SSC CGL 2020 objective Ques (15 results)
SSC CGL 20201)In the figure, PA is a tangent from an external point P to the circle with centre O. If \(\angle POB = 110 ^{\circ}\), then measure of \(\angle APO\) is:
\(20 ^{\circ}\)
PA is a tangent
\(\angle POA =180^{\circ}- 110 ^{\circ}=70^{\circ}\)
\(\angle PAO =90^{\circ}\)
\(\angle APO =180^{\circ}- 70 ^{\circ}-90^{\circ}=20^{\circ}\)
SSC CGL 20202)In a circle, chords PQ and TS are produced to meet at R. If RQ = 14.4 cm, PQ = 11.2 cm, and SR = 12.8 cm, then the length of chord TS is:
16 cm
RQ X PR = RS X TR; 14.4 X (14.4+11.2) = 12.8 X TR; TR = \({14.4\times25.6\over12.8}=28.8 cm\) ; TS = TR - SR = (28.8 - 12.8) cm = 16 cm
SSC CGL 20203)From an external point P, a tangent PQ is drawn to a circle, with the centre O, touching the circle at Q. If the distance of P from the centre is 13 cm and length of the tangent PQ is 12 cm, then the radius of the circle is:
5 cm
\triangle OPQ is a right angle triangle because \(\angle Q = 90^0\),
By Pythagoras,
\((OQ)^2 + (PQ)^2 = (OP)^2\);
\((OQ)^2 = (13)^2 - (12)^2\);
\((OQ)^2\) = 169 - 144;
\((OQ)^2\) = 25;
OQ = 5 cm.
SSC CGL 20204)In a circle, AB is a the diameter and CD is a chord. AB and CD produced meet at a point P, outside the circle. If PD = 15.3 cm, CD = 11.9 cm and AP = 30.6 cm,then the radius of the circle is is:
8.5 cm
From the property,
\(PA \times PB = PC \times PD\);
\(30.6 \times PB = (PD + CD) \times 15.3\);
\(30.6 \times PB = (15.3 + 11.9) \times 15.3\);
\(30.6 \times PB = 27.2 \times 15.3\);
PB = 416.16/30.6 = 13.6 cm;
Diameter (AB) = PA - PB = 30.6 - 13.6 = 17 cm;
Radius = AB/2 = 17/2 = 8.5 cm
SSC CGL 20205)PAQ is a tangent to circle with centre O, at a point A on it. AB is a chord such that \(\angle BAQ=x^0 (x<90)\). C is a point on the major arc AB such that \(\angle ACB = y^0\). If \(\angle ABO = 32^0\), then the value of x + y is:
116
OA = OB; \(\angle ABO = \angle BAO =32^0\); \(\angle AOB = 116^0\); so \(y ={ 1\over2}\angle AOB = 58^0\); and \(x= 90^0-32^0= 58^0\); x + y = \(58^0 +58^0 = 116^0\)
SSC CGL 20206)AB is a diameter of a circle with centre O. The tangent at a point C on the circle meets AB produced at Q. If \(\angle CAB=42^0\), then what is the measure of \(\angle CQA\)?
\(6^0\)
\(\angle OAC = \angle OCA = 42^0 ( \because OA = OC)\); \(\angle OCQ = 90^0 \) (since CQ is tangent). Now in \(\triangle ACQ\), \(\angle A+\angle C+\angle Q =180^0\); ⇒ \(42^0+(42^0+90^0)+x = 180\); x = \(6^0\)
SSC CGL 20207)In the given figure, MP is a tangent to a circle with center A and NQ is a tangent to a circle with center B. If MP = 15 cm, NQ = 8 cm, PA = 17 cm and BQ = 10 cm, then AB is:
14 cm.
PA = 17 ; PM = 15 cm ; \(\angle PMA= 90^0\) ; In \(\triangle AMP,\) \(\therefore MA = 8 cm\); Again, BQ = 10 cm. ; NQ = 8 cm; \(\angle BNQ = 90^0\) ; So BN = 6 cm ; Ab = AC + CB = AM + BN = (8 + 6) = 14 cm.
SSC CGL 20208)A, B and C are three points on a circle such that the angle subtended by the chords AB and AC at the centre O are \(80^0\) and \(120^0\), respectively. The value of \(\angle BAC\) is:
\(80^0\)
\(\angle AOB = 80^0\); \(\angle AOC = 120^0\); \(\therefore \angle BOC = 360^0-(80^0+120^0)=160^0\) \(\therefore \angle BAC = {\angle BOC\over2}={160^0\over2} = 80^0\)
SSC CGL 20209)In the given figure, if AB = 10 cm, CD = 7 cm, SD = 4 cm and AS = 5 cm, then BC = ?
8 cm
SD = 4cm; So DR = 4 cm ; CR = CQ = 7 - 4 = 3 cm; AS = 5 cm ; AS = AP = 5 cm ; AB = 10 cm ; BP = BQ = 10 - 5= 5 cm; So BC = BQ + CQ = 5 + 3 = 8 cm
SSC CGL 202010)In the given figure, if \(\angle APO = 35^0\), then which of the following options is correct?
\(\angle BPO = 35^0\)
Join OA and OB. Since OA = OB ; AP = PB and OP is common. Therefore, triangle AOP is similar to triangle BOP. Since angle OAP = 90 & OBP = 90. Therefore, \({OAP\over OBP} ={APO\over BPO}\); ⇒ \({90\over90}={35\over BPO}\); ⇒ BPO = \(35^0\)