Plane Geometry
- Two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other. The length of the common chord is :
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According to question , we draw a figure of two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other ,
OC = 2cm
OA = 4cm
∴ AC = √OA² - OC²
∴ AC = √4² - 2² = √16 - 4Correct Option: B
According to question , we draw a figure of two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other ,
OC = 2cm
OA = 4cm
∴ AC = √OA² - OC²
∴ AC = √4² - 2² = √16 - 4
AC = √12 = 2√3
∴ AB = 4√3cm
- At least two pairs of consecutive angles are congruent in a _____.
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As per the given in question , we draw a figure cyclic trapezium ABCD
In an isosceles trapezium,
AB || CD ; AD = BC
AC = BD
The angles of base are equal. i.e.Correct Option: B
As per the given in question , we draw a figure cyclic trapezium ABCD
In an isosceles trapezium,
AB || CD ; AD = BC
AC = BD
The angles of base are equal. i.e.
∠A = ∠B ; ∠C = ∠D ; ∠A + ∠D = ∠B + ∠C = 180°
- ABCD is a trapezium in which AD || BC and AB = DC = 10 m. then the distance of AD from BC is :
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According to question , we draw a figure of trapezium ABCD in which AD || BC and AB = DC = 10 m
AE ⊥ BC; DF ⊥ BC
∴ ∠DCB = 45°
In ∆ CDF,sin 45° = DF DC ⇒ 1 = DF √2 10
Correct Option: C
According to question , we draw a figure of trapezium ABCD in which AD || BC and AB = DC = 10 m
AE ⊥ BC; DF ⊥ BC
∴ ∠DCB = 45°
In ∆ CDF,sin 45° = DF DC ⇒ 1 = DF √2 10 ⇒ DF = 10 = 5√2 metre. √2
- If the parallel sides of a trapezium are 8 cm. and 4 cm., M and N are the mid-points of the diagonals of the trapezium, then length of MN is
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As per the given in question , we draw a figure cyclic trapezium ABCD in which M and N are the mid-points of the diagonals ,
The line segment joining the mid– points of the diagonals of a trapezium is parallel to each of parallel sides and is equal to half the difference of these sides.
Here , AB = 8 cm. and CD = 4 cm∴ MN = 1 (AB - CD) 2
Correct Option: D
As per the given in question , we draw a figure cyclic trapezium ABCD in which M and N are the mid-points of the diagonals ,
The line segment joining the mid– points of the diagonals of a trapezium is parallel to each of parallel sides and is equal to half the difference of these sides.
Here , AB = 8 cm. and CD = 4 cm∴ MN = 1 (AB - CD) 2 MN = 1 (8 - 4) cm. = 2 cm. 2
- ABCD is a cyclic trapezium with AD || BC. If ∠A = 105°, then other three angles are
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On the basis of question we draw a figure of trapezium ABCD with AD || BC ,
AD || BC
In concyclic quadrilateral ABCD,
∠A + ∠C = 180°
⇒ 105° + ∠C = 180°
⇒ ∠C = 180° – 105° = 75°
Again,
∠A + ∠B = 180°
⇒ 105° + ∠B = 180°Correct Option: A
On the basis of question we draw a figure of trapezium ABCD with AD || BC ,
AD || BC
In concyclic quadrilateral ABCD,
∠A + ∠C = 180°
⇒ 105° + ∠C = 180°
⇒ ∠C = 180° – 105° = 75°
Again,
∠A + ∠B = 180°
⇒ 105° + ∠B = 180°
⇒ ∠B = 180° – 105° = 75°
∴ ∠D = 180° – 75° = 105°
