Plane Geometry
- ABCD is a trapezium where AD|| BC. The diagonal AC and BD intersect each other at the point O. If AO = 3, CO = x – 3, BO = 3x – 19 and DO = x – 5, the value of x is
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As per the given in question , we draw a figure trapezium ABCD
Given that , AO = 3, CO = x – 3, BO = 3x – 19 and DO = x – 5,
We have , AO × BO = OD × OC
⇒ 3 (3x – 19) = (x – 5) (x – 3)
⇒ 9x – 57 = x2 – 8x + 15
⇒ x2 – 17x + 72 = 0
⇒ x2 – 9x – 8x + 72 = 0
⇒ x (x – 9) – 8 (x – 9) = 0Correct Option: D
As per the given in question , we draw a figure trapezium ABCD
Given that , AO = 3, CO = x – 3, BO = 3x – 19 and DO = x – 5,
We have , AO × BO = OD × OC
⇒ 3 (3x – 19) = (x – 5) (x – 3)
⇒ 9x – 57 = x2 – 8x + 15
⇒ x2 – 17x + 72 = 0
⇒ x2 – 9x – 8x + 72 = 0
⇒ x (x – 9) – 8 (x – 9) = 0
⇒ (x – 8) (x – 9) = 0
⇒ x = 8 or 9
- In a quadrilateral ABCD, the bisectors of ∠A and ∠B meet at O. If ∠C = 70° and ∠D = 130°, then measure of ∠AOB is
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On the basis of question we draw a figure of quadrilateral ABCD
Here , ∠C = 70° and ∠D = 130°
We know that , ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + 70° + 130° = 360°
⇒ ∠A + ∠B = 360° – 70° – 130° = 160°
In ∆ AOB,
∠OAB + ∠OBA + ∠AOB = 180°⇒ ∠A + ∠B + ∠AOB = 180° 2 2 ⇒ 1 (∠A + ∠B) + ∠AOB = 180° 2
Correct Option: D
On the basis of question we draw a figure of quadrilateral ABCD
Here , ∠C = 70° and ∠D = 130°
We know that , ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + 70° + 130° = 360°
⇒ ∠A + ∠B = 360° – 70° – 130° = 160°
In ∆ AOB,
∠OAB + ∠OBA + ∠AOB = 180°⇒ ∠A + ∠B + ∠AOB = 180° 2 2 ⇒ 1 (∠A + ∠B) + ∠AOB = 180° 2 ⇒ 1 × 160° + ∠AOB = 180° 2
⇒ ∠AOB = 180° – 80° = 100°
- ABCD is a trapezium whose side AD is parallel to BC. Diagonals AC and BD intersect at O. If AO = 3 , CO = x - 3 , BO = 3x - 19 and DO = x - 5 , the value(s) of x will be :
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As per the given in question , we draw a figure trapezium ABCD
Clearly,
∆AOD ~ ∆BOC∴ BO = OD OC OA ⇒ 3x - 19 = x - 5 x - 3 3
Correct Option: D
As per the given in question , we draw a figure trapezium ABCD
Clearly,
∆AOD ~ ∆BOC∴ BO = OD OC OA ⇒ 3x - 19 = x - 5 x - 3 3
⇒ 9x – 57 = x2 – 8x + 15
⇒ x2 – 17x + 72 = 0
⇒ x2 – 8x – 9x + 72 = 0
⇒ x (x – 8) – 9 (x – 8) = 0
⇒ (x – 8) (x – 9) = 0
⇒ x = 8 or 9
- ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB = AE = BF. Then :
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According to question , we draw a figure of rhombus ABCD
We know that diagonals of a rhombus are perpendicular bisector of each other.
∴ OA = OC; OB = OD
∠AOD = ∠COD = 90°
∠AOB = ∠COB = 90°
In ∆ BDE,
OA || DE
⇒ OC || DG
In ∆CFA,
OB || CF
⇒ OD || GC
In quadrilateral DOCG ,Correct Option: B
According to question , we draw a figure of rhombus ABCD
We know that diagonals of a rhombus are perpendicular bisector of each other.
∴ OA = OC; OB = OD
∠AOD = ∠COD = 90°
∠AOB = ∠COB = 90°
In ∆ BDE,
OA || DE
⇒ OC || DG
In ∆CFA,
OB || CF
⇒ OD || GC
In quadrilateral DOCG ,
OC || DG
⇒ OD || GC
∴ DOCG is a parallelogram.
∴ ∠DGC = ∠DOC
⇒ ∠DGC = 90°
EG ⊥ GF or ED ⊥ CF
- ABCD is a rhombus whose side AB = 4 cm and ∠ABC = 120°, then the length of diagonal BD is equal to :
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According to question , we draw a figure of rhombus ABCD
Here , AB = 4 cm and ∠ABC = 120°
From ∆BOC,cos 60° = BO 4
Correct Option: D
According to question , we draw a figure of rhombus ABCD
Here , AB = 4 cm and ∠ABC = 120°
From ∆BOC,cos 60° = BO 4 ⇒ BO = 1 × 4 = 2 cm 2
∴ BD = 2BO = 2 × 2 = 4 cm
