5 Marks Questions

Question 15 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°.

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°
and $\angle B Q O$ = 90° [$\because$ 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 25 Marks

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer

Let O be the common centre of the two concentric circles.

Let $A B$ be a chord of the larger circle which touches the smaller circle at $P$.
Join OP and OA
Then, $\angle O P A=90^{\circ}[\because$ The tangent at any point of a circle is perpendicular to th radius through the point of contact $]$
$\therefore O A^2=O P^2+A P^2 \ldots \ldots .$. By Pythagoras theorem
$\Rightarrow$ $(5)^2 = (3)^2 + AP^2$
$\Rightarrow$ $25 = 9 + AP^2$
$\Rightarrow$ $P^2 = 25 - 9$
$\Rightarrow$ $AP^2 = 16$
$\Rightarrow$ AP = $\sqrt{16}$ = 4 cm
SInce the perpendicular from the centre of a circle to a chord bisects the chord, therfore,
AP = BP = 4 cm
$\therefore$ $AB = AP + BP = AP + AP = 2AP = 2(4) = 8 cm$
Hence, the required length is 8 cm.

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Question 35 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 $= 180^o$
$\angle$BOC + $\angle$AOD $= 180^o$

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 $360^o.$
$\therefore$ $\angle$1 + $\angle$2 + $\angle$3 + $\angle$4+ $\angle$5 + $\angle$6 + $\angle$7 + $\angle$8 =$360^o.$
$\Rightarrow$ $\angle$1 + $\angle$1 + $\angle$4 + $\angle$4+ $\angle$5 + $\angle$5 + $\angle$8 + $\angle$8 = $360^o.$
$\Rightarrow$ 2 ($\angle$1 + $\angle$4 + $\angle$5 + $\angle$8) = $360^o.$
$\Rightarrow$ $\angle$1 + $\angle$4 + $\angle$5 + $\angle$8 $= 180^o$
$\Rightarrow$ ($\angle$1 + $\angle$5) + ($\angle$4 + $\angle$8) $= 180^o$
$\Rightarrow$ $\angle$AOB + $\angle$COD $= 180^o$
Similarly we can prove that
$\angle$BOC + $\angle$AOD $= 180^o$

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Question 45 Marks

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

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)}$$cm^2$
Now, Area of ΔABC = Area of $\triangle$OBC + Area of $\triangle$OCA + Area of $\triangle$OAB
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$=$\frac{(6+8)4}2+\frac{\displaystyle(x+6)4}{\displaystyle2}+\frac{\displaystyle(x+8)4}{\displaystyle2}$
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$= 28 + 2x + 12 + 2x + 16
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$= 4x + 56
$\Rightarrow$$\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 55 Marks

Prove that parallelogram circumscribing a circle is a rhombus.

Answer

Given ABCD is a parallelogram in which all the sides touch a given circle
To prove:- ABCD is a rhombus
Proof:-
$\because$ ABCD is a parallelogram
$\therefore$ AB = DC and AD = BC
Again AP, AQ are tangents to the circle from the point A
$\therefore$ AP = AQ
Similarly, BR = BQ
CR = CS
DP = DS
$\therefore$(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS)
$\Rightarrow$ AD + BC = AB + DC
$\Rightarrow$ BC + BC = AB + AB [$\because$ AB = DC, AD = BC]
$\Rightarrow$ 2BC = 2AB
$\Rightarrow$ BC = AB
Hence, parallelogram ABCD is a rhombus

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Question 65 Marks

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Answer

Steps of Construction:

Draw a circle with any radius and center O. Here xy is given line.
Choose any point P on the circumference of the circle, and draw a line passing through P, Let's name it AB.
Draw a line AB parallel to xy, such that AB intersects the circle at two points P and A.Here, AB and xy are two parallel lines. AB intersects the circle at exactly two points, P and Q. Therefore, line AB is the secant of this circle.
CD intersects the circle at exactly one point, R, line CD is the tangent to the circle.

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