Advertisements
Advertisements
प्रश्न
In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.
Advertisements
उत्तर
Given:
∠BAD = 65°
∠ABD = 70°
∠BDC = 45°
(i) In Δ ABD,
∠ BAD + ∠ABD + ∠ADB = 180°
65° + 70° + ∠ADB = 180° ....(Sum of three angles of a)
∠ADB = 180° - (65° + 70°)
∠ADB = 45°.
∠ ADC = ∠ADB + ∠BDC
45° + 45° = 90°
AC is the diameter of the circle. ....(Angle in a semi circle is 90°)
Proved.
(ii) ACB = ADB = 45° ....(Angles in the same segment of a circle)
संबंधित प्रश्न
Calculate the area of the shaded region, if the diameter of the semicircle is equal to 14 cm. Take `pi = 22/7`
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 parallelogram, inscribed in a circle, is a rectangle.
Prove that the rhombus, inscribed in a circle, is a square.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°, calculate:
- ∠DAB,
- ∠BDC.
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
- ∠DAB,
- ∠DBA,
- ∠DBC,
- ∠ADC.
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC
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, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
