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CAT 2019 Slot 1 — QA Question 7

Basics of CirclesEasy

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

Answer & solution

Correct answer: 0.5

  • 0.5

  • B

    2.5

  • C

    3.5

  • D

    1.5

Solution

Easy

Two chords cross at EE, so use the intersecting-chords theorem: AE⋅EB=CE⋅EDAE\cdot EB = CE\cdot ED. We know the diameter CD=22CD=22 and CE=7CE=7, so EDED is fixed. With AE+EB=AB=20.5AE+EB=AB=20.5 known too, solve for the two segment lengths and take their difference.

C D A B E
1

Lengths along the diameter. Radius 1111, so CD=22CD=22. Given CE=7CE=7.

ED=CD−CE=22−7=15\begin{aligned} &ED = CD - CE = 22 - 7 = 15 \end{aligned}
2

Apply the intersecting-chords theorem. Let AE=xAE=x, so EB=20.5−xEB = 20.5 - x.

AE⋅EB=CE⋅ED(intersecting chords)⇒ x(20.5−x)=7×15=105\begin{aligned} &AE\cdot EB = CE\cdot ED \quad\text{(intersecting chords)}\\ &\Rightarrow\ x(20.5 - x) = 7\times 15 = 105 \end{aligned}
3

Solve the quadratic. Rearrange and factor.

x2−20.5x+105=0⇒ x=20.5±20.52−4202=20.5±0.52⇒ x=10.5 or 10\begin{aligned} &x^2 - 20.5x + 105 = 0\\ &\Rightarrow\ x = \frac{20.5 \pm \sqrt{20.5^2 - 420}}{2} = \frac{20.5 \pm 0.5}{2}\\ &\Rightarrow\ x = 10.5 \ \text{or}\ 10 \end{aligned}
4

Take the difference. The two segments are 10.510.5 and 1010 (in some order).

∣BE−AE∣=∣10.5−10∣=0.5\begin{aligned} &|BE - AE| = |10.5 - 10| = 0.5 \end{aligned}
∣BE−AE∣=0.5 cm|BE - AE| = 0.5 \text{ cm}

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CAT 2019 Slot 1 QA Q7: In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a — Solution | TheCATExam