Advertisements
Advertisements
प्रश्न
Prove that the rhombus, inscribed in a circle, is a square.
Prove that the cyclic rhombus is a square.
Advertisements
उत्तर
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.
APPEARS IN
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
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
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 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 figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find 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 : ∠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 given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
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 given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
