RD Chapter 11 Co ordinate Geometry Ex MCQS Solutions
Question - 21 : - In the figure, the value of x is
(a) 65°
(b) 80°
(c) 95°
(d) 120°
Answer - 21 : -
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°
Now in ∆ABD,
Ext. ∠DBC = ∠A + ∠D
= 55° + 25° = 80°
Similarly, in ∆BCE,
Ext. ∠DEC = ∠EBC + ∠ECB
= 80° + 40° = 120° (d)
Question - 22 : - In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20°
(b) 25°
(c) 30°
(d) 35°
Answer - 22 : -
In the figure, AC = BC, BP || CQ
∵ BP || CQ
∴ ∠PBC – ∠QCD
⇒ 20° + ∠ABC = 70°
⇒ ∠ABC = 70° – 20° = 50°
∵ BC = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides)
= 50°
Now in ∆ABC,
Ext. ∠ACD = ∠B + ∠A
⇒ x + 70° = 50° + 50°
⇒ x + 70° = 100°
∴ x = 100° – 70° = 30° (c)
Question - 23 : - In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
(a) x = 55°, y = 40°
(b) x = 50°, y = 45°
(c) x = 60°, y = 35°
(d) x = 35°, y = 60°
Answer - 23 : -
In the figure,
∵ AB || CD, EF intersects them at P and Q respectively,
∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°
∵ AB || CD
∴ ∠APQ = ∠CQF (Corresponding anlges)
⇒ y + 25° = 65°
⇒ y = 65° – 25° = 40°
and APQ + PQC = 180° (Co-interior angles)
y + 25° + ∠1 +30°= 180°
40° + 25° + ∠1 + 30° = 180°
⇒ ∠1 + 95° = 180°
∴ ∠1 = 180° – 95° = 85°
Now, ∆PQR,
∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)
⇒ 40° + x + 85° = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° = 55°
∴ x = 55°, y = 40° (a)
Question - 24 : - The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠BAC = ?
(a) 94°
(b) 54°
(c) 40°
(d) 44°
Answer - 24 : -
In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 94° = ∠BAC + ∠ABC
Similarly, ∠ABE = ∠BAC + ∠ACB
⇒ 126° = ∠BAC + ∠ACB
Adding,
94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC
220° = 180° + ∠BAC
∴ ∠BAC = 220° -180° = 40° (c)
Question - 25 : - If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is
(a) 45°
(b) 95°
(c) 135°
(d) 90°
Answer - 25 : -
In right ∆ABC, ∠A = 90°
Bisectors of ∠B and ∠C meet at O, then 1
∠BOC = 90° + 
∠A = 90°+ x 90° = 90° + 45°= 135° (c)
Question - 26 : - The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=
Answer - 26 : -
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.
Question - 27 : - In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
(a) 25°
(b) 50°
(c) 100°
(d) 75°
Answer - 27 : -
In ∆ABC, ∠A = 50°
BC is produced
Bisectors of ∠ABC and ∠ACD meet at ∠E
∴ ∠E = ∠A = x 50° = 25° (a)
Question - 28 : - The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD =115°,then ∠ALC =
(a) 85°
(b) 72 ° (c) 145°
(d) none of these
Answer - 28 : -
In ∆ABC, BC is produced to D
∠B = 30°, ∠ACD = 115°
Question - 29 : - In the figure , if l1 || l2, the value of x is
(a) 22 (b) 30
(c) 45
(d) 60
Answer - 29 : -
In the figure, l1 || l2
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB
∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively
∴ ∠CEB = 90°
∴ a + b = 90° ,
and ∠GEB = 90° (∵ ∠CEB = 90°)
2x = 90° ⇒ x =
= 45 (c)
Question - 30 : - In ∆RST (in the figure), what is the value of x?
(a) 40°
(b) 90°
(c) 80°
(d) 100°
Answer - 30 : -