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CAT 2017 Slot 1 — QA Question 16

Geometric CentersEasy

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:

Answer & solution

Correct answer: 500/√3

  • A

    225√3

  • 500/√3

  • C

    275√3

  • D

    250√3

Solution

Easy

The three medians split a triangle into 66 equal-area pieces. Triangle GBCGBC (vertex = centroid, base = BCBC) is exactly 13\tfrac13 of the whole, so the remaining piece is 23\tfrac23 of the area. Get the total area by Heron's formula.

A B C G
1

Total area by Heron. Sides a=40,b=25,c=35a=40,b=25,c=35, semi-perimeter s=40+25+352=50s=\tfrac{40+25+35}{2}=50:

Area=s(s−a)(s−b)(s−c)⇒ 50⋅10⋅25⋅15⇒ 187500=2503\begin{aligned} &\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\\ &\Rightarrow\ \sqrt{50\cdot 10\cdot 25\cdot 15}\\ &\Rightarrow\ \sqrt{187500}=250\sqrt{3} \end{aligned}
2

Area of the cut-off triangle. Medians give 66 equal cells; △GBC\triangle GBC spans two of them.

[GBC]=13 [ABC]=13⋅2503\begin{aligned} &[GBC]=\tfrac13\,[ABC]=\tfrac13\cdot 250\sqrt{3} \end{aligned}
3

Remaining area. Subtract step 2 from the whole, i.e. take 23\tfrac23:

[remaining]=23⋅2503=50033=5003\begin{aligned} &[\text{remaining}]=\tfrac23\cdot 250\sqrt{3}=\frac{500\sqrt{3}}{3}=\frac{500}{\sqrt{3}} \end{aligned}
Remaining area=5003 sq ft\text{Remaining area}=\dfrac{500}{\sqrt{3}}\ \text{sq ft}

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CAT 2017 Slot 1 QA Q16: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G — Solution | TheCATExam