Advertisements
Advertisements
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.
Advertisements
उत्तर
- Draw a line segment BC = 7 cm.
- Taking the midpoint of BC as centre O, draw a semicircle.
with radius = 3.5 cm. - Now, the semicircle circumscribes the ΔABC.
- Draw angle bisector of ∠ABC and make it intersect the
semi-circle at D. - Measure the angle ∠DBC, which comes out to be 22.5°
APPEARS IN
संबंधित प्रश्न
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.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.
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 : ∠DBC
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 following figure, AD is the diameter of the circle with centre O. chords AB, BC and CD are equal. If ∠DEF = 110°, Calculate: ∠FAB.
In Fig, Chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, Calculate ∠ DEC.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
