Mathematics for SSC, DSSSB, KVS TGT PGT, UP LT GRADE
Math
In this deck I've shared maths questions which were asked in various competitive exams in india like SSC, KVS PGT, KVS TGT, DSSSB TGT MATHS, DSSSB PGT MATHS, RPSC SR GRADE 2.
Sample Data
| Text | {{c1::2sinA.cosB}} = {{c2::sin(A + B) + sin(A – B)]}} |
| Extra |
| Question_ID | 98958015738 |
| Question |
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| Option1 |
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| Option2 |
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| Option3 |
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| Option4 |
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| Correct_Option | 3 |
| Solution |
| Question | As shown in the figure, AB is the chord of the smaller circle and PQ is a chord of the larger circle. PQ also touches the smaller circle at point B and BA is extended such that it touches the bigger circle at point C. Both the circles are concentric. Find the length of PQ, if CB = 50 cm and 4AB = 21CA.
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| Option1 | 40 cm |
| Option2 | 42 cm |
| Option3 | 44 cm |
| Option4 | 36 cm |
| Option5 | |
| CorrectOption | 1 |
| Solution | Note: isko khud karo. Maine ye bahut jaldi solve kiya thaPT2=PA x PB ka istemaal karkePQ ko slide karke P ko C ko upar le aaoPhir PT wala formula laga do
 In the figure, O is the centre of both the circles and OX is the perpendicular dropped to chord AB.Suppose ‘R’ & ‘r’ are the radius of bigger and smaller circles;Since CB = 50 cm and 4AB = 21CA∴ AB + CA = 50⇒ CA + (21/4)CA = 50⇒ 25CA/4 = 50⇒ CA = 8 cmAnd AB = 50 – 8 = 42 cm∴ AX = 42/2 = 21 cmCX = 8 + 21 = 29 cmIn ΔOAX∴ OX2 = r2 – 212⇒ OX2 = r2 – 441 ---- (1)And in ΔOCXOX2 = R2 – 292⇒ OX2 = R2 – 841 ---- (2)Equating both the equations⇒ r2 – 441 = R2 – 841∴ R2 – r2 = 841 – 441 = 400⇒ R2 – r2 = 400
 In the figure: OB will be perpendicular to the tangent PQ.∴ In ΔOPBR2 – r2 = PB2∴ PB2 = 400⇒ PB = 20 cm∴ PQ = 2PB = 40 cm |