Exercise 14 E
Question 1
Ans: if $A B$ and $C D$ are two chords which intersect at $P$ inside the circle
So $A P \times P B=C P \times P D$
(i) Now $A P=8 \mathrm{~cm}$, $C P=6 \mathrm{~cm}$ and $P D=4 \mathrm{~cm}$
So $8 \times P B=6 \times 4 \Rightarrow P B=\frac{6 \times 4}{8} \mathrm{~cm}$
So $P B=3 \mathrm{~cm}$
(ii) $A B=12 \mathrm{~cm}, A P=2 \mathrm{~cm}$ and $P D=4 \mathrm{~cm}$
So $P B=A B-A P=12-2=10 \mathrm{~cm}$
Now $A P \times P B=C P \times P D$
$\Rightarrow 2 \times 10=C P \times 4$
$\Rightarrow c p=\frac{2 \times 10}{4}=5 \mathrm{~cm}$
So $c p=5 \mathrm{~cm}$
(iii) $A P=6 \mathrm{~cm}, P B=5 \mathrm{~cm}$ and $C D=13 \mathrm{~cm}$
let $C P=x$, then $P D=C D-C P$
$=(13-x) \mathrm{cm}$
So NOW AP ｘPB $=C P \times P D$
$\begin{aligned}&\Rightarrow 6 \times 5=x(13-x) \Rightarrow 30=13 x-x^{2} \\&\Rightarrow x^{2}-13 x+30=0 \\&\Rightarrow x^{2}-10 x-3 x+30=0\end{aligned}$
$\Rightarrow x(x-10)-3(x-10)=0$
$\Rightarrow(x-10)(x-3)=0$
Either $x-10=0$, then $x=10$
or $x-3=0$ then $x=3$
so CP $=10 \mathrm{~cm}$ or $3 \mathrm{~cm}$
Question 2
Ans: if chords $A B$ and $C D$ of a circle intersect each other at $P$ intersect the circle
So $A P \times P B=C P \times P D$
(i) $P A=10 \mathrm{~cm}, P B=4 \mathrm{~cm}$ and $P C=8 \mathrm{~cm}$
Now $A P \times P B=C P \times P D$
$\begin{gathered}\Rightarrow 10 \times 4=8 \times P D \\\Rightarrow P D=\frac{10 \times 4}{8}=5 \\\text { SO } P D=5 \mathrm{~cm}\end{gathered}$
(ii) $P C=15 \mathrm{~cm}, C D=7 \mathrm{~cm}$ and $P A=12 \mathrm{~cm}$
$P D=C P-C D=15-7=8 \mathrm{~cm}$
Now $P A \times P B=P C \times P D$
$12 \times P B=15 \times 8$
$\Rightarrow P B=\frac{15 \times 8}{12}=10$
So $A B=P A-P B=12-10=2 \mathrm{~cm}$
(iii) $P A=16 \mathrm{~cm}, P C=10 \mathrm{~cm}$ and $P D=8 \mathrm{~cm}$
Now $P A \times P B=P C \times P D$
$\Rightarrow 16 \times P B=10 \times 8$
$\Rightarrow P B=\frac{10 \times 8}{16}=5$
So $A B=P A-P B=16-5=11 \mathrm{~cm}$
Question 3
Ans: (i) In the figure, $P T$ is tangent and $P B A$ is a secant to the circle $P B=4 \mathrm{~cm} \quad 1 A B=12 \mathrm{~cm}$
(IMAGE TO BE ADDED)
So $\quad P A=P B+A B=4+12=16 \mathrm{~cm}$
So $P T^{2}=P A \times P B$
$=16 \times 4=64=\left(81^{2}\right.$
So PT $=8 \mathrm{~cm}$
(ii) From an external point P,PT is the tangent to the circle and $P A B$ is a secant
(IMAGE TO BE ADDED)
So and $P A=9.6 \mathrm{~cm}, P B=2.4 \mathrm{~cm}$
$\begin{aligned}\text { Now } & P T^{2}=P A \times P B=9.6 \times 2.4 \\=& 23.04=14.812 \\& \text { So } P T=4.8 \mathrm{~cm}\end{aligned}$
Question 4
Ans: In $\triangle A B C , \angle A=90^{\circ}$
A circle is drawn on AC as diameter which intersects $B C$ at $D$
AD is Joined
$B D=9$ and $D C=7$
(IMAGE TO BE ADDED)
We see that in the circle $A B$ is the tangent at $A$ and $B D C$ is the secant of the circle
$\text { So } \begin{aligned}& A B^{2}=B D \times B C \\=& 9 \times 16 \\=& 144=(12)^{2} \\\text { So } A B=12\end{aligned}$
Question 5
Ans: Steps of constructions :
(i) Draw a line segment $A B=9$
At $A$ draw $\angle B A C=90^{\circ}$ and cut off $A C=12$
(ii) Join $B C$. Take $F$ as mid point of $A C$.
