Area of the largest triangle that can be inscribed in a semicircle of radius \(r\) is
\(r^2\) sq. units
\(\dfrac{1}{2}r^2\) sq. units
\(2r^2\) sq. units
\(\sqrt{2}\, r^2\) sq. units
Step 1: When a triangle is drawn inside a semicircle with the diameter as its base, the triangle will always be a right-angled triangle (this is a property of a semicircle).
Step 2: The diameter of the semicircle is \(2r\). This diameter becomes the hypotenuse of the right triangle.
Step 3: Let the other two sides of the right triangle (the legs) be \(a\) and \(b\). By Pythagoras’ theorem:
\(a^2 + b^2 = (2r)^2 = 4r^2.\)
Step 4: The area of a right triangle is given by:
\( \text{Area} = \dfrac{1}{2} \times a \times b. \)
Step 5: For a fixed value of \(a^2 + b^2\), the product \(ab\) is maximum when \(a = b\). (This is a standard result from algebra or you can check by AM–GM inequality).
Step 6: If \(a = b\), then:
\(a^2 + b^2 = 2a^2 = 4r^2 \implies a^2 = 2r^2 \implies a = b = \sqrt{2}r.\)
Step 7: Now calculate the area:
\( \text{Area} = \dfrac{1}{2} \times a \times b = \dfrac{1}{2} (\sqrt{2}r)(\sqrt{2}r).\)
\( = \dfrac{1}{2} (2r^2) = r^2. \)
Final Answer: The largest area = \(r^2\) sq. units. So the correct option is (A).