$
$32x = 192$
$x = \frac{192}{32} = 6$.
Now, find the base of the triangle:
Base BC = $2x = 2 \times 6 = 12$ cm.
Calculate the Area of the triangle:
Area = $\frac{1}{2} \times \text{Base} \times \text{Height}$
Area = $\frac{1}{2} \times 12 \text{ cm} \times 8 \text{ cm}$
Area = $6 \times 8 = 48 \text{ cm}^2$.

Step 4: Final Answer:
Option (B) 48 is the correct answer