Question

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD,in cm,is

• A. $\sqrt{6}$
• B. $\sqrt{5}$
• C. $\sqrt{8}$
• D. $\sqrt{7}$

Answer 1

Answer: (D)

$\sqrt{7}$

Answer 2

Given that ABC is an equilateral triangle with side length 3 cm, let's find the length of AD such that the area of triangle ADC is half the area of triangle ABD.
1. First, let's calculate the area of triangle ABC using the formula for the area of an equilateral triangle: $\text{Area}_{ABC} = \frac{\sqrt{3}}{4} \times (\text{side length})^2$
Plugging in the side length ($3$ cm), we get:
$\text{Area}_{ABC} = \frac{\sqrt{3}}{4} \times 3^2 = \frac{9\sqrt{3}}{4} \text{ cm}^2$

2. Since triangle ADC has half the area of triangle ABD, we have:
$\text{Area}_{ADC} = \frac{1}{2} \times \text{Area}_{ABD}$
Now, let's find the area of triangle ABD. For this, we can use the formula for the area of a triangle: $\text{Area}_{ABD} = \frac{1}{2} \times \text{base} \times \text{height}$
The base of triangle ABD is AD, and the height can be found using the Pythagorean theorem in triangle ABD. Let h be the height of triangle ABD:
$h^2 = (\text{side length of } \triangle ABC)^2 - (\text{half of side length of } \triangle ADC)^2$
$h^2 = 3^2 - \left(\frac{3}{2}\right)^2$
$h^2 = 9 - \frac{9}{4} = \frac{27}{4}$

Therefore, $h = \frac{\sqrt{27}}{2} = \frac{3\sqrt{3}}{2}$.

Now, we can find the area of triangle ABD:
$\text{Area}_{ABD} = \frac{1}{2} \times AD \times \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} \times AD \text{ cm}^2$

3. Substituting this into the equation for the area of triangle ADC, we get:
$\frac{3\sqrt{3}}{4} \times AD = \frac{1}{2} \times \frac{9\sqrt{3}}{4}$
$AD = \frac{1}{3} \times 3 = \sqrt{7} \text{ cm}$

Hence, the length of AD is $√7$ cm.