Exercise-9.2, Class 9th, Maths, Chapter 9, NCERT

Q1

Problem. Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $4$ cm. Find the length of the common chord.

Solution.
Let the centres be $O_1$ (radius $R_1=5$) and $O_2$ (radius $R_2=3$), and the distance $O_1{O}_2=d=4$. Let the common chord be $XY$. Let the perpendicular from $O_1$ to chord $XY$ meet it at $M$. Let $O_{1}M=x$. Then for the two right triangles:

$$\{\begin{array}{l}{x}^2+{\left(\frac{XY}{2}\right)}^2=R_1^2=25\\ (d-x{)}^2+{\left(\frac{XY}{2}\right)}^2=R_2^2=9\end{array}$$

Subtract the second from the first:

$$25-9=x^2-(d-x{)}^2=2dx-d^2\mathrm{}$$

So

$$16=2\cdot 4\cdot x-16\Rightarrow 16=8x-16$$
$$\Rightarrow 8x=32\Rightarrow x=4$$

Then half–chord length $\frac{XY}{2}=\sqrt{25-x^2}=\sqrt{25-16}=\sqrt{9}=3$
Hence the common chord length $XY=2\times 3=\boxed{{\displaystyle 6\text{ cm}}}\mathrm{<span\; class="mord"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist"><span\; class="boxpad"><span\; class="mspace"> </span><span\; class="mord\; text">cm</span></span></span><span\; class="vlist-s"></span></span></span></span><span\; class="mord"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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Q2

Problem. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.

Solution.
Let chords $AB$ and $CD$ be equal and intersect at $E$. Put $AE=x,EB=s-x$ where $s=AB$. Put $CE=y,ED=s-y$ (since $CD=AB=s$). By the intersecting-chords theorem (power of a point):

$$AE\cdot EB=CE\cdot ED\Rightarrow x(s-x)=y(s-y)\mathrm{}$$

Rearrange:

$$(x-y)(s-(x+y))=0.$$

The factor $s-(x+y)$ would be zero only if $x+y=s$, i.e. $AE+CE=AB$, which would place $C$ on segment $AB$ (not possible for two distinct chords intersecting inside). Hence $x-y=0$, so $x=y$. Therefore

$$AE=CE\text{and similarly}EB=ED\mathrm{}$$

So corresponding segments are equal.

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Q3

Problem. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the two chords.

Solution.
With the notation of Q2, let $O$ be the circle’s centre and $AE=CE$ and $BE=DE$ (from Q2). Consider triangles $\mathrm{△}OAE$ and $\mathrm{△}OCE$. They have

$$OA=OC(\text{radii}),AE=CE,OE=OE\mathrm{}$$

So $\mathrm{△}OAE\cong \mathrm{△}OCE$ (SSS). Corresponding angles at $E$ are equal:

$$\mathrm{\angle }OEA=\mathrm{\angle }OEC\mathrm{.}$$

Thus the line $OE$ makes equal angles with chords $AB$ and $CD$(Similarly $\mathrm{\angle }OEB=\mathrm{\angle }OED$.)

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Q4

Problem. If a line intersects two concentric circles (same centre $O$) at $A,B,C,D$ (in that order along the line), prove that $AB=CD$. (See Fig. 9.12.)

Solution.
Let the line meet the two concentric circles so that along the line, from left to right, the points are $A$ (outer circle), $B$(inner circle), $O$ (centre), $C$ (inner circle), $D$ (outer circle). Denote outer radius $R$ and inner radius $r$. Then

$$AB=OB-OA=(\text{distance }O\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}B)-(\text{distance }O\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}A)\mathrm{.}$$

But because of symmetry about $O$ the distances satisfy $OA=OD=R$ and $OB=OC=r$ (on appropriate signed scale). Hence

$$AB=OA-OB=R-r,$$

and

$$CD=OD-OC=R-r\mathrm{.}$$

Therefore $AB=CD$. (Geometrically: the two segments from outer to inner circle on opposite sides are equal by symmetry about $O$.)

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Q5

Problem. Three girls Reshma, Salma and Mandip stand on a circle of radius $5$ m. Reshma–Salma distance $=6$, Salma– Mandip distance $=6$. Find the distance between Reshma and Mandip.

Solution.
All three lie on the same circle radius $R=5$. Chord length $L$ corresponding to a central angle $\theta$ satisfies

$$L=2R\sin \frac{\theta }{2}\mathrm{}$$

Given chord $=6$, so

$$6=2\cdot 5\sin \frac{\theta }{2}\Rightarrow \sin \frac{\theta }{2}=\frac{6}{10}=\mathrm{0.6.}$$

Thus $\cos \frac{\theta }{2}=\sqrt{1-{0.6}^2}=\sqrt{0.64}=0.8$Then

$$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cdot 0.6\cdot 0.8=\mathrm{0.96.}$$

The triangle formed by the three girls has two equal sides (6,6) so the central angles subtending those equal chords are equal (call each $\theta$). The remaining central angle is ${360}^{\circ }-2\theta$. The chord between Reshma and Mandip corresponds to central angle ${360}^{\circ }-2\theta$; its length is

$$\text{RM}=2R\sin \frac{{360}^{\circ }-2\theta }{2}=2R\sin ({180}^{\circ }-\theta )=2R\sin \theta \mathrm{.}$$

So

$$\text{RM}=2\cdot 5\cdot 0.96=10\cdot 0.96=\boxed{{\displaystyle 9.6\text{ m}}}\mathrm{}$$

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Q6

Problem. A circular park has radius $20$. Three boys sit at equal distances on its boundary (i.e., they are vertices of an inscribed equilateral triangle). Find the length of the string of each toy telephone (distance between any two boys).

Solution.
Three equal points on a circle divide the circumference into three equal arcs; each central angle is ${120}^{\circ }$. The chord length for central angle ${120}^{\circ }$ is

$$L=2R\sin \frac{{120}^{\circ }}{2}=2R\sin {60}^{\circ }=2R\cdot \frac{\sqrt{3}}{2}=R\sqrt{3}\mathrm{}$$

With $R=20$, string length $=20\sqrt{3}\text{ m}\approx 34.64\text{ m}\mathrm{}$
So exact answer: $\boxed{{\displaystyle 20\sqrt{3}\text{ m}}}$