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

Answer 2

Rather than setting up and solving equations from scratch, we can use the fact that the altitude of 8 cm strongly suggests a well-known Pythagorean triple is at play, and then verify it against the given perimeter.

The multiple of the 3-4-5 triple that contains an 8 is 6-8-10 (scaling 3-4-5 by 2). If the altitude, one leg of the right triangle formed by the altitude, half the base, and the equal side, is 8, and the equal side (hypotenuse) is 10, then half the base is 6, making the full base 12.

Let's check whether this fits the given perimeter: the two equal sides would be 10 each, and the base 12, giving a perimeter of \(10+10+12=32\) cm, which matches the given perimeter exactly.

So the base is 12 cm and the altitude is 8 cm. The area is:

\[ \text{Area} = \frac{1}{2}\times\text{base}\times\text{height} = \frac{1}{2}\times12\times8 = 48 \text{ cm}^2 \]

Checking the other options: 60, 70, and 80 would all require a larger base than 12 for the same altitude of 8, but a larger base pushes the perimeter above 32 cm once the equal sides are recalculated using the Pythagorean theorem, so none of those areas are consistent with the given perimeter.

Therefore, the correct answer is 48.