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

The triangle's altitude to the base is fixed at 8 cm, so once we know the area we immediately know the base, and once we know the base we can check whether the two equal sides implied by it add up to the given perimeter of 32 cm. This gives a way to test each option directly.

1. 48: Base \( = \frac{2 \times 48}{8} = 12 \) cm, so half the base is 6 cm. Each equal side is \( \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10 \) cm. Perimeter \( = 12 + 10 + 10 = 32 \) cm, matching the given perimeter exactly.
2. 60: Base \( = \frac{2 \times 60}{8} = 15 \) cm, half-base 7.5 cm. Each equal side \( = \sqrt{8^2 + 7.5^2} = \sqrt{64+56.25} = \sqrt{120.25} \approx 10.97 \) cm. Perimeter \( \approx 15 + 10.97 + 10.97 \approx 36.94 \) cm, which overshoots 32 cm.
3. 70: Base \( = \frac{2 \times 70}{8} = 17.5 \) cm, half-base 8.75 cm. Each equal side \( = \sqrt{8^2+8.75^2} = \sqrt{64+76.5625} \approx \sqrt{140.56} \approx 11.86 \) cm. Perimeter \( \approx 17.5 + 11.86 + 11.86 \approx 41.22 \) cm, far above 32 cm.
4. 80: Base \( = \frac{2 \times 80}{8} = 20 \) cm, half-base 10 cm. Each equal side \( = \sqrt{8^2+10^2} = \sqrt{64+100} = \sqrt{164} \approx 12.8 \) cm. Perimeter \( \approx 20 + 12.8 + 12.8 \approx 45.6 \) cm, well beyond 32 cm.

Only an area of 48 produces a base and pair of equal sides whose lengths add up to exactly the given perimeter of 32 cm; every larger area option overshoots the perimeter.

Therefore, the correct answer is 48.