Question

${\huge{\fcolorbox{aqua}{navy}{\fcolorbox{yellow}{purple}{\bf{\color{yellow}{Question}}}}}}$
ABC is a right angled ∆.Find the area of shaded region if AB=6cm, BC=10 cm and l is centre of incircle of ∆ABC.


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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the concept of Pythagoras Theorem, Area of Circle and Area of Triangle has been used We are given that ABC is a right angled triangle. So first we will find the third side of this triangle. After that, we will find the area of the Triangle. We know that area of bigger triangle is equal to sum of areas of all triangles Inscribing it. So using that we will find the radius of circle and then find the area of shaded region.

Let's do it

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★ Equations Used :-

$\\\;\boxed{\sf{\green{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}}$

$\\\;\boxed{\sf{\green{Area\;of\;Triangle\;=\;\bf{\dfrac{1}{2}\;\times\;Base\;\times\;Height}}}}$

$\\\;\boxed{\sf{\green{Area\;of\;Circle\;=\;\bf{\pi r^{2}}}}}$

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★ Solution :-

Given (from figure),

» AB = 6 cm

» BC = 10 cm

» ∠BAC = 90° (since triangle is right angled)

» Radius of Circle = Pl = Ql = Rl = r

» Centre of Circle = l

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~ For length of AC ::

We know that ∆ABC is right angled. Thus,

• Height = AB = 6 cm
• Hypotenuse = BC = 10 cm
• Base = AC

By using Pythagoras Theorem here, we get

$\\\;\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(BC)^{2}\;=\;\bf{(AC)^{2}\;+\;(AB)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(AC)^{2}\;=\;\bf{(BC)^{2}\;-\;(AB)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(AC)^{2}\;=\;\bf{(10)^{2}\;-\;(6)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(AC)^{2}\;=\;\bf{100\;-\;36}}$

$\\\;\;\sf{:\rightarrow\;\;(AC)^{2}\;=\;\bf{64}}$

$\\\;\;\sf{:\rightarrow\;\;(AC)\;=\;\bf{\sqrt{64}}}$

$\\\;\;\underline{\bf{:\rightarrow\;\;AC\;=\;\bf{8\;\;cm}}}$

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~ For Area of ∆ABC ::

$\\\;\;\sf{:\rightarrow\;\;Area\;of\;Triangle\;=\;\bf{\dfrac{1}{2}\;\times\;Base\;\times\;Height}}$

$\\\;\;\sf{:\rightarrow\;\;Area\;of\;\Delta\:ABC\;=\;\bf{\dfrac{1}{2}\;\times\;8\;\times\;6}}$

$\\\;\;\underline{\bf{:\rightarrow\;\;Area\;of\;\Delta\:ABC\;=\;\bf{24\;\;cm^{2}}}}$

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~ For the value of r (radius of circle) ::

From figure, we see that ∆ClB , ∆ClA and ∆AlB are in the triangle ABC which completely inscribe it. Thus,

✒ Ar.(∆ABC) = Ar.(∆CIB) + Ar.(∆CIA) + Ar.(∆AIB)

Also we see that,

• In ∆CIB -

• Base = BC = 10 cm
• Height = Pl = r

• In ∆AIC -

• Base = AC = 8 cm
• Height = Ql = r

• In ∆AIB -

• Base = AB = 6 cm
• Height = Rl = r

By applying these values, we get,

$\\\sf{:\Longrightarrow\;Ar.(\Delta\:ABC)\:=\:\bf{\bigg(\dfrac{1}{2}\:\times\:10\:\times\:r\bigg)\:+\:\bigg(\dfrac{1}{2}\:\times\:8\:\times\:r\bigg)\:+\:\bigg(\dfrac{1}{2}\:\times\:6\:\times\:r\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;24\;=\;\bf{5r\;+\;4r\;+\;3r}}$

$\\\;\sf{:\Longrightarrow\;\;12\:r\;=\;\bf{24}}$

$\\\;\sf{:\Longrightarrow\;\;r\;=\;\bf{\dfrac{24}{12}}}$

$\\\;\underline{\bf{:\Longrightarrow\;\;r\;=\;\bf{2\;\;cm}}}$

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~ For Area of the Circle ::

$\\\;\;\sf{:\odot\;\;Area\;of\;Circle\;=\;\bf{\pi r^{2}}}$

$\\\;\;\sf{:\odot\;\;Area\;of\;Circle\;=\;\bf{\dfrac{22}{7}\;\times\;(2)^{2}}}$

$\\\;\;\sf{:\odot\;\;Area\;of\;Circle\;=\;\bf{\dfrac{22}{7}\;\times\;4}}$

$\\\;\;\sf{:\odot\;\;Area\;of\;Circle\;=\;\bf{\dfrac{88}{7}}}$

$\\\;\;\underline{\bf{:\odot\;\;Area\;of\;Circle\;=\;\bf{12.56\;\;cm^{2}}}}$

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~ For the Area of Shaded Region ::

This is given as,

$\\\;\;\sf{:\mapsto\;\;Area\;of\;Shaded\;Region\;=\;\bf{Area\;of\;\Delta\:ABC\;-\;Area\;of\;Circle}}$

$\\\;\;\sf{:\mapsto\;\;Area\;of\;Shaded\;Region\;=\;\bf{24\;-\;12.56}}$

$\\\;\;\sf{:\mapsto\;\;Area\;of\;Shaded\;Region\;=\;\bf{11.44\;\;cm^{2}}}$

$\\\;\underline{\boxed{\tt{\red{Area\;\;of\;\;shaded\;\;region}\;=\;\bf{\blue{11.44\;\;cm^{2}}}}}}$

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★ More Formulas to Know :-

$\\\;\sf{\pink{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}}$

$\\\;\sf{\pink{\leadsto\;\;Area\;of\;Rectangle\;=\;Length\;\times\;Breadth}}$

$\\\;\sf{\pink{\leadsto\;\;Area\;of\;Parallelogram\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}}$

Answer 2

Answer:

Hope this helps you

Step-by-step explanation:

Hence, the area of the shaded region is 11.44 cm2. Sol: ABC is a right angled triangle at right angled at A. BC = 10 cm, AB = 6 cm.