Area of triangle ACD using a hypotenuse segment

Key concept: Triangle $ACD$ shares the same altitude from $C$ to line $AB$, so its area follows directly from the base ratio.[1]

Step by step

Identify the full right triangle

In right triangle $ABC$, the hypotenuse is $AB=25$ and one leg is $AC=7$. Since $AB$ is the hypotenuse, the right angle is at $C$. That means $BC$ can be found by the Pythagorean theorem:

$BC^2=AB^2-AC^2=25^2-7^2=625-49=576$

so

$BC=24$.

The area of triangle $ABC$ is therefore

$\frac{1}{2}\cdot 7\cdot 24=84$.

Use the shared altitude

Point $D$ lies on side $AB$ with $AD=10$. Triangle $ACD$ has the same height from $C$ to line $AB$ as triangle $ABC$, because both triangles use a base on the same line $AB$.

That means their areas are proportional to their bases:

$\frac{[ACD]}{[ABC]}=\frac{AD}{AB}=\frac{10}{25}=\frac{2}{5}$

So the area of triangle $ACD$ is

$[ACD]=\frac{2}{5}\cdot 84=\frac{168}{5}$.

Why this works

The entire method depends on the fact that triangles with the same altitude have areas in the same ratio as their bases. Here, the altitude from $C$ to the line $AB$ is shared by both triangles, so you do not need to find the exact location of $D$ on the hypotenuse.

The final area is $\frac{168}{5}$ square units.

Pitfall alert

The mistake that often appears in this $25$-$7$ right triangle is treating $AD=10$ as if it were a leg of a new right triangle and then trying to use the Pythagorean theorem again. Segment $AD$ is only part of the hypotenuse, not a perpendicular side, so it cannot be used that way. Another error is forgetting that the area ratio depends on the shared altitude from $C$ to line $AB$. Because both triangles sit on the same line, the altitude is identical, and the bases alone determine the ratio. It is also easy to compute the whole triangle area incorrectly by mixing up the hypotenuse with a leg. First find $BC=24$, then use $\frac{AD}{AB}$ to scale the area from $ABC$ to $ACD$. That keeps the geometry clean and avoids unnecessary coordinate work.

Try different conditions

If the point were placed so that $AD=15$ instead of $10$, while the triangle still had $AB=25$ and $AC=7$, the same area-scaling idea would give $[ACD]=\frac{15}{25}\cdot 84=\frac{252}{5}$. If the hypotenuse were changed to $AB=26$ with the same leg $AC=10$ and $AD=13$, you would first find the other leg using the Pythagorean theorem, then scale the full triangle area by $AD/AB$. This variation shows that the key step is not memorizing one number, but recognizing that any triangle cut off by a segment on the same base line keeps the same altitude and therefore has an area proportional to its base.

Further reading

Pythagorean theorem, triangle area ratio, shared altitude