Question

ABCD is a quadrilateral such that ZD=90°. Ac rcle C (0.r) touches the sides AB, BC. CD an
DA at P, Q, R and S respectively. If BC = 38 cm. CD = 25 cm and BP = 27 cm. then find r.
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Answer 1

Step-by-step explanation:

this your answer.udhdhf

Answer 2

⠀

$\large\bf\underline{given:}$

• ABCD as a quadrilateral
• BC = 38cm
• CD = 25cm
• BP = 27cm
• ∠D = 90°

$\large\bf\underline{to\:find:}$

$\tt\underline\green{"r"}$

$\large\bf\underline{solution:}$

So,BP = BQ (tangent from an external point B)

But BP = 27cm

➔$\tt\underline\red{BQ \:= \:27cm}$

it is given that BC = 38cm

➔ BQ + CQ = 38cm

➔27 + CQ = 38cm

➔$\tt\underline\red{CQ \:= \:11 cm}$

➔CQ = CR (tangent from an external point C)

But CQ = 11cm

➔$\tt\underline\red{CR\: = \:11cm}$

it is given that: CD = 25cm

➔ CR + DR = 25

➔11 + DR = 25

➔$\tt\underline\red{DR \:= \:14cm}$

Since tangent to a circle is perpendicular to the radius thorough the point of contact.

∴ ∠ORD = ∠OSD = 90°

it is given that

$\tt\underline\red{∠\:D\: =\: 90°}$

now in quadrilateral ORDS,

∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°

and OR = OS (radii of circle)

therefore,ORDS is a square

So,OR = DR = 14cm

$\tt\underline\red{hence\:"r"\:\:=\:14}$