Question

100 points question✔️✔️

In the figure ABC is a right angled triangle withAB=6cm and AB=8cm.A circle with centre O has been inscribed inside the triangle.Calculate the value of r ,the radius of the inscribed circle.

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#Solve with proper explanation​

Answer 1

By Pythagoras theorem,

AC^2+AB^2=BC^2

BC=√6²+8²

BC=√100

BC=10 cm

Join OA,OB,OC.

We know,

Ar( △ABC)=Ar( △AOB+ △BOC+ △COA)

=>  (1/2*8*r)+(1/2*6*r)+(1/2*10*r)=1/2*6*8

=>  4r+3r+5r=24

=>  12r=24

=>  r=24/12=2 cm

Answer 2

$\huge{\boxed{\boxed{\tt{Answer:}}}}$

Given:

• In the figure ABC is a right angled triangle with AB= 6cm and AB = 8cm.

• A circle with centre O has been inscribed inside the triangle.

Find:

Calculate the value of r, the radius of the inscribed circle. Note: R = Radius.

Calculations:

Given: We know that, ABC given is a right angled triangle. AC = 6 cm and AB = 8 cm.

By using Pythagorean theorem:

$\sf\rightarrow {\sqrt{AB^2 + BC^2}}$

$\sf\rightarrow {\sqrt{AB {}^{2} + AC {}^{2} }}$

$\sf\rightarrow \sqrt{6 {}^{2} + 8 {}^{2} } \\ \sf\rightarrow \sqrt{36 + 64} \\ \sf\rightarrow \sqrt{100} \\ \sf\rightarrow10$

Now, let's join OA, OB, OC, we get;

$\sf\rightarrow \: ar( AOC) + ar(OAB) +(BOC) = (