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प्रश्न
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.
उत्तर
i. Draw a line segment BC = 7 cm
ii. Taking mid point of BC as centre O , draw a semi-circle
with radius = 3.5 cm
iii. Now, the semicircle circumscribes the ΔABC
iv. Draw angle bisector of ∠ABC and make it intersect the
semi-circle at D.
v. Measure the angle ∠DBC which comes out to be 22.5°
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संबंधित प्रश्न
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
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.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
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.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
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, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
