4 Marks Questions

Question 14 Marks

In Figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersects $XY$ at A and $X'Y'$ at $B.$ Prove that $\angle A O B = 90^\circ .$

Answer

According to the question, $XY$ and $X 'Y '$ are $x$ two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact C intersects $XY $ at $A$ and $X 'Y '$ at $B.$

In quad. $APQB,$ we have
$\angle A P O =90^\circ $
and $\angle B Q O = 90^\circ [ \because$ tangent at any point is perpendicular to the radius through the point of contact$]$
Now, $\angle A P O + \angle B Q O + \angle Q B C + \angle P A C = 360 ^ { \circ }$
$\Rightarrow \angle P A C + \angle Q B C = 360 ^ { \circ }$$- ( \angle A P O + \angle B Q O ) = 180 ^ { \circ } ...(i)$
We have,
$\angle C A O = \frac { 1 } { 2 } \angle P A C$
and $\angle C B O = \frac { 1 } { 2 } \angle Q B C [ \because$ tangents from an external point are equally inclined to the line segment joining the centre to that point$]$
$\therefore \angle \mathrm { CAO } + \angle \mathrm { CBO } = \frac { 1 } { 2 } ( \angle P A C + \angle Q B C ) = \frac { 1 } { 2 } \times 180 ^ { \circ } = 90 ^ { \circ }.....(ii)$
In $\triangle A O B$, we have
$\angle C A O + \angle A O B + \angle C B O = 180 ^ { \circ }$
$\Rightarrow 90 ^ { \circ }+\angle A O B = 180 ^ { \circ }$
$\Rightarrow \quad \angle A O B = 90 ^ { \circ }$

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Question 24 Marks

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer

Given: $ABCD$ is a quadrilateral circumscribing a circle whose centre is $O.$
To prove:

$\angle AOB + \angle COD =180^{\circ} $
$\angle BOC + \angle AOD = 180^{\circ} $

Construction: Join $OP, OQ, OR$ and $OS.$

Proof: Since tangents from an external point to a circle are equal.
$\therefore AP = AS,$
$BP = BQ ........ (i)$
$CQ = CR$
$DR = DS$
In $\triangle$OBP and $\triangle$OBQ,
$OP = OQ$ [Radii of the same circle]
$OB = OB$ [Common]
$BP = BQ$ [From eq. $(i)]$
$\therefore$ $\triangle$OPB $\cong$ $\triangle$OBQ [By $SSS$ congruence criterion]
$\therefore$ $\angle$1 = $\angle$2 [By $C.P.C.T.]$
Similarly, $\angle$3 = $\angle$4, $\angle$5 = $\angle$6, $\angle$7 = $\angle$8
Since, the sum of all the angles round a point is equal to $360^{\circ}.$
$\therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ} $
$\Rightarrow \angle 1+\angle 1+\angle 4+\angle 4+\angle 5+\angle 5+\angle 8+\angle 8=360^{\circ} $
$\Rightarrow 2(\angle 1+\angle 4+\angle 5+\angle 8)=360^{\circ} $
$\Rightarrow \angle 1+\angle 4+\angle 5+\angle 8=180^{\circ} $
$\Rightarrow(\angle 1+\angle 5)+(\angle 4+\angle 8)=180^{\circ} $
$\Rightarrow \angle A O B+\angle C O D=180^{\circ}$
Similarly we can prove that
$\angle$BOC + $\angle$AOD $180^{\circ}$

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Question 34 Marks

A triangle $A B C$ is drawn to circumscribe a circle of radius $4 \ cm$ such that the segments $B D$ and $D C$ into which $B C$ is divided by the point of contact $D$ are of lengths $8 \ cm$ and $6 \ cm$ respectively (see figure). Find the sides $A B$ and $A C$.

Answer

Join $OE$ and $OF$. Also join $OA, OB$ and $OC.$

Since $BD = 8 cm$
$\therefore BE = 8 cm$
[Tangents from an external point to a circle are equal]
Since $CD = 6 cm$
$\therefore CF = 6 cm$
[Tangents from an external point to a circle are equal]
Let $AE = AF = x$
Since $OD = OE = OF = 4 cm$ [Radii of a circle are equal]
$\therefore$ Semi-perimeter of $\triangle$ $ABC$ = $\frac{(x\;+6)(x\;+8)\;+\;(6+8)}2 = \frac{(2x\;+28)}2$= (x + 14)cm
$\therefore$ Area of $\triangle$ABC =$\sqrt{s(s-a)(s-b)\;(s-c)}$
= $\sqrt{(x\;+14)(x\;+14\;-14)(x\;+14-\overline{x+8})(x\;+14-\overline{x+6})}$
= $\sqrt{(x\;+14)(x\;)(8)(6)}cm^2$
Now, Area of $\triangle ABC =$ Area of $\triangle OBC +$ Area of $\triangle OCA +$ Area of $\triangle OAB$
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$=$\frac{(6+8)4}2+\frac{\displaystyle(x+6)4}{\displaystyle2}+\frac{\displaystyle(x+8)4}{\displaystyle2}$
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}= 28 + 2x + 12 + 2x + 16$
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}= 4x + 56$
$\Rightarrow$$\sqrt{(x\;+14)(x\;)(8)(6)}=4(x + 14)$
Squaring both sides,
$(x + 14) (x) (6) (8) = 16(x + 14)^2$
$\Rightarrow 3x = x + 14$
$\Rightarrow 2x = 14$
$\Rightarrow x = 7$
$\therefore AB = x + 8 = 7 + 8 = 15 cm$
And $AC = x + 6= 7 + 6 = 13 cm$

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Question 44 Marks

$PQ$ is a chord of length $8 \ cm$ of a circle of radius $5 \ cm$. The tangents at $P$ and $Q$ intersect at a point $T.$ Find the length $TP.$

Answer

Let TR be $x \ cm$ and $TP$ be y cm
$OT$ is perpendicular bisector of $PQ$
So $PR = 4 \ cm ( PR$ = $\frac {PQ} {2} = \frac {8 } {2 } $)
In $\triangle$OPR, $OP^2= PR^2+ OR^2 5^2= 4^2+ OR^2 OR = \sqrt{25 - 16} $
$\therefore OR = 3 cm$
In $\triangle PRT, PR^2+RT^2= PT^2$
$y^2= x^2+ 4^2.....(1)$
In $\triangle OPT, OP^2+ PT^2= OT^2$
$(x + 3)^2= 5^2+ y^2( OT = OR + RT = 3 + x)$
$\therefore (x + 3)^2= 5^2+ x^2+ 16 $ [using $(1)]$
Solving, we get x = $\frac{16}{3}$ cm
From (1), $y^2= \frac{256}{9}$ + 16 = $\frac{400}{9}$
So, y = $\frac{20}{3}$cm $= 6.667 cm$

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