Geometry

Triangles

Medium

From a triangle $ABC$ with sides of lengths $40$ ft, $25$ ft and $35$ ft, a triangular portion $GBC$ is cut off where $G$ is the centroid of $ABC$ . The area, in sq ft, of the remaining portion of triangle $ABC$ is

$225/\sqrt{3}$

$500 / \sqrt{3}$

$275/\sqrt{3}$

$250 / \sqrt{3}$

✅ Correct Option: 2

We need to find the area of triangle ABC first, then determine what portion remains after removing triangle GBC.

We have triangle ABC with sides 40 ft, 25 ft, and 35 ft. The centroid G divides the triangle, and we're removing triangle GBC (the triangle formed by connecting the centroid to vertices B and C).

Here's something important to remember: when you connect the centroid of a triangle to each of its three vertices, you create three smaller triangles. Each of these triangles has exactly $\tfrac{1}{3}$ of the original triangle's area.

So triangle GBC has area = $\tfrac{1}{3} \times$ (Area of triangle ABC)

With sides a = 40, b = 25, c = 35:

Semi-perimeter: s = (40 + 25 + 35) ÷ 2 = 50

Heron's Formula: Area = $\sqrt{s(s-a)(s-b)(s-c)}$

Area = $\sqrt{50(50-40)(50-25)(50-35)}$

= $\sqrt{50 \times 10 \times 25 \times 15}$

= $\sqrt{187500}$

Let us simplify this:

187500 = 625 × 300

625 = 25² and 300 = 100 × 3

So $\sqrt{187500} = \sqrt{25^2 \times 100 \times 3} = 25 \times 10 \times \sqrt{3} = 250\sqrt{3}$

Therefore, Area of triangle ABC = $250\sqrt{3}$ sq ft

Area of triangle GBC = $\tfrac{1}{3} \times 250\sqrt{3} = \dfrac{250\sqrt{3}}{3}$

Remaining area = Total area - Area removed

= $250\sqrt{3} - \dfrac{250\sqrt{3}}{3}$

= $\dfrac{3 \times 250\sqrt{3}}{3} - \dfrac{250\sqrt{3}}{3}$

= $\dfrac{750\sqrt{3} - 250\sqrt{3}}{3}$

= $\dfrac{500\sqrt{3}}{3}$

To express this in the form given in the reference:

$\dfrac{500\sqrt{3}}{3} = \dfrac{500\sqrt{3}}{3} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{500 \times 3}{3\sqrt{3}} = \dfrac{1500}{3\sqrt{3}} = \dfrac{500}{\sqrt{3}}$

Therefore, the area of the remaining portion is $\dfrac{500}{\sqrt{3}}$ sq ft.

The centroid property is a powerful shortcut - instead of calculating coordinates and complex geometry, we use the fact that the centroid always creates three equal-area triangles. This makes the problem much simpler!

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From a triangle ABCABCABC with sides of lengths 404040 ft, 252525 ft and 353535 ft, a triangular portion GBCGBCGBC is cut off where GGG is the centroid of ABCABCABC. The area, in sq ft, of the remaining portion of triangle ABCABCABC is

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From a triangle $ABC$ with sides of lengths $40$ ft, $25$ ft and $35$ ft, a triangular portion $GBC$ is cut off where $G$ is the centroid of $ABC$ . The area, in sq ft, of the remaining portion of triangle $ABC$ is