A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a:b. If the radius of the circle is $r , then the area of the triangle is
Official Correct Answer: C. ### Core Concept
When a triangle is inscribed in a circle with one side as the diameter, it is a right-angled triangle (Thales' theorem). The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
### Step-by-Step Solution
1. **Identify the Triangle Type**:
- The triangle is right-angled with the hypotenuse as the diameter of the circle.
- Hypotenuse length = $2r$.
2. **Assign Variables to Sides**:
- Let the other two sides be $a \cdot k$ and $b \cdot k$ where $k$ is a scaling factor.
3. **Apply Pythagoras Theorem**:
\[
(a \cdot k)^2 + (b \cdot k)^2 = (2r)^2
\]
\[
k^2(a^2 + b^2) = 4r^2
\]
\[
k = \frac{2r}{\sqrt{a^2 + b^2}}
\]
4. **Calculate the Area**:
\[
\text{Area} = \frac{1}{2} \times (a \cdot k) \times (b \cdot k)
\]
\[
= \frac{1}{2} \times a \cdot b \cdot \left(\frac{2r}{\sqrt{a^2 + b^2}}\right)^2
\]
\[
= \frac{1}{2} \times a \cdot b \cdot \frac{4r^2}{a^2 + b^2}
\]
\[
= \frac{2abr^2}{a^2 + b^2}
\]
### 30-Second Shortcut
- Recognize the triangle is right-angled.
- Use the area formula for right-angled triangles.
- Substitute the sides using the given ratio and solve.
### Final Answer
Correct Answer: Option C
Answer and solution
Answer:C) a2+b22abr2
📌Core Concept
When a triangle is inscribed in a circle with one side as the diameter, it is a right-angled triangle (Thales' theorem). The area of a right-angled triangle is given by 21×base×height.
🔢Step-by-Step Solution
1
Identify the Triangle Type:
The triangle is right-angled with the hypotenuse as the diameter of the circle.
Hypotenuse length = 2r$.
2
Assign Variables to Sides:
Let the other two sides be a⋅k and b⋅k where k is a scaling factor.
3
Apply Pythagoras Theorem:
(a⋅k)2+(b⋅k)2=(2r)2k2(a2+b2)=4r2k=a2+b22r
4
Calculate the Area:
Area=21×(a⋅k)×(b⋅k) = 21 × a \cdot b \cdot \left(\frac{2r}{\
a2+b2
}\right)^2=21×a⋅b⋅a2+b24r2 = a2+b22abr2$
⚡30-Second Shortcut
Recognize the triangle is right-angled.
Use the area formula for right-angled triangles.
Substitute the sides using the given ratio and solve.
