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
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`

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.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
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.
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, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM

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.

