The area of triangle ABC is 39 cm². D and E are two points on BC such that BD = DE = EC, then what is the area of triangle ADC?

Correct option is A

Given:
The area of triangle ABC is 39 cm².
Points D and E divide BC into three equal parts, BD = DE = EC.
Concept used:
Since D and E divide BC into three equal segments, we can say that triangles ABD, ADE, and AEC have equal bases and share the same height from point A.

Solution:
The total area of triangle ABC is 39 cm²,

​$$\text{Area of } \triangle ABD = \text{Area of } \triangle ADE = \text{Area of } \triangle AEC = \frac{1}{3} \times 39 = 13 \, \text{cm}^2$$​​

The area of triangle ADC is the sum of the areas of triangles ADE and AEC:


Area of triangle ADC = 13 + 13 = 26$$\, \text{cm}^2$$​​


Thus, the area of triangle ADC =26$$cm^2$$​​