The length of each side of an equilateral triangle ABC is $3 \mathrm{~cm}$. Let D be a point on BC such that the area of triangle ADC is half the area of triangle $A B D$. Then the length of $A D$, in cm , is

We need to find the length of AD in an equilateral triangle where point D creates specific area relationships.

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Given: Area of triangle ADC = $\tfrac{1}{2} \times$ Area of triangle ABD

This means: Area of triangle ABD = $2 \times$ Area of triangle ADC

Key Insight: When two triangles share the same height, their areas are proportional to their bases.

Since triangles ABD and ADC both have the same height from vertex A to line BC, we have:

Area of ABD ∝ BD (base length)

Area of ADC ∝ DC (base length)

If Area of ABD : Area of ADC = 2 : 1, then BD : DC = 2 : 1

This means D divides BC in the ratio 2:1.

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Since BC = 3 cm and BD : DC = 2 : 1:

BD = $\tfrac{2}{3} \times 3 = 2$ cm

DC = $\tfrac{1}{3} \times 3 = 1$ cm

Let us verify: BD + DC = 2 + 1 = 3 cm

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To find AD, we'll use the height of the equilateral triangle.

Height of Equilateral Triangle Formula: For an equilateral triangle with side length 'a', the height = $\tfrac{\sqrt{3}}{2} \times a$

For our triangle with side length 3 cm:

Height AL = $\tfrac{\sqrt{3}}{2} \times 3 = \tfrac{3\sqrt{3}}{2}$ cm

Where L is the foot of the perpendicular from A to BC.

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In an equilateral triangle, the height bisects the base. So:

BL = LC = $\tfrac{3}{2}$ cm

Since BD = 2 cm and BL = $\tfrac{3}{2}$ cm:

LD = BD - BL = $2 - \tfrac{3}{2} = \tfrac{1}{2}$ cm

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Now we have a right triangle ALD where:

AL = $\tfrac{3\sqrt{3}}{2}$ cm (height)

LD = $\tfrac{1}{2}$ cm (horizontal distance)

AD = ? (hypotenuse)

Using the Pythagorean theorem:

$AD^2 = AL^2 + LD^2$

$AD^2 = \left(\tfrac{3\sqrt{3}}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2$

$AD^2 = \tfrac{9 \times 3}{4} + \tfrac{1}{4}$

$AD^2 = \tfrac{27}{4} + \tfrac{1}{4}$

$= \tfrac{28}{4}$

$= 7$

Therefore: $AD = \sqrt{7}$ cm

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Final Answer: $AD = \sqrt{7}$ cm