Question

The bottom side of the quadrilateral in the picture is diameter of the circle and the top side is a chord parallel to it. AB =10 centimetres.CD= 6 centimetres. (a) What is the radius of the circle? (b) Find the area of the quadrilateral ABCD.​

Answer 1

Answer:

O is the centre of the circle.

CD is the diameter, CD = 5 cm

EF is the chord parallel to CD, EF = 3 cm

OG is perpendicular drawn on EF from O.

In ΔOGE we have,

∠OGE = 90° [∵ OG is perpendicular on EF]

OE = 5/2 = 2.5 cm [∵ Diameter = 5 cm]

EG = 3/2 = 1.5 cm [∵perpendicular drawn from centre bisects chord]

∴ Area of the quadrilateral,

= 8 cm2

Answer 2

The answers are Radius of circle 5 cm

               Area of Quadrilateral  = 20 cm²

Given: Bottom side of the quadrilateral is diameter of the circle

From given picture bottom of the quadrilateral AB = 10cm

and CD = 6 cm

To find: Radius of the circle and area of the quadrilateral

Solution:

a) Radius of circle

given that diameter of circle = bottom of quadrilateral  

⇒ diameter of circle 2r = 10 cm

⇒ Radius of circle r = 10/2 = 5 cm

∴ Radius of circle 5 cm

b) Area of the quadrilateral

Draw a line by joining mid points of AB and CD

and join mid point of AB and vertice C then a right angle triangle POC  is formed as show figure

In triangle POC  ⇒  OC² = OP²+ PC²  

[ here OC = radius = 5 cm and PC = $\frac{1}{2} (CD)$ = 3 cm ]

⇒  5 ² = OP²+ 3²  

⇒ OP² = 5² - 3² = 25 - 9 = 16

⇒ OP² = 4²

⇒ OP = 4  is the altitude/height of trapezium

Therefore, area of trapezium =  $\frac{1}{2} (base)(height)$  

⇒ $\frac{1}{2} (10)(4)$ = 5(4) = 20 cm²

∴ Area of triangle = 20 cm²

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