In a triangle $ABC$ , medians $AD$ and $BE$ are perpendicular to each other, and have lengths $12 \mathrm{~cm}$ and $9 \mathrm{~cm}$, respectively. Then, the area of triangle $ABC$, in sq cm, is

We need to work with medians of a triangle and use the special properties when they're perpendicular.

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A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

All three medians of a triangle intersect at a single point called the centroid. The centroid divides each median in a 2:1 ratio from vertex to midpoint.

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Let's say the medians AD and BE intersect at point F (the centroid).

Since F is the centroid:

AF : FD = 2 : 1

BF : FE = 2 : 1

Given that AD = 12 cm:

AF = $\dfrac{2}{3} \times 12 = 8$ cm

FD = $\dfrac{1}{3} \times 12 = 4$ cm

Given that BE = 9 cm:

BF = $\dfrac{2}{3} \times 9 = 6$ cm

FE = $\dfrac{1}{3} \times 9 = 3$ cm

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The centroid divides each median such that the distance from vertex to centroid is twice the distance from centroid to the midpoint of the opposite side.

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Since medians AD and BE are perpendicular, we can use F as a right angle to calculate areas.

Area of triangle ABE:

Triangle ABE has:

Base = BE = 9 cm

Height = AF = 8 cm (perpendicular distance from A to line BE)

Area of triangle ABE = $\dfrac{1}{2} \times 9 \times 8 = 36$ cm²

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Since E is the midpoint of AC, triangle ABE has exactly half the area of triangle ABC.

Both triangles ABE and ABC share the same vertex B, but triangle ABE has base AE while triangle ABC has base AC. Since E is the midpoint, AE = $\dfrac{1}{2} \times AC$.

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Since Area of triangle ABE = 36 cm²

And Area of triangle ABE = $\dfrac{1}{2} \times$ Area of triangle ABC

Therefore: Area of triangle ABC = $2 \times 36 = 72$ cm²

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The area of triangle ABC is 72 cm².