objective Ques (149 results)
SSC CGL 202011)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 202012)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 = \(6^0\)
13)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:
SSC CGL 2020
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 202014)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:
\(80^0\)
\(\angle AOB = 80^0\); \(\angle AOC = 120^0\); \(\therefore \angle BOC = 360^0-(80^0+120^0)=160^0\) \(\therefore \angle BAC = {\angle BOC\over2}={160^0\over2} = 80^0\)
15)In the given figure, if AB = 10 cm, CD = 7 cm, SD = 4 cm and AS = 5 cm, then BC = ?
SSC CGL 2020
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
16)In the given figure, if \(\angle APO = 35^0\), then which of the following options is correct?
SSC CGL 2020
\(\angle BPO = 35^0\)
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 = \(35^0\)
SSC CGL 202017)Two tangents PA and PB are drawn to a circle with centre O from an external point P. If \(\angle OAB = 30^0\), then \(\angle APB\) is:
\(60^0\)
\angle OAB = \(30^0\); OA = OB = radii of circle; \(\angle OAB = \angle OBA = 30^0\); \(\angle AOB = 180^0-(30^0+30^0)=120^0\); In quadrilateral OAPB, \(\therefore\angle OAP = \angle OBP = 90^0\); \(\therefore \angle AOB +\angle APB = 180^0\); ⇒\(\angle APB = 180-120 =60^0\)
18)In the figure, two circles with centres P and Q touch externally at R. Tangents AT and BT meet the common tangent TR at T. If AP = 6 cm and PT = 10 cm, then BT = ?
SSC CGL 2020
8 cm
In \(\triangle PAT\) \(\angle PAT = 90^0\);
\(\therefore AT= \sqrt{ PT^2-AP^2} = \sqrt{10^2-6^2}= 8cm\);
Tangent drawn from an external point to a circle are equal.
\(\therefore\) AT = BT = 8 cm
19)In the given figure, \(\angle KLN = 58^0\), then \(\angle KMN=\space?\)
SSC CGL 2020
\(58^0\)
\(\angle KLN = 58^0\) ; \(\therefore \angle KMN = 58^0\) ;
Because angles in the same segment are equal.
SSC CGL 202020)Diameter AB of a circle with centre O is produced to a point P such that PO = 16.8 cm. PQR is a secant which intersects the circle at Q and R, such that PQ = 12 cm and PR = 19.2 cm.The length of AB (in cm.) is :
14.4
PQ = 12 cm ; PR = 19.2 cm ; PO = 16.8 cm ; Let, OA = OB = x ;
BP x AP = PQ x PR ; ⇒ (16.8 - x)(16.8 + x) = 12 x 19.2 ; x = 7.2 cm.
\(\therefore AB=2\times7.2 = 14.4cm\)
