Let $D$ and $E$ be points on sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $A D: B D=2: 1$ and $A E: C E=2: 3$. If the area of the triangle $A D E$ is $8 \mathrm{ sq } \mathrm{cm}$, then the area of the triangle $ABC$ , in sq cm , is

We understand what we're given:

Triangle ABC with points D on side AB and E on side AC

AD:BD = 2:1 (meaning D divides AB in the ratio 2:1)

AE:CE = 2:3 (meaning E divides AC in the ratio 2:3)

Area of triangle ADE = 8 sq cm

Since D and E are on the sides of triangle ABC, triangle ADE is a smaller triangle inside ABC that shares the same angle A.

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From the given ratios:

Let AD = 2x and BD = x

Therefore, AB = AD + BD = 2x + x = 3x

Let AE = 2y and CE = 3y

Therefore, AC = AE + CE = 2y + 3y = 5y

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Both triangles ADE and ABC share the same angle A. For any triangle with two sides and the included angle:

Area = $\tfrac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{angle between them})$

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Area of triangle ADE = $\tfrac{1}{2} \times AD \times AE \times \sin(A)$

= $\tfrac{1}{2} \times 2x \times 2y \times \sin(A)$

= $\tfrac{1}{2} \times 4xy \times \sin(A)$

= $2xy \sin(A)$

Since this area equals 8:

$2xy \sin(A) = 8$

Therefore: $xy \sin(A) = 4$

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Area of triangle ABC = $\tfrac{1}{2} \times AB \times AC \times \sin(A)$

= $\tfrac{1}{2} \times 3x \times 5y \times \sin(A)$

= $\tfrac{1}{2} \times 15xy \times \sin(A)$

Since $xy \sin(A) = 4$:

Area of triangle ABC = $\tfrac{1}{2} \times 15 \times 4 = 30$ sq cm

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Notice how the ratio of areas depends only on the ratios of the sides:

Triangle ADE uses sides of length 2x and 2y

Triangle ABC uses sides of length 3x and 5y

The ratio of areas = $\tfrac{2 \times 2}{3 \times 5} = \tfrac{4}{15}$

So: Area of ABC = $\tfrac{15}{4} \times$ Area of ADE = $\tfrac{15}{4} \times 8 = 30$

Answer: 30 sq cm

Whenever you see triangles sharing a common angle with points on the sides, think about using the sine area formula. It's often the fastest route to the solution.