Advertisements
Advertisements
Questions
Prove that the rhombus, inscribed in a circle, is a square.
Prove that the cyclic rhombus is a square.
Advertisements
Solution
Let ABCD be a rhombus, inscribed in a circle
Now, ∠BAD + ∠BCD
(Opposite angles of a parallelogram are equal)
And ∠BAD + ∠BCD =180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
∴ ∠BAD + ∠BCD = `(180^circ)/2` = 90°
The other two angles are 90°, and all the sides are equal.
∴ ABCD is a square.
RELATED QUESTIONS
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
Calculate the area of the shaded region, if the diameter of the semicircle is equal to 14 cm. Take `pi = 22/7`
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM
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, 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.
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 : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
