Geometry based Mathematics Quiz For CTET Exam : 17th Feb 2020

Maths Questions

Q1. The sum of all Interior angles of a regular polygon is 1080°. The number of sides is
एक नियमित बहुभुज के सभी आंतरिक कोणों का योग 1080 ° है। भुजाओं की संख्या है
(a) 2
(b) 8
(c) 6
(d) 10

Q2. The sum of all interior angles of a regular convex polygon is 540°. The measure of each of its interior angle is
एक नियमित उत्तल बहुभुज के सभी आंतरिक कोणों का योग 540° है। इसके प्रत्येक आंतरिक कोण का माप है
(a) 108°
(b) 90°
(c) 100°
(d) 94°

Q3. In ∆ABC and ∆PQR, if AB = PR, BC = QR and AC = PQ, then
∆ABC और ∆PQR में, यदि AB = PR, BC = QR और AC = PQ, तो
(a) ∆ABC = ∆PQR
(b) ∆ABC = ∆PRQ
(c) ∆ABC = ∆QPR
(d) None of these / इनमें से कोई नहीं

Q4. What is the value of ∠BOC in adjoining figure?
आसन्न आकृति में ∠BOC का मान क्या है?

(a) 80°
(b) 70°
(c) 85°
(d) 75°

Q5. In the given figure. find the ∠1 +∠2 + ∠3 + ∠4 + ∠5 + ∠6 =?
दिए गए चित्र में, ज्ञात करें ∠1 +∠2 + ∠3 + ∠4 + ∠5 + ∠6 =?

(a) 360°
(b) 540°
(c) 180°
(d) None of these / इनमें से कोई नहीं

Q6. In the given figure, find the value of x
दिए गए चित्र में. X का मान ज्ञात करें

(a) 110°
(b) 80°
(c) 150°
(d) 90°

Q7. In the figure PQ ∥ SR and ST ∥ QU, find the value of x?
दिए गए चित्र में PQ ∥ SR और ST ∥ QU, x का मान ज्ञात करें?

(a) 130°
(b) 110°
(c) 100°
(d) 90°

Q8. PQ ∥ RS. AB cuts PQ and RS at T and U respectively TC is the bisector of ∠ QTU. It ∠ CTQ = 45° then ∠RUB will be:
PQ ∥ RS. AB क्रमशः PQ और RS को T और U पर काटती है. TC ∠ QTU का द्विभाजक है. यदि ∠ CTQ = 45° तो ∠RUB होगा:


(a)80°
(b) 90°
(c) 95°
(d) 100°

Q9. In the figure AB ⊥BC and EB ⊥ BD. Find ∠EBA if ∠DBC = 35°.
दिए गए चित्र में AB ⊥BC और EB ⊥ BD. ∠EBA ज्ञात करें यदि ∠DBC = 35°.


(a) 45°
(b) 40°
(c) 35°
(d) 50°

Q10. In the figure ∠AOC and ∠BOC Form a Linear pair If x – y = 90°, find the value of x and y.
दिए गए चित्र में ∠AOC और ∠BOC एक रैखिक युग्म बनाते है यदि x – y = 90°, तो x और y का मान ज्ञात करें.


(a) 135, 45°
(b) 130,50°
(c) 45,135°
(d) 120,60°

Solutions

S1. Ans.(B)
Sol. Sum of all interior angles = (n – 2) ×180
1080° = (n – 2) × 180
6 = n – 2
n=8
the no. of sides = 8

S2. Ans.(a)
Sol. no. sides of convex polygon = 5
so, each Interior angle = 540/5
= 108°

S3. Ans.(b)
Sol.

AB = PR
BC = QR
AC = PQ
So, ∆ABC = ∆PRQ

S4. Ans.(b)
Sol. ∠AOC + ∠BOC = 180°
6x + 5 + 4x = 180°
10x = 175°
x = 17.5°
∠BOC = 4 × x
= 4 × 17.5
= 70°

S5. Ans.(a)
Sol. Sum of All angles of two triangle = 180 + 180° ⟹ 360°
So, ∠1 +∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

S6. Ans.(c)
Sol. ∠ PQR + ∠ QRP = 110° (sum of interior opposite angles equals to their exterior angle)
45° + ∠ QRP = 110°
∠ QRP = 65°
∠TRS = 180° – ∠QRP
∠TRS = 180°- 65°
∠TRS= 115°
∠ STR + ∠ TRS = x°
35° + 115° = x
x = 150°

S7. Ans.(a)
Sol. PQ ∥ SR and ST ∥ QU (given)
∠ PQR + ∠ QPR + ∠ PRQ = 180°
∠ PQR + 60° +70° = 180°
∠ PQR = 180° – 130°
∠ PQR = 50°
∠ PQR = ∠ SRU = 50°
∠ TSR + 50° = 180°
∠ TSR = 130°
So, the value of x = 130°

S8. Ans.(b)
Sol. PQ ∥ RS (given)
TC is the bisector of ∠ QTU (given)
So, the value of ∠ QTU = 90°
∠ QTU + ∠ TUS = 180°
90° + ∠ TUS = 180°
∠ TUS = 90°
∠ TUS = ∠ RUB = 90° (Vertical opposite angle)
∠ RUB = 90°

S9. Ans.(c)
Sol. AB ⊥ BC (given)
∠ ABD + ∠ DBC = 90°
∠ ABD + 35° = 90°
∠ ABD = 55°
EB ⊥ BD (given)
∠ EBA + ∠ ABD = 90°
∠ EBA = 90° – 55°
∠ EBA = 35°

S10. Ans.(a)
Sol. ∠ AOC + ∠ BOC = 180° (Linear pair)


y = 45°
x = 180° – 45°
x = 135°

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