Question

In adjoining figure a circle is inscribed in the quadrilateral ABCD given that BC is equal to 38 cm cube is equal to 27 cm and BC equal to 25 cm and that ad is perpendicular to DC find the radius of the circle.​

Answer 1

Answer:

14 cm

Step-by-step explanation:

Given

• BQ = BR, AQ = AP, CR = CS, DP = DS = OP = OS
• BQ = 27 cm
• BC = 38 cm
• DC = 25 cm

Solving

• BQ = BR (given)
• ⇒ BR = 27 cm
• BC = BR + CR
• 38 = 27 + CR
• CR = 11 cm
• ⇒ CS = 11cm
• DC = DS + CS
• 25 = 11 + DS
• ⇒ DS = 14 cm
• But, DS = OS and OP (radii)
• ⇒ Radius = 14 cm

Answer 2

Radius = 14cm

Step-by-step explanation:

Let O be the centre of the circle. Since P and S are the points of contact of tangents AD and DC respectively, OP ⊥ AD and OS ⊥ DC

Also, AD ⊥ DC (given), therefore, OPDS is a square.

BR = BQ = 27cm...(tangents from an external point to a circle are equal in length)

therefore, CR = CB - BR = (38 - 27)cm = 11cm

SC = CR = 11 cm...(tangents from an external point) therefore, DS = DC - SC = (25 - 11) cm = 14 cm.

therefore,

Radius of a circle = OP = DS = 14 cm...(therefore, OPDS is a square).

Hope it helps you!!