Question

In Right angle ∆ABC the two small sides are 9cm and 12cm. In it a circle inscribed that touches all the side of ∆ABC . Find out (i) radius of circle. (ii)Area of shaded region. Solve this question in full details.​

Answer 1

$\sf\small\underline\blue{Correct\:Question:-}$

In Right angle ∆ABC the two small sides are 9cm, 12cm and longer side is 15cm . In it a circle inscribed that touches all the side of ∆ABC . Find out (i) radius of circle. (ii)Area of shaded region.

$\sf\small\underline\blue{Given:-}$

$\rm{\implies Two\:small\:sides\:_{(\triangle\:ABC)}=9cm\:,12cm}$

$\sf\small\underline\blue{To\:Find:-}$

$\rm{\implies Radius\:of\:_{(circle)}=?}$

$\rm{\implies Area\:_{(shaded\: region)}=?}$

$\sf\small\underline\blue{Solution:-}$

As per the construction figure. AB, BC and AC are 9cm, 12cm and 15cm . Draw perpendicular OE_|_AB , OF_|_AC , OG_|_BC. Here we have to assume the radius of circle be x so, OE=OG=OF=x.

By joining O to Points E, G and F which lies on the sides of ∆ ABC. A square form having sides x ( here x is the radius of circle). So square EBGO form.

AE = AB - EB (9 - x)

AE = AF = (9 - x)

GC = BC - BG ( 12 - x)

GC = CF = ( 12 - x)

$\rm{\implies AE=AF\:(tangent\:of\: circle)}$

$\rm{\implies GC=CF\:(tangent\:of\: circle)}$

$\sf\small\underline\blue{Calculation\:begins:-}$

$\tt{\leadsto AC=AF+CF}$

$\tt{\leadsto 9-x+12-x=15}$

$\tt{\leadsto 21-2x=15}$

$\tt{\leadsto 2x=21-15}$

$\tt{\leadsto 2x=6}$

$\tt{\leadsto x=3cm}$

$\tt{\leadsto \therefore\: Radius\:(x)=3cm}$

• Now calculate Shaded region :-]

$\tt{\leadsto Area\:_{(shaded\: region)}= Area\:_{(triangle)}-Area\:_{(circle)}}$

$\tt{\leadsto Area\:_{(shaded\: region)}=\bigg(\dfrac{bh}{2}-(\pi\:r^2)\bigg)}$

$\tt{\leadsto Area\:_{(shaded\: region)}=\bigg(\dfrac{9*12}{2}-(\pi\:3^2)\bigg)}$

$\tt{\leadsto Area\:_{(shaded\: region)}=9*6-9\pi}$

$\tt{\leadsto Area\:_{(shaded\: region)}=54-\dfrac{9*22}{7}}$

$\tt{\leadsto Area\:_{(shaded\: region)}=\dfrac{378-198}{7}}$

$\tt{\leadsto Area\:_{(shaded\: region)}=\dfrac{180}{7}}$

$\tt{\leadsto Area\:_{(shaded\: region)}=25.71\:cm^2}$

$\sf\large{Hence'}$

$\bf{\implies Area\:_{(shaded\: region)}=25.71\:cm^2}$

$\rm{\implies Radius\:of\:_{(circle)}=3cm}$

Answer 2

Given :-

In  Right angle ∆ABC the two small sides are 9cm and 12cm.

To Find :-

(i) radius circle. (ii)Area of shaded region.

Solution :-

Let the center of circle be O

And the onscribed circle be PQR

Now

A square may be formed as PBRO

Let the radii be r

Tagnets of circle are equal. So,

AP = AB - PE

AP = 9 - r

Also

AP = AQ

AQ = 9 - r

RC = BC - BR

RC = 12 - x

According to the question

Length of radius = AP + CP

15 = 9 - r + 12 - r

15 = 21 - 2r

15 - 21 = -2r

-6 = -2r

-6/-2 = r

6/2 = r

3 = r

Radius is 3 cm

Finding area of shaded region

Area of triangle = 1/2 × Base × Height

Area of circle = π r²

Area of shaded region = 1/2 × Base × Height - πr²

Area = 1/2 × 9 × 12 - π × (3)²

Area = 1/2 × 9 × 12 - π × 9

Area = 6 × 9 - 9π

Area = 54 - 9π

Area = 54 - 9(3.14)

Area = 25.74 cm²

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