(iii) Now draw perpendicular bisects of $B C$ and $B C$ in intersecting each other at 0 .
(iv)with center 0 , and radius $O C$ draw a circle with Passes through B,F and $C$ and intersects $A B$ at $E$ on measuring $B E=1$.
(IMAGE TO BE ADDED)
Question 6
Ans: $A B C$ is a right angled triangle in which $\angle A=90^{\circ}$
$A B \perp B C$
A B=4, A C=3$
(IMAGE TO BE ADDED)
So $\begin{aligned} B C^{2}=& A B^{2}+A C^{2} \\=&(4)^{2}+(3)^{2}=16+9=25=(5)^{2} \\ & \text { So } B C=S \\ & \text { let } B D=x, \text { then } D C=5-x \end{aligned}$
Now $A B^{2}=B D \times B C$
$(4)^{2}=x \times 5$
$\Rightarrow 16=5 x$
$\Rightarrow x=\frac{16}{5}=3 \frac{1}{5}$
So $B D=3 \frac{1}{5}$
Question 7
Ans: from a point $P$,
outside the circle PA is tangent and P BC is the secant PA =7
(IMAGE TO BE ADDED)
Power of P with respect to the circle $=P A^{2}=(7)^{2}=49$
is $P A$ is tangent and $P B C$ is the secant
$\text { so } \begin{aligned}P A^{2} &=P B \times P C \\& \Rightarrow 49=P B \times P C\end{aligned}$
Hence $P B \cdot P C=49$.
Question 8
Ans: $\triangle A B C$ in which $A B=A C=10 \mathrm{~cm}$ and $B C=16 \mathrm{~cm}$ is inscribed in a circle AE is a chord which is at right angle to $B C$ at $D$
In $\triangle A B C$.
$A B=A C$ and $A D \perp B C$
So $B D=D C=\frac{16}{2}=8 \mathrm{~cm}$
Now in right $\triangle A B U$
$A B^{2}=B D^{2}+A D^{2} \Rightarrow(10)^{2}=(8)^{2}+A D^{2}$
$\Rightarrow 100=64+A D^{2}$
$\Rightarrow A D^{2}=100-64=36=(6)^{2}$
So $A D=6$
Now two chords $A E$ and $B C$ intersect each other at $D$
So $B D \times D C=A D \times D E$
$\begin{aligned} \Rightarrow & 8 \times 8=6 \times D E \Rightarrow D E=\frac{8 \times 8}{6}=\frac{64}{6}=\frac{32}{3} \\ & \text { SO DE }=\frac{32}{3}=10 \frac{2}{3} \mathrm{~cm} \end{aligned}$
Now $A E=A D+D E=6+10 \frac{2}{3}=16 \frac{2}{3} \mathrm{~cm}$
So Radices $=\frac{1}{2} A E$
$=\frac{1}{2} \times 16 \frac{2}{3}=8 \frac{1}{3} \mathrm{~cm}$
Question 9
Ans: In the figure $x y$ is a tangent to the circle with center $O \cdot X C D$ is a secant
$O N⊥ D X$
Join $O C$ and Join $O B$
$CX=4 \text { and } X B=6$
If XB is tangent and XCD is the secant
Let CD = X
So $\begin{aligned} & XB^{2}=\times C \times \times D \\ \Rightarrow &(6)^{2}=4(4+x) \\=& 4(4+x)=36 \\ \Rightarrow & 4+x=\frac{36}{4}=9 \end{aligned}$
$x=9-4=5$
So $C D=5$
if on $\perp C D$
So $C N=N D=\frac{5}{2}$
So Radius $O B=N X=C N+X C$
$=\frac{5}{2}+4=6 \frac{1}{2} \mathrm{~cm}$
Question 10
Ans: In the figure, $P B A$, $P D C$ are the secants $P T$ is the tangents to the circle with center $O A D$ and $B C$ are Joined which intersect each other at X
(i) $A X=5 \mathrm{~cm}, X D=7 \mathrm{~cm}, C x=10 \mathrm{~cm}$
if Two chords $A D$ and $B C$ intersects each other at $x$
So $A X \cdot X D=C X \cdot \times B$
$\begin{aligned}&\Rightarrow 5 \times 7=10 \times \times \beta \\&\Rightarrow \times B=\frac{5 \times 7}{10}=\frac{7}{2}=3.5 \mathrm{~cm}\end{aligned}$
(ii) $O A=6 \mathrm{~cm}, B X=5 \mathrm{~cm}, O X=4 \mathrm{~cm}$ Join $O A, 0 x$ and produce it to both sides meeting the circle at $P$ and $Q$
(IMAGE TO BE ADDED)
$O A=O P=O Q$
$X P=O P-O X=O A-O X=6-4=2 \mathrm{~cm}$
$X Q=O X+Q Q=O X+O A=6+4=10 \mathrm{~cm}$
if Chord $B C$ and $P Q$ intersect each other X
So XC $\times XB = XP \times$ XQ
Question 11
Ans: Two circles intersect each other at p and Q
$A B$ and $A C$ are the tangents on $Q P$ produced trom $A$
To prove: $A B=A C$
Proof: if $A B$ is the Tangent and $A P Q$ is a secant
