Plane Geometry
- ABCD is a cyclic quadrilateral and O is the centre of the circle. If ∠COD = 140° and ∠BAC = 40°, then the value of ∠BCD is equal to
-
View Hint View Answer Discuss in Forum
As per the given in question , we draw a figure cyclic quadrilateral ABCD ,
The angle subtended at the centre by an arc is twice to that of angle subtended at the circumference.∴ ∠CAD = 1 ∠COD = 70° 2
Correct Option: A
As per the given in question , we draw a figure cyclic quadrilateral ABCD ,
The angle subtended at the centre by an arc is twice to that of angle subtended at the circumference.∴ ∠CAD = 1 ∠COD = 70° 2
∴ ∠BAD = 70° + 40° = 110°
∴ ∠BCD = 180° – 110° = 70°
- ABCD is a cyclic trapezium with AB || DC and AB = diameter of the circle. If ∠CAB = 30°, then ∠ADC is
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure of cyclic trapezium ABCD
∠ACB = 90° (Angle of semi-circle)
∠CAB = 30°
∴ ∠CBA = 180°– 90° – 30° = 60°Correct Option: B
According to question , we draw a figure of cyclic trapezium ABCD
∠ACB = 90° (Angle of semi-circle)
∠CAB = 30°
∴ ∠CBA = 180°– 90° – 30° = 60°
Again, ∠ADC + ∠ABC = 180°
∴ ∠ADC = 180° – 60° = 120°
- Inside a square ABCD, ∆ BEC is an equilateral triangle. If CE and BD intersect at O, then ∠BOC is equal to
-
View Hint View Answer Discuss in Forum
According to question , we draw a figure of square ABCD in which ∆BEC is an equilateral triangle ,
∠ABC = 90°
⇒ ∠OBC = 45°
[∵ ∠ABC = 2∠OBC]
∠OCB = 60°
[∵ ∠BEC is equilateral]Correct Option: B
According to question , we draw a figure of square ABCD in which ∆BEC is an equilateral triangle ,
∠ABC = 90°
⇒ ∠OBC = 45°
[∵ ∠ABC = 2∠OBC]
∠OCB = 60°
[∵ ∠BEC is equilateral]
∴ ∠BOC + ∠OCB + ∠OBC = 180°
∴ ∠BOC = 180° – 60° – 45° = 75°
- The medians CD and BE of a triangle ABC intersect each other at O. The ratio ∆ ODE : ∆ ABC is equal to
-
View Hint View Answer Discuss in Forum
On the basis of question we draw a figure of triangle ABC in which the medians CD and BE intersect each other at O ,
In ∆ ADE and ∆ ABC,
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ AED ~ ∆ ABC∴ AD = DE AB BC ⇒ AD = 1 DB ⇒ DB + 1 = 2 AD ⇒ DB + AD = 2 AD ⇒ AB = 2 ⇒ AD = 1 AD AB 2 ⇒ DE = 1 BC 2 ∴ ∆ ODE = 
1 
² = 1 ∆ BOC 2 4
Correct Option: D
On the basis of question we draw a figure of triangle ABC in which the medians CD and BE intersect each other at O ,
In ∆ ADE and ∆ ABC,
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ AED ~ ∆ ABC∴ AD = DE AB BC ⇒ AD = 1 DB ⇒ DB + 1 = 2 AD ⇒ DB + AD = 2 AD ⇒ AB = 2 ⇒ AD = 1 AD AB 2 ⇒ DE = 1 BC 2 ∴ ∆ ODE = 1 ² = 1 ∆ BOC 2 4 ∴ ∆ ODE = 1 = 1 : 12 ∆ ABC 12
[∵ 3 ∆ BOC = ∆ ABC]
- ∆ ABC and ∆DEF are two similar triangles and the perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm respectively. If the length of DE = 36 cm, then length of AB is
-
View Hint View Answer Discuss in Forum
As per the given in question , we draw figures of two similar triangles ABC and DEF
Given , The perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm
and DE = 36 cm
∆ ABC ~ ∆ DEF∴ AB = BC = AC DE EF DF = AB + BC + CA DE + EF + FA ⇒ AB = 30 DE 18 ⇒ AB = 30 36 18
Correct Option: A
As per the given in question , we draw figures of two similar triangles ABC and DEF
Given , The perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm
and DE = 36 cm
∆ ABC ~ ∆ DEF∴ AB = BC = AC DE EF DF = AB + BC + CA DE + EF + FA ⇒ AB = 30 DE 18 ⇒ AB = 30 36 18 ⇒ AB = 30 × 36 = 60 cm. 18
