Question

On a plate in the shape of an equilateral triangle \(ABC\) with area \(16\sqrt{3}\) sq cm, a rod \(GD\), of height 8 cm, is fixed vertically at the centre of the triangle. \(G\) is a point on the plate. If the areas of the triangles \(AGD\) and \(BGD\) are both equal to \(4\sqrt{19}\) sq cm, find the area of the triangle \(CGD\) (in sq cm).

• A. \(3\sqrt{19}\)
• B. \(4\sqrt{19}\)
• C. \(12\sqrt{3}\)
• D. None of these

Hint:

In composite 3D geometry problems, subtract known triangle areas from the total to get unknown parts. Use symmetry if areas are equal.

Answer 1

The Correct Option is D

Step 1: Understand the configuration
We are given: - Triangle \(ABC\) is equilateral with area \(16\sqrt{3}\) sq cm. - Rod \(GD\) of height 8 cm is fixed vertically at the centre of triangle. - Point \(G\) is on the plate, and \(GD\) is perpendicular to the plane. - Area of triangles \(AGD\) and \(BGD\) = \(4\sqrt{19}\) each. We are to find the area of triangle \(CGD\). Step 2: Use 3D Triangle Area Formula
For triangle \(AGD\), we use the formula for area when a vertex is above the plane: \[ \text{Area of triangle} = \frac{1}{2} \times \text{base length} \times \text{height} \] But here, since \(GD\) is vertical and \(AG\) lies in the plane, we can use the 3D triangle area formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}_\perp \] Let the perpendicular height from \(D\) to triangle base in projection = \(h\), which is constant for all three triangles. Each triangle's area is: \[ \text{Area of } AGD = \frac{1}{2} \times AG \times 8 \times \sin\theta = 4\sqrt{19} \Rightarrow AG \times \sin\theta = \frac{8\sqrt{19}}{8} = \sqrt{19} \] Same applies to \(BGD\). Step 3: Total area approach
The total area of triangle \(ABC\) is partitioned among triangles \(AGD\), \(BGD\), and \(CGD\) with the same vertical height 8 cm from point \(D\) to base \(ABC\). Let the area of triangle \(CGD = x\) Then total 3D volume: \[ \text{Sum of areas} = 4\sqrt{19} + 4\sqrt{19} + x = \text{Area of pyramid with triangular base and vertical height} \] But note: Since all three sub-triangles together form the volume of the upright triangular prism above \(ABC\), and given the triangle lies flat and rod \(GD = 8\) cm, the total volume is not needed — just the remaining area. \[ \text{Total area of plate} = \text{area of ABC} = 16\sqrt{3} \] \[ \text{Area of triangle } CGD = 16\sqrt{3} - 2 \times 4\sqrt{19} = 16\sqrt{3} - 8\sqrt{19} \] Now test options. Only one matches: Try (a): \(3\sqrt{19}\) Try computing: \[ \text{Is } 4\sqrt{19} + 4\sqrt{19} + 3\sqrt{19} = 11\sqrt{19} ? \Rightarrow \text{No} \] So reverse logic is wrong. Instead, the sum of the three triangle areas in 3D: \[ \text{Total volume-like interpretation} \Rightarrow \text{No, better approach: Use geometry} Since \(AGD = BGD = 4\sqrt{19}\), assume same height 8. Use formula for area of triangle with vertical height: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{vertical height} \Rightarrow 4\sqrt{19} = \frac{1}{2} \times b \times 8 \Rightarrow b = \frac{8\sqrt{19}}{4} = 2\sqrt{19} \]