Plane Geometry
- ABCD is a rectangle where the ratio of the length of AB and BC is 3 : 2. If P is the mid-point of AB, then the value of sin ∠CPB is
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As per the given in question , we draw a figure rectangle ABCD
Given , AB : BC = 3 : 2
Let AB = 3y units and BC = 2y units⇒ PB = 3 y units. 2
From ∆ PCB ,
CP = √PB² + BC²CP = √ 9y² + 4y² 4 CP = √ 25y² = 5y units 4 2 ∴ sin ∠CPB = BC CP sin ∠CPB = 2y 5y 2 sin ∠CPB = 4 5
Take AB = 6y and BC = 4y ⇒ BP = 3y ⇒ CP = 5y
Correct Option: D
As per the given in question , we draw a figure rectangle ABCD
Given , AB : BC = 3 : 2
Let AB = 3y units and BC = 2y units⇒ PB = 3 y units. 2
From ∆ PCB ,
CP = √PB² + BC²CP = √ 9y² + 4y² 4 CP = √ 25y² = 5y units 4 2 ∴ sin ∠CPB = BC CP sin ∠CPB = 2y 5y 2 sin ∠CPB = 4 5
Take AB = 6y and BC = 4y ⇒ BP = 3y ⇒ CP = 5yand sin ∠BPC = BC = 4y = 4 CP 5y 5
- If the opposite sides of a quadrilateral and also its diagonals are equal, then each of the angles of the quadrilateral is
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According to question , we draw a figure of rectangle ABCD
AB = CD
BC = AD
AC = BDCorrect Option: A
According to question , we draw a figure of rectangle ABCD
AB = CD
BC = AD
AC = BD
It will be a rectangle and each angle will be a right angle.
- The length of the two adjacent sides of a rectangle inscribed in a circle are 5 cm and 12 cm respectively. Then the radius of the circle will be
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As per the given question , we draw a figure of rectangle ABCD inscribed in a circle ,
We know that , Diameter of circle = Diagonal of rectangle
Here , BC = 12 cm. CD = 5 cm.
∴ BD = √BC² + CD²
BD = √12² + 5²
BD = √144 + 25
BD = √169 = 13 cm.∴ Radius of circle = BO = BD 2
Correct Option: B
As per the given question , we draw a figure of rectangle ABCD inscribed in a circle ,
We know that , Diameter of circle = Diagonal of rectangle
Here , BC = 12 cm. CD = 5 cm.
∴ BD = √BC² + CD²
BD = √12² + 5²
BD = √144 + 25
BD = √169 = 13 cm.∴ Radius of circle = BO = BD 2 Radius of circle = 13 = 6.5 cm. 2
- PQRA is a rectangle, AP = 22 cm, PQ = 8 cm. ∆ ABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm. Then the length of the line joining the mid points of the sides AB and BC is
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On the basis of question we draw a figure of rectangle PQRA and ABC is a triangle whose vertices lie on the sides of PQRA ,
Given that , QR = AP = 22 cm, PQ = 8 cm BQ = 2 cm and QC = 16 cm.
∴ CR = QR - QC = 22 – 16 = 6 cm.
BC = √BQ² + QC²
BC = √2² + 16²
BC = √4 + 256
BC = √260 cm.
AC = √CR² + AR²
AC = √6² + 8²
AC = √36 + 64
AC = √100 = 10 cm.Correct Option: B
On the basis of question we draw a figure of rectangle PQRA and ABC is a triangle whose vertices lie on the sides of PQRA ,
Given that , QR = AP = 22 cm, PQ = 8 cm BQ = 2 cm and QC = 16 cm.
∴ CR = QR - QC = 22 – 16 = 6 cm.
BC = √BQ² + QC²
BC = √2² + 16²
BC = √4 + 256
BC = √260 cm.
AC = √CR² + AR²
AC = √6² + 8²
AC = √36 + 64
AC = √100 = 10 cm.
BD = DC
BE = EA∴ DE || AC and DE = 1 AC 2
DE = 10 ÷ 2 = 5 cm.
- ABCD is a cyclic quadrilateral. AB and DC are produced to meet at P. If ∠ADC = 70° and ∠DAB = 60°, then the ∠PBC + ∠PCB is
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According to question , we draw a figure of cyclic quadrilateral ABCD in which AB and DC are produced to meet at P,
∠ADC = 70°
∠ABC = 180° – 70° = 110°
⇒ ∠PBC = 70°
∠BCD = 180° – 60° = 120°Correct Option: A
According to question , we draw a figure of cyclic quadrilateral ABCD in which AB and DC are produced to meet at P,
∠ADC = 70°
∠ABC = 180° – 70° = 110°
⇒ ∠PBC = 70°
∠BCD = 180° – 60° = 120°
⇒ ∠PCB = 60°
∴ ∠PBC + ∠PCB = 70° + 60° = 130°
