Plane Geometry
- If ABCD be a cyclic quadrilateral in which ∠A = 4x°, ∠B = 7x°, ∠C = 5y°, ∠D = y°, then x : y is
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According to question , we draw a figure of cyclic quadrilateral ABCD ,
Given that , ∠A = 4x°, ∠B = 7x°, ∠C = 5y°, ∠D = y°,
We know that the sum of opposite angles of a concyclic quadrilateral is 180°.
∴ ∠A + C = 180°
⇒ 4x + 5y = 180° ...(i)
∠B + ∠D = 180°
⇒ 7x + y = 180° ...(ii)
By equation (ii) × 5 – (i),
x = 720 31
From equation (ii),
7x + y = 180°⇒ 7 × 720 + y = 180° 31 y = 180° - 5040 31 y = 5580 - 5040 = 540 31 31
Correct Option: B
According to question , we draw a figure of cyclic quadrilateral ABCD ,
Given that , ∠A = 4x°, ∠B = 7x°, ∠C = 5y°, ∠D = y°,
We know that the sum of opposite angles of a concyclic quadrilateral is 180°.
∴ ∠A + C = 180°
⇒ 4x + 5y = 180° ...(i)
∠B + ∠D = 180°
⇒ 7x + y = 180° ...(ii)
By equation (ii) × 5 – (i),x = 720 31
From equation (ii),
7x + y = 180°⇒ 7 × 720 + y = 180° 31 y = 180° - 5040 31 y = 5580 - 5040 = 540 31 31 ∴ x : y = 720 : 540 = 4 : 3 31 31
- In a cyclic quadrilateral ∠A + ∠C = ∠B + ∠D = ?
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As per the given in question , we draw a figure cyclic quadrilateral ABCD ,
The sum of opposite angles of a concyclic quadrilateral = 180°Correct Option: D
As per the given in question , we draw a figure cyclic quadrilateral ABCD ,
The sum of opposite angles of a concyclic quadrilateral = 180°
∴ ∠A + ∠C = ∠B + ∠D = 180°
- ABCD is a cyclic quadrilateral. The side AB is extended to E in such a way that BE = BC. If ∠ADC = 70°, ∠BAD = 95°, then ∠DCE is equal to
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According to question , we draw a figure of cyclic quadrilateral ABCD in which side AB is extended to E in such a way that BE = BC ,
In concyclic quadrilateral ABCD,
∠ADC + ∠ABC = 180°
⇒ 70° + ∠ ABC = 180°
∴ ∠ ABC = 180° – 70° = 110°
∴ ∠CBE = 180° – 110° = 70°
BC = BE∴ ∠ BCE = ∠ BEC = 110 = 55° 2
Correct Option: A
According to question , we draw a figure of cyclic quadrilateral ABCD in which side AB is extended to E in such a way that BE = BC ,
In concyclic quadrilateral ABCD,
∠ADC + ∠ABC = 180°
⇒ 70° + ∠ ABC = 180°
∴ ∠ ABC = 180° – 70° = 110°
∴ ∠CBE = 180° – 110° = 70°
BC = BE∴ ∠ BCE = ∠ BEC = 110 = 55° 2
∠ BAD = 95°
∴ ∠ BAD + ∠ BCD = 180°
⇒ ∠ BCD = 180° – 95° = 85°
∴ ∠ DCE = ∠ BCD + ∠ BCE
∠ DCE = 85° + 55°= 140°
- In a cyclic quadrilateral ABCD m∠A + m∠B + m∠C + m∠D =?
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The sum of opposite angles of a concyclic quadrilateral is 180°.
Correct Option: B
As we know that the sum of opposite angles of a concyclic quadrilateral is 180°.
i.e. m∠A + m∠B + m∠C + m∠D = 180°.
- A quadrilateral ABCD circumscribes a circle and AB = 6 cm, CD = 5 cm and AD = 7 cm. The length of side BC is
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According to question , we draw a figure of quadrilateral ABCD circumscribes a circle
We know tangents drawn to a circle from same external point are equal
⇒ AM = AQ = x (say)
∴ MB = BN = 6 – x, QD = DP = 7 – x,
Let NC = PC = ZCorrect Option: A
According to question , we draw a figure of quadrilateral ABCD circumscribes a circle
We know tangents drawn to a circle from same external point are equal
⇒ AM = AQ = x (say)
∴ MB = BN = 6 – x, QD = DP = 7 – x,
Let NC = PC = Z
Now let us Consider side DC = 7 – x + z = 5
BC = 6 – x + z = (7 – x + z) – 1
BC = 5 – 1 = 4
