CAT 2025

Slot 3

QA

Question & SolutionQues & Sol

In triangle ABC, (AB=AC=12) cm and D is a point on side BC such that (AD=8) cm. If AD is extended to a point E such that ∠ ACB = ∠ AEB, then the length, in cm, of AE is

1. Concept Used

• Topic: Geometry – Isosceles Triangles, Angle Conditions, and Trigonometric Ratios
• Formula:

$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}, \quad AE = AD + DE$$

2. Calculation

Let the side of the isosceles triangle be $AB = AC = 12$ cm, with $D$ on $BC$ and $AD = 8$ cm.

Step 1: Establish perpendicularity.

Since $\triangle ABC$ is isosceles with $AB = AC$, the line $AD$ is both the altitude and the median from $A$ to $BC$. Thus $AD \perp BC$, so $\angle ADB = \angle ADC = 90°$.

Step 2: Find CD using the Pythagorean theorem in $\triangle ACD$.

$$CD^2 = AC^2 - AD^2 = 12^2 - 8^2 = 144 - 64 = 80$$

$$CD = \sqrt{80} \ \text{cm}$$

Since $AD$ is the median, $BD = CD = \sqrt{80}$ cm.

Step 3: Set up the tangent equation using the equal angle condition.

In $\triangle ACD$:

$$\tan(\angle ACB) = \frac{AD}{CD} = \frac{8}{\sqrt{80}}$$

Since $AD$ is extended to $E$ and $\angle BDE = 90°$, in $\triangle BDE$:

$$\tan(\angle AEB) = \frac{BD}{DE} = \frac{\sqrt{80}}{DE}$$

Step 4: Equate the two expressions since $\angle ACB = \angle AEB$.

$$\frac{8}{\sqrt{80}} = \frac{\sqrt{80}}{DE}$$

$$DE = \frac{\sqrt{80} \times \sqrt{80}}{8} = \frac{80}{8} = 10 \ \text{cm}$$

Step 5: Compute $AE$.

$$AE = AD + DE = 8 + 10 = 18 \ \text{cm}$$

3. Solution

Answer = Option C ✅

The length of $AE$ is $\mathbf{18}$ cm.