Q1. In the given figure, if EC∥AB,∠ECD=70°,∠BDO=20°, then ∠OBD is-
(a) 70°
(b) 60°
(c) 50°
(d) 20°
Sol.
∠AOD=∠ECO
⇒ ∠AOD=90°
So, ∠BOD=110°
In ∆BOD,
∠OBD+∠BOD+∠ODB=180°
∠OBD=180°-(110°-20°)
∴∠OBD=50°
Q2. In ∆PQR,PS is the bisector of ∠P and PT⊥QR, then ∠TPS is equal to-
(a) ∠Q+∠R
(b) 90°+1/2∠Q
(c) 90°-1/2∠R
(d) 1/2 (∠Q-∠R)
Sol.
∠1+∠2=∠3
In ∆PQT,∠Q=90°-∠1
In ∆PTR,∠R=90°-∠2-∠3
So, ∠Q-∠R=(90°-∠1)-(90°-∠2-∠3)
∠Q-∠R=∠2+∠3-∠1
=∠2+(∠1+∠2)-∠1
∠Q-∠R=2∠2
1/2 (∠Q-∠R)=∠TPS
Q3. In figure sides AB and AC of ∆ABC are produced to P and Q, respectively. The bisectors of ∠PBC and ∠QCB intersect at O. Then, ∠BOC is equal to-
(a) 90°-1/2∠BAC
(b) 1/2 (∠PBC+∠QCB)
(c) 90°+1/2∠BAC
(d) None of these
Sol.
∠1=90°-1/2∠3
∠2=90°-1/2∠4
Now, in ∆BOC,
∠1+∠2+∠BOC=180°
∠BOC=180°-(∠2+∠1)
=180°-[90°-1/2∠4+90°-1/2∠3]
∠BOC=1/2 (∠3+∠4)
∠BOC=1/2 (180°-∠A)
∴∠A+∠3+∠4=180°
∠BOC=90°-1/2∠A
Q4. In figure, AB, EF and CD are parallel lines. Given that GE = 5 cm, GC = 10 cm and DC = 18 cm, then EF is equal to-
(a) 11 cm
(b) 5 cm
(c) 6 cm
(d) 9 cm
Sol.
In, ∆GEF and ∆GCD, we have
∠EFG=∠GDC (∴ Alternate angles)
∠EGF=∠CGD (∴ Vertically opposite angles)
∆GEF∼∆GCD
GE/CG=EF/CD⇒5/10=EF/18
⇒EF=(5 × 18)/10=9 cm
Q5. In figure, ∆ADE and ∆ABC are similar, if AC : BC = 3 : 2, then the ratio DE/AE is –
(a) 3 : 2
(b) 2 : 3
(c) 1 : 3
(d) 1 : 2
Sol.
As, ∆ADE∼∆ABC
So, AD/AB=DE/BC=AE/AC
DE/AE=BC/AC=2/3
Hence, DE : AE = 2 : 3
Q6. A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then –
(a) OB + OD = OC + OA
(b) OB^2+OA^2=OC^2+OD^2
(c) OB.CD = OC.OA
(d) OB^2+OD^2=OC^2+OA^2
Q7. ∆ABC is a right angled triangle with BC = 6cm and AB = 8 cm. A circle with A centre O and radius x cm has been inscribed in ∆ABC. Then, the value of x is –
(a) 2 cm
(b) 4 cm
(c) 3 cm
(d) 2.5 cm
Sol.
AF = AD = 8 – x, CF = CE = 6 – x
∴AC=AF+FC=√(8^2+6^2 )
⇒ 8 – x + 6 – x = 10 ⇒ 14 – 2x = 10
⇒ 2x = 4 ⇒ x = 2 cm
Q8. In the adjoining figure, ∆ABC is isosceles triangle with AB = AC and ∠ABC=50°. The ∆BDC is –
(a) 110°
(b) 90°
(c) 80°
(d) 70°
Sol.
As, AB = AC
⇒ ∠ACB=∠ABC=50°
∴∠BAC=180°-(50°+50°)=80°
∴∠BDC=∠BAC=80°
(Angle lying in same segment)
Q9. If one angle of a cyclic trapezium is triple of the other, then the greater one measures –
(a) 90°
(b) 105°
(c) 120°
(d) 135°
Sol.
Let ∠A=3∠C
∴∠A+ ∠C=180°
(∴ ABCD being a cyclic trapezium)
Or 3∠C+∠C=180°
Or 4∠C=180°
∴∠C=45°
∴ Measure of greater angle
∠A=3×45=135°
Q10. In the figure, XAY is a tangent to the circle with centre at O if ∠BAX=70°,∠BAQ=40°, then ∠ABQ is equal to –
(a) 20°
(b) 30°
(c) 35°
(d) 40°
Sol.
∠QAX=∠BAX-∠BAQ
=70°-40°=30°
∠BAY=180°-∠BAX=180°-70°=110°
∠EBA=90° [Angle in semi-circle]
∠BAY=∠AQB=110°
∠ABQ=180°-(∠BAQ+∠AQB)
180°-(40°+110°)=30°
