Maths 4 Marks Questions

Thus we have proved.

Given,
$\angle\text{POQ}=180^{\circ}-(\angle\text{POR})=180^{\circ}-130^{\circ}=50^{\circ}$
Since, $PQ$ is a tangent.
$\angle\text{POQ}=90^{\circ}$
Now, In
$\angle\text{POQ}+\angle\text{PQO}+\angle\text{QPO}=180^{\circ}$
$\Rightarrow50^{\circ}+90^{\circ}+\angle1=180^{\circ}$
$\Rightarrow\angle1=180^{\circ}-140^{\circ}$
$\Rightarrow\angle1=40^{\circ}$
Now, In
$\angle\text{RST}=\frac{1}{2}\angle\text{ROT}$ [Angle which is subtended by on arc at the centre of a circle is double the size of the angle subtended at any point on the circumference.]
$\Rightarrow\angle2=\frac{1}{2}\times130^{\circ}=65^{\circ}$
Now$\angle1+\angle2=40^{\circ}+65^{\circ}=105^{\circ}$

Given: From a point P. Out side the circle with centre $O, PA$ and $PB$ are tangerts drawn and $\angle\text{APB}=120^{\circ}$
$OP$ is joined.
To prove: $OP = 2AP$
Const: Take mid point $M$ of $OP$ and join $AM$, join also $OA$ and $OB.$

Proof: In right $\triangle\text{OAP},$

In the figure,

$\angle\text{PQO}=90^{\circ}.$ Therefore we can use pythagoras theorem to find the side $PO.$
$P O^2=P Q^2+O Q^2...(1)$
In the problem it is given that,
$\frac{\text{OQ}}{\text{PQ}}=\frac{3}{4}$
$\text{OQ}=\frac{3}{4}\text{PQ}...(2)$
Substituting this in equation $(1)$, we have,
$\text{PO}^2=\frac{9\text{PQ}^2}{16}+\text{PQ}^2$
$\text{PO}^2=\frac{25\text{PQ}^2}{16}$
$\text{PO}=\sqrt{\frac{25\text{PQ}^2}{16}}$
$\text{PO}=\frac{5}{4}\text{PQ}...(3)$
It is given that the perimeter of $\triangle\text{POQ}$is $60\ cm$. Therefore,
$PQ + OQ + PO = 60$
Substituting $(2)$ and $(3)$ in the above equation, we have,
$\text{PQ}+\frac{3}{4}\text{PQ}+\frac{5}{4}\text{PQ}=60$
$\frac{12}{4}\text{PQ}=60$
$3PQ = 60$
$PQ = 20$
Substituting for $PQ$ in equation $(2)$, we have,
$\text{PO}=\frac{5}{4}\times20$
$\text{OQ}=\frac{3}{4}\times20$
$OQ = 15$
$OQ$ is the radius of the circle and $QR$ is the diameter.
Therefore,
$QR = 2 × OQ$
$QR = 30$
Substituting for $PQ$ in equation $(3)$, we have,
$\text{PO}=\frac{5}{4}\times20$
$PO = 25$

⇒ Chord $AM$ = Chord $MB$
In $\triangle\text{AMB}, AM = MB$
$\Rightarrow\angle\text{MAB}=\angle\text{MBA}...\text{(i)}$[equal sides corresponding to the equal angle]
Since, $TMT’$ is a tangent line.
$\angle\text{AMT}=\angle\text{MAB}$ [angle in alternate segment are equal]
$\angle\text{AMT}=\angle\text{MAB}$ [from Eq. $(i)]$
But $\angle\text{AMT}$ and $\angle\text{MAB}$ are alternate angles, which is possible only when $AB || TMT’$
Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence proved.

Two equal circles with centre $O$ and $O’$ touch each other externally at $X OO’$ produced to meet at $A.$

AC is the tangent of circle with centre $O,$
$\text{O'D}\bot\text{AC}$ is drawn OC is joined
AC is tangent and OC is the radius.
$\text{OC}\bot\text{AC}$
$\text{O'D}\bot\text{AC}$
$\text{OC}\parallel\text{O'D}$
Now $\text{O'A}=\frac{1}{2}\text{A}\times\text{or}\frac{1}{2}\text{AO}$
Now in $\triangle\text{O’AD}$ and $\triangle\text{OAC}$
$\angle\text{A}=\angle\text{A}$ (common)
$\angle\text{AO'D}=\angle\text{AOC}$ (corresponding angles.)

In right $\triangle\text{ABC},\angle\text{B}=90^{\circ}$
$BC = 6cm, AB = 8cm$
Let $r$ be the radius of incircle whose centre is $O$ and touches the sides $A B, B C$ and $C A$ at $P, Q$ and $R$ respectively.
$A P$ and $A R$ are the tangents to the circle $A P=A R$
Similarly $C R=C Q$ and $B Q=B P$
$O P$ and $O Q$ are radii of the circle
$O P \backslash$ perp AB and $OQ \backslash$ perp BC and $\angle B =90^{\circ}$ (given)
$BPOQ$ is a square
$BP = BQ = r$
$AR = AP = AB – BD = 8 – r$
and $CR = CQ = BC – BQ = 6 – r$
But $A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$=(8)^2+(6)^2=64+36=100=(10)^2$
$AC = 10cm$
$\Rightarrow AR + CR = 10$
$\Rightarrow 8 – r + 6 – r = 10$
$\Rightarrow 14 – 2r = 10$
$\Rightarrow 2r = 14 – 10 = 4$
$\Rightarrow r = 2$
Radius of the in circle $= 2cm$

Proof: AM and AN are the tangents to the circle from A.
$AM = AN$
$But AB = AC (given)$
$AB – AN = AC – AM$
$BN = CM$
Now $BL$ and $BN$ are the tangents from $B.$
$BL = BN$
Similarly $CL$ and $CM$ are tangents.
$CL = CM$
But $BM = CM$ (proved)
$BL = CL$
$L$ is midpoint of $BC.$

Suppose op meets the circle at $Q.$
then join $AQ.$
we have,
$OP =$ diameter.
$\Rightarrow OQ + QP =$ diameter
$\Rightarrow QP =$ diameter - $OQ$ [$\therefore OQ =$ radius]
$\Rightarrow QP =$ radius
Thus, $OQ = PQ =$ radius
Thus,OP is the hypoteneus of right triangle $OAP$ and $Q$ is the mid point of $OP.$
$\therefore$ $OA = AQ = OQ$
$\triangle\text{OAQ}$ is equilateral.
$\Rightarrow\angle\text{AOQ}=60^{\circ}$
So,$\angle\text{APO}=30^{\circ}$
$\therefore\angle\text{APB}=2\angle\text{APO}=60^{\circ}$
Also,
$PA = PB$
$\Rightarrow\angle\text{PAB}=\angle\text{PBA}$
But, $\angle\text{APB}=60^{\circ}$
Therefore, $\angle\text{PAB}=\angle\text{PBA}=60^{\circ}$
Hence, $\triangle\text{APB}$ is equilateral.

In the given figure,
$AB$ is the diameter, $AT$ is the tangent,
and $\angle\text{AOQ}=58^{\circ}$
To find $\angle\text{ATQ}$
Arc $AQ$ subtends $\angle\text{AOQ}$ at the centre and $\angle\text{ABQ}$ at the remaining part of the circle.
$\angle\text{ABQ}=\frac{1}{2}\angle\text{AOQ}=\frac{1}{2}\times58^{\circ}=29^{\circ}$
Now in $\triangle\text{ABT},$
$\angle\text{BAT}=90^{\circ}(\text{OA}\bot\text{AT})$
$\angle\text{ABT}+\angle\text{ATB}=90^{\circ}$
$\Rightarrow\angle\text{ABT}+\angle\text{ATQ}=90^{\circ}$
$\Rightarrow29^{\circ}+\angle\text{ATQ}=90^{\circ}$
$\Rightarrow\angle\text{ATQ}=90^{\circ}-29^{\circ}=61^{\circ}$

To prove: $PQ$ and $OT$ are right bisector of each other.
Proof: $PT$ and $QT$ are tangents to the circle.
$PT = QT$
$OP$ and $OQ$ are radii of the circle and $\angle\text{POQ}=90^{\circ}(\text{PO}\bot\text{OQ})$
$OQTP$ is a square Where $PQ$ and $OT$ are diagonals.
Diagonals of a square bisect each other at right angles.
$PQ$ and $OT$ bisect each other at right angles.
Hence $PQ$ and $QT$ are right bisectors of each other.