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CAT 2020 Slot 3 — QA Question 18

Basics of CirclesEasy

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is

Answer & solution

  • A

    123π/7

  • 205π/9

  • C

    14π/3

  • D

    12π/5

Solution

Medium

The circle through the three vertices is the circumcircle. Use the relation Area=abc4R\text{Area}=\dfrac{abc}{4R}: find the triangle's area and its three side lengths, solve for the circumradius RR, then compute πR2\pi R^2.

(0,0) (4,0) (3,9)
1

Area of the triangle. The base lies on the xx-axis from (0,0)(0,0) to (4,0)(4,0), so base =4=4 and the height is the yy-coordinate of (3,9)(3,9), namely 99.

Area=12×4×9=18\begin{aligned} &\text{Area}=\frac{1}{2}\times 4\times 9=18 \end{aligned}
2

Side lengths. Use the distance formula between the three vertices.

a=(4−0)2+02=4b=(4−3)2+(0−9)2=1+81=82c=(3−0)2+(9−0)2=9+81=90\begin{aligned} &a=\sqrt{(4-0)^2+0^2}=4\\ &b=\sqrt{(4-3)^2+(0-9)^2}=\sqrt{1+81}=\sqrt{82}\\ &c=\sqrt{(3-0)^2+(9-0)^2}=\sqrt{9+81}=\sqrt{90} \end{aligned}
3

Solve for the circumradius. Apply Area=abc4R\text{Area}=\dfrac{abc}{4R} with the area from step 1 and the sides from step 2.

18=4⋅82⋅904R⇒ 18=473804R=7380R⇒ R=738018=620518=2053\begin{aligned} &18=\frac{4\cdot\sqrt{82}\cdot\sqrt{90}}{4R}\\ &\Rightarrow\ 18=\frac{4\sqrt{7380}}{4R}=\frac{\sqrt{7380}}{R}\\ &\Rightarrow\ R=\frac{\sqrt{7380}}{18}=\frac{6\sqrt{205}}{18}=\frac{\sqrt{205}}{3} \end{aligned}
4

Area of the circumcircle.

πR2=π(2053)2=205π9\begin{aligned} &\pi R^2=\pi\left(\frac{\sqrt{205}}{3}\right)^2=\frac{205\pi}{9} \end{aligned}
Area=205π9\text{Area}=\frac{205\pi}{9}
CAT 2020 Slot 3 QA Q18: The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three p — Solution | TheCATExam