Question: Here's a clean, step-by-step solution.Given $AD=100$ cm, $CD=64$ cm, $D$ lies on $AC$. $\triangle ABD \sim \triangle BCD$. From the similarity, the corresponding sides are: $AD \leftrightarrow BD$, $BD \leftrightarrow CD$. Therefore, $\frac{AD}{BD} = \frac{BD}{CD}$. (a) Find $\angle BDC$ Step 1: Find $BD$ Using the proportion, $\frac{100}{BD} = \frac{BD}{64}$. Cross multiply: $BD^2 = 100 \times 64 = 6400$. $BD = \sqrt{6400} = 80$ cm Step 2: Show that $\triangle ABD$ is right-angled Check Pythagoras: $100^2 = 80^2 + 60^2$. $10000 = 6400 + 3600$. This is true, so $\angle ADB = 90^\circ$. Since $A$, $D$, and $C$ lie on the same straight line, $\angle ADB + \angle BDC = 180^\circ$. Hence, $90^\circ + \angle BDC = 180^\circ$. $\boxed{\angle BDC = 90^\circ}$. (b) Find the area of $\triangle ABC$ Step 1: Find $AC$ $AC = AD + CD = 100 + 64 = 164$ cm Step 2: Use the area formula Since $BD \perp AC$, $BD$ is the height. Area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 164 \times 80 = 82 \times 80 = \boxed{6560}$ cm$^2$. Final Answers (a) $\boxed{90^\circ}$ (b) $\boxed{6560}$ cm$^2$
1. **Problem Statement:** Given $AD=100$ cm, $CD=64$ cm, and point $D$ lies on line segment $AC$. Triangles $\triangle ABD$ and $\triangle BCD$ are similar. We are asked to find (a) the angle $\angle BDC$ and (b) the area of $\triangle ABC$.
2. **Similarity and Proportion:** From the similarity $\triangle ABD \sim \triangle BCD$, the corresponding sides satisfy:
$$\frac{AD}{BD} = \frac{BD}{CD}$$
This proportion will help us find $BD$.
3. **Step 1: Find $BD$**
Using the proportion:
$$\frac{100}{BD} = \frac{BD}{64}$$
Cross-multiplied:
$$BD^2 = 100 \times 64 = 6400$$
Taking the square root:
$$BD = \sqrt{6400} = 80 \text{ cm}$$
4. **Step 2: Show $\triangle ABD$ is right-angled**
Check using the Pythagorean theorem:
$$100^2 \stackrel{?}{=} 80^2 + 60^2$$
Calculate each side:
$$10000 = 6400 + 3600$$
$$10000 = 10000$$
This confirms $\triangle ABD$ is right-angled at $D$, so:
$$\angle ADB = 90^\circ$$
5. **Step 3: Find $\angle BDC$**
Since points $A$, $D$, and $C$ are collinear, angles on a straight line sum to $180^\circ$:
$$\angle ADB + \angle BDC = 180^\circ$$
Substitute $\angle ADB = 90^\circ$:
$$90^\circ + \angle BDC = 180^\circ$$
Solve for $\angle BDC$:
$$\boxed{\angle BDC = 90^\circ}$$
6. **Step 4: Find $AC$**
Since $D$ lies on $AC$:
$$AC = AD + CD = 100 + 64 = 164 \text{ cm}$$
7. **Step 5: Calculate the area of $\triangle ABC$**
Since $BD$ is perpendicular to $AC$, $BD$ is the height of $\triangle ABC$ with base $AC$.
Area formula for a triangle:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
Substitute values:
$$\text{Area} = \frac{1}{2} \times 164 \times 80 = 82 \times 80 = 6560 \text{ cm}^2$$
8. **Final Answers:**
(a) $\boxed{90^\circ}$
(b) $\boxed{6560}$ cm$^2$