Maths 4 Marks Questions

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.
$PO^2 = PQ^2 + OQ^2...(1)$
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 60cm. 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,
$QR = 2 \times OQ$
$QR = 30$
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, BC and CA at P, Q and R respectively.
AP and AR are the tangents to the circle AP = AR
Similarly CR = CQ and BQ = BP
OP and OQ are radii of the circle
OP ⊥ AB and OQ ⊥ BC and $\angle\text{B}=90^{\circ}$ (given)
BPOQ is a square
$BP = BQ = r$
$AR = AP = AB – BD = 8 – r$
and $CR = CQ = BC – BQ = 6 – r$
But $AC^2 = AB^2 + BC^2$​​​​​​​ (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.
⇒ OQ + QP = diameter
⇒ QP = diameter - OQ [$\therefore$ OQ = radius]
⇒ 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.

Thus we have proved.