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CAT 2018 Slot 2 — QA Question 15

TrianglesEasy

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0, 0) is

Answer & solution

Correct answer: 4 units

  • A

    8 units

  • 4 units

  • C

    2√2 units

  • D

    2√4 units

Solution

Easy

Fix the base BCBC (length 88) on the vertical line x=4x=4. The area pins the horizontal distance of AA from that line. Then choose AA's position to sit as close to the origin as possible.

1

Find the height from AA to line BCBC. Area =12⋅base⋅height=\tfrac12\cdot \text{base}\cdot\text{height} with base BC=8BC=8.

32=12⋅8⋅h⇒ h=2⋅328=8\begin{aligned} &32=\frac12\cdot 8\cdot h\\ &\Rightarrow\ h=\frac{2\cdot 32}{8}=8 \end{aligned}
2

Locate AA. BCBC lies on x=4x=4 (a vertical line), so the perpendicular distance from AA to it is the horizontal gap. Thus AA lies on a vertical line 88 units away from x=4x=4: either x=12x=12 or x=−4x=-4.

A=(12, t)orA=(−4, t),  t∈R\begin{aligned} &A=(12,\,t)\quad\text{or}\quad A=(-4,\,t),\ \ t\in\mathbb{R} \end{aligned}
3

Minimise distance to the origin. Distance =x2+t2=\sqrt{x^2+t^2}. Take the line nearer the origin, x=−4x=-4, and set t=0t=0 to kill the second term.

nearest A=(−4,0)⇒ OA=(−4)2+02=4\begin{aligned} &\text{nearest } A=(-4,0)\\ &\Rightarrow\ OA=\sqrt{(-4)^2+0^2}=4 \end{aligned}
x y O x = 4 B C A(-4,0) OA = 4
OAmin⁡=4 unitsOA_{\min}=4\text{ units}

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CAT 2018 Slot 2 QA Q15: A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the short — Solution | TheCATExam