Advertisements
Advertisements
प्रश्न
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.

Advertisements
उत्तर
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm
QR = 3RS = 6 cm
3RS = 6 cm
RS = 2 cm
Now in ∆PQR,
∠Q = 90° ...[Angles in a semi-circle]
∴ PR2 = PQ2 + QR2 ...[Pythagoras theorem]
= 72 + 62
= 49 + 36
= 85
Again in right ΔPSQ,
PR2 = PS2 + RS2
`=>` 85 = PS2 + 22
`=>` PS2 = 85 – 4 = 81 = (9)2
∴ PS = 9 cm
Now, perimeter of quad PQRS
= PQ + QR + RS + SP
= (7 + 9 + 2 + 6) cm
= 24
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.

Prove that the parallelogram, inscribed in a circle, is a rectangle.
Prove that the rhombus, inscribed in a circle, is a square.
In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.

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 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.

In the given figure, AC is the diameter of the circle with center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED

