Advertisements
Advertisements
प्रश्न
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate : ∠BCD.
उत्तर
AB || ED
Therefore ∠DEB = EBA = 27° (Alternate angles)
Therefore BCDE is a cyclic quadrilateral
Therefore ∠DEB + ∠BCD = 180°
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Therefore ∠BCD = 180° – 27° = 153°
APPEARS IN
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Prove that the rhombus, inscribed in a circle, is a square.
In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate: ∠AEF
In the figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, fins the distance MN between the two chords of lengths 24 cm and 18 cm respectively.
Using ruler and a compass only construct a semi-circle with diameter BC = 7cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠ADC
Also, show that the ΔAOD is an equilateral triangle.
In the given figure AC is the diameter of the circle with centre O. CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°.
Calculate
- ∠BEC
- ∠BCD
- ∠CED