So $A B^{2}=A P \times A Q$
similarly $A C$ is the tangent and $A P Q$ is the secant
So $A B^{2}=A P \times A Q$.........(i)
Similarly AC is the tangent and APQ is the secant
So $A C^{2}=A P \times A Q$.........(ii)
from ii and (ii)
$\begin{aligned}& A B^{2}=A C^{2} \\\Rightarrow & A B=A C\end{aligned}$
Question 12
Ans: In a circle $A B$ is a chord of a circle with center O
$P$ is any point on $A B$ and $P X \perp O P$
To prove $\because A P, P B=P X^{2}$
construction: produce XP to meet the circle at Y
(IMAGE TO BE ADDED)
Proof: IF XY is a chord and OP ⊥ XY
So P is the mid point of XY
So PX= PY
Now chord AB and XY intersect each other at P
So $A P \times P B=X P \times P Y$
$\begin{aligned}&\Rightarrow A P \cdot P B=x P \times x P \\&\Rightarrow A P \cdot P B=x P^{2} \text { or } A P \cdot P B=P \times 2\end{aligned}$
Hence proved
Question 13
Ans: Two circles Intersect each other at T and S.STP, BSC , BAP and CDP are straight lines Drawn.
To prove:
(i) Quad . PATD is a cyclic
(ii) PA. PB = PD.PC
Construction : Join AT and TD
Proof: (i) IF ABST is a cyclic quad
SO Ext $\angle A T P=$ int. OPP. $\angle B$.......(i)
Similarly in cyclic quad CDTS
Ext. $\angle D T P=\angle C$
Adding (i) and (ii)
$\angle A T P+\angle O T P=\angle B+\angle C$........(ii)
But in $\triangle P B C$
$\begin{aligned}& \angle B P C+\angle B+\angle C=180^{\circ} \\\Rightarrow & \angle B+\angle C=180^{\circ}-\angle B P C\end{aligned}$
But these are sum of opposite angles of quad PATD
So quad PATD is a cyclic quad.
(ii) If chords BA and ST intersect at P outside the circle
So $P A \times P B=P T \times P S$ .......(i)
Similarly CD and ST chords Meet at P
So P$D \times P C=P T \times P S$.........(ii)
from (i) and (ii)
$P A \cdot P B=P D \cdot P C$
Question 14
Ans: Two circles intersect each other at A
To prove: $P A \cdot A Q=R A \cdot A S$
construction : Join PR and QS
(IMAGE TO BE ADDED)
Proof: in $\triangle A P R$ and $triangle A S Q$
$\angle P=\angle S$
$\angle P A R=\angle Q A S$
So $\triangle A P R \backsim \triangle A S Q$
So $\frac{P A}{A S}=\frac{R A}{A Q}$
So $P A \cdot A Q=R A \cdot A S$ Hence proved
Question 15
Ans: Two circle Intersect each other at A and B
P is a point on BA produced through P, PCD and PEH are Secant drawn to each circle
(IMAGE TO BE ADDED)
To prove: C,D H,E are concyclic
Proof : IF DC and BA are chords which intersect at P outside the first Circle
So $P C \times P D=P A \times P B$ .........(i)
Similarly Chords BA and HE intersect each other at P outside the second Circle
So $P A \times P B=P E \times P H$ .......(ii)
from( ii) and (ii)
$P C \times P D=P E \times P H$
But there are two chordl $D C$ and $H E$ which meet at p outside the circle
so $\mathrm{C}, \mathrm{D}, \mathrm{H}$ and $\mathrm{E}$ are concyclic
Question 16
Ans: In the figure $P B=B T$
$P T$ is the tangent to the circle $P B$ is produced to meet the circle at $C$
$T C$ is Joined
To prove:
(i) $\triangle P T C$ an isosceles
(ii) $P B \cdot P C=T C^{2}$
Proof:
(i) In $\triangle P B T$
$P B=B T$
So $\angle 1=\angle 2$......(i)
So $\angle 2=\angle 3$
from (i) and (iii)
$\angle 1=\angle 3$
So In $\triangle T P C$
$T B=T C$
So $\triangle P T C$ is an isosceles triangle
(ii) If $P T$ is tangent and $P B C$ is secant of the circle
So $P T^{2}=P B \cdot P C$
$\Rightarrow P B \cdot P C=P T^{2}=T C^{2}$
$\Rightarrow P B \cdot P C=T C^{2}$ Hence proved
No comments:
Post a Comment