Question

The perimeter of an isosceles triangle is 42 cm. It's base is 2/3 times the sum of equal sides. Find the length of each side and the area of the triangle.

Answer 1

Question #44041 – Math – Geometry

The perimeter of an isosceles triangle is $42 \, \text{cm}$. It's base is $2/3$ times the sum of equal sides. Find the length of each side and the area of the triangle.

Solution:

Let $x$ is the length of equal sides ($AB = CB = x$), then $AC = 2x \cdot \frac{2}{3} = \frac{4x}{3}$ is the length of a base.

The perimeter of an isosceles triangle is:

$$P = AB + CB + AC = 2x + \frac{4x}{3}$$

Given $P = 42$

$$2x + \frac{4x}{3} = 42$$ $$\frac{10x}{3} = 42$$ $$AB = CB = x = 12.6 \, \text{cm}$$ $$AC = \frac{4x}{3} = \frac{4 \cdot 12.6}{3} = 16.8 \, \text{cm}$$

Now given a triangle with sides $AB = CB = 12.6 \, \text{cm}$ and $AC = 16.8 \, \text{cm}$.

The formula for the area is: $A = \sqrt{p(p - AB)(p - CB)(p - AC)}$

Where $p = \frac{P}{2} = \frac{42}{2} = 21$

$$A = \sqrt{21(21 - 12.6)(21 - 12.6)(21 - 16.8)} \approx 79 \, \text{cm}^2$$

Second method:

If $BH$ is the altitude of the triangle, then the area of the triangle is:

$$A = \frac{AC \cdot BH}{2}$$

Using Pythagoras' theorem:

$$B H ^ {2} + A H ^ {2} = A B ^ {2}$$

If $\Delta ABC$ is an isosceles triangle ( $AB = CB$ ), then:

$$A H = H C = \frac {A C}{2} = \frac {1 6 . 8}{2} = 8. 4 c m$$ $$B H = \sqrt {A B ^ {2} - A H ^ {2}} = \sqrt {1 2 . 6 ^ {2} - 8 . 4 ^ {2}} \approx 9. 4 c m$$ $$\boldsymbol {A} = \frac {1 6 . 8 \cdot 9 . 4}{2} \approx \boldsymbol {7 9 c m ^ {2}}$$

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