Question

A quadrilateral abcd is drawn so that d = 90, bc = 38 cm and cd = 25 cm. A circle is inscribed in the quadrilateral and it touches the side ab, bc, cd and da at p, q, r and s respectively. If bp = 27 cm, find the radius of the inscribed circle.

Answer 1

Question

A quadrilateral ABCD is drawn so that angle D = 90, BC = 38 cm and CD = 25 cm. A circle is inscribed in the quadrilateral and it touches the side AB, BC, CD and DA at P, Q, R and S respectively. If BP = 27 cm. Find the radius of the inscribed circle.

Answer

Radius of the inscribed circle is 14 cm.

Explanation-

BP and BQ are the tangents drawn from a point B.

So, BP = BQ = 27 cm

Now,

→ BC = BQ + QC

→ 38 = 27 + QC

→ 38 - 27 = QC

→ 11 = QC

→ QC = 11 cm

Similarly, CQ and CR are tangents drawn from point C.

So, CQ = CR = 11 cm

Therefore,

→ CD = CR + RD

→ 25 = 11 + RD

→ 25 - 11 = RD

→ 14 = RD

→ RD = 4 cm

From D; SD and RD are the tangents drawn. So, SD = RD = 14 cm

Also,

OS perpendicular AD and OR perpendicular CD.

And angle ADC = 90°.

Therefore, OSRD is a square. (In square all sides are equal and makes an angle of 90°).

So, OS = SD = RD = OR = 14 cm.

•°• Radius of the inscribed circle is 14 cm.

Answer 2

||✪✪ QUESTION ✪✪||

A quadrilateral abcd is drawn so that Angle D = 90°, BC = 38cm and CD = 25cm. A circle is inscribed in the quadrilateral and it touches the side AB, BC, CD and DA at p, q, r and s respectively. If BP = 27 cm, find the radius of the inscribed circle. ?

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

Concept used :-

☛ Property ❶ :- Tangents From same External Point ,to tje circle are Equal in Length.

☛ Property ❷ :- The tangent at a point on a circle is at right angles to the radius.

From Image , we have :-

➪ BP = 27cm

➪ BC = 38cm.

According to Property ❶ , we can say That, since B is External points of Both tangents QB and PB,

So,

⇒ QB = PB = 27cm.

Hence ,

⇒ QC = BC - QB

⇒ QC = 38 - 27

⇒ QC = 11cm.

_____________________________

Now, Again, From Property ❶ , since Point C is External point of Two tangents QC ans CR .

So,

➼ QC = CR

➼ CR = 11 cm.

Hence,

➼ RD = CD - CR

Given, now, CD is 25cm.

So,

➼ RD = 25 - 11

➼ RD = 14cm.

______________________________

Again, with same Property ❶, This Time D is External Point ,

So,

⟿ RD = DS

⟿ RD = DS = 14cm.

______________________________

Now, we Have Given That,

⟼ ∠CDA = 90°

or,

⟼ ∠RDS = 90°

Now, with property ❷ , since, O is the centre of circle, Hence, OR and OS are radius of the circle , So, we can say That :-

➻ ∠OSD = ∠ORD = 90° Each.

______________________________

Conclusion ❶ :- Now, we have 3 angles of a Quadrilateral OSDR, which are 90° each, That Means we can say That, our 4th angle is also 90°.. { 360- (90*3)}

Conclusion ❷ :- And, we also have two adjacent sides RD and DS are Equal .

with These Two conclusions we can say That, OSDR is a Square . { with Adjacent sides Equal and each angle as 90°}.

______________________________

⃞ OSDR now, we have :-

☛ OS = SD = RD = OR

since,

☞ RD = SD = 14cm.

Hence,

☛ OS = OR = 14cm = Radius of The circle.

Hence, Radius of The inscribed Circle is 14cm.