Examkin

Question

Difficulty: MediumTriangles: Properties, Perimeter, and Area

In isosceles triangle ABCABC, side ABAB is equal in length to side ACAC. The perimeter of triangle ABCABC is 3636, and the length of the altitude from vertex AA to base BCBC is 1212. What is the area of triangle ABCABC?

Answer: 60

Answer

The area of triangle ABCABC is 6060.
Let xx be the length of the two equal sides ABAB and ACAC, and bb be the length of base BCBC. The perimeter is 2x+b=362x + b = 36, yielding x=18−b2x = 18 - \frac{b}{2}. The altitude from AA to BCBC has length 1212 and bisects BCBC into two segments of length b2\frac{b}{2}. Applying the Pythagorean theorem to one of the right triangles gives x2=122+(b2)2x^2 = 12^2 + \left(\frac{b}{2}\right)^2. Substituting x=18−b2x = 18 - \frac{b}{2} gives (18−b2)2=144+b24  ⟹  324−18b+b24=144+b24  ⟹  18b=180  ⟹  b=10\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10. The area is 12×10×12=60\frac{1}{2} \times 10 \times 12 = 60.

Step-by-Step Solution

1
Set up an equation for the side lengths using the perimeter.
Let bb be the length of base BCBC, and xx be the length of sides ABAB and ACAC. Since the perimeter is 3636, 2x+b=362x + b = 36, which gives x=18−b2x = 18 - \frac{b}{2}.
An isosceles triangle has two sides of equal length, and perimeter is the sum of all three side lengths.
2
Apply the Pythagorean theorem to the right triangle formed by the altitude.
The altitude of length 1212 drops perpendicularly to base BCBC, bisecting it into two equal segments of length b2\frac{b}{2}. Thus, x2=122+(b2)2=144+b24x^2 = 12^2 + \left(\frac{b}{2}\right)^2 = 144 + \frac{b^2}{4}.
In an isosceles triangle, the altitude to the base bisects the base and creates two congruent right-angled triangles.
3
Solve for the base length bb.
Substitute x=18−b2x = 18 - \frac{b}{2} into the equation: (18−b2)2=144+b24  ⟹  324−18b+b24=144+b24  ⟹  18b=180  ⟹  b=10\left(18 - \frac{b}{2}\right)^2 = 144 + \frac{b^2}{4} \implies 324 - 18b + \frac{b^2}{4} = 144 + \frac{b^2}{4} \implies 18b = 180 \implies b = 10.
Expanding the squared binomial allows the b24\frac{b^2}{4} terms to cancel out, resulting in a linear equation for bb.
4
Calculate the area of the triangle.
\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 10 \times 12 = 60.
The area of a triangle is evaluated using half the product of its base and corresponding altitude.

Key Concept

Isosceles triangle properties, altitude-to-base bisector property, Pythagorean theorem, and triangle area calculation.
Rate this question

Topics

Triangles: Properties, Perimeter, and Area

More questions from this topic