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CAT 2018 Slot 1 — QA Question 16

Basics of CirclesEasy

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

Answer & solution

Correct answer: √13

  • A

    √12

  •  √13

  • C

     √11

  • D

     √14

Solution

Easy

Drop a perpendicular from the centre to each chord; it bisects the chord. Each half-chord with its distance from the centre forms a right triangle whose hypotenuse is the radius. Set the two radius expressions equal to find the distances.

O 6 cm 4 cm x 1
1

Half the chords. The perpendicular from the centre bisects each chord, so the half-lengths are 33 (for the 66 cm chord) and 22 (for the 44 cm chord).

62=3,42=2\begin{aligned} &\tfrac{6}{2}=3,\qquad \tfrac{4}{2}=2 \end{aligned}
2

Distances from the centre. Let the 66 cm chord be xx cm from the centre. The chords are on the same side and 11 cm apart, so the shorter (44 cm) chord is farther, at x+1x+1 cm. Both right triangles share the radius:

32+x2=22+(x+1)2(both equal the radius)\begin{aligned} &\sqrt{3^2+x^2} = \sqrt{2^2+(x+1)^2} \quad\text{(both equal the radius)} \end{aligned}
3

Solve for xx. Square both sides:

9+x2=4+x2+2x+1⇒ 9=5+2x⇒ x=2\begin{aligned} &9+x^2 = 4 + x^2 + 2x + 1\\ &\Rightarrow\ 9 = 5 + 2x\\ &\Rightarrow\ x = 2 \end{aligned}
4

Radius. Substitute x=2x=2 into the 66 cm chord's triangle:

r=32+22=9+4=13\begin{aligned} &r = \sqrt{3^2+2^2} = \sqrt{9+4} = \sqrt{13} \end{aligned}
Radius=13 cm\text{Radius} = \mathbf{\sqrt{13}}\ \text{cm}

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CAT 2018 Slot 1 QA Q16: In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance be — Solution | TheCATExam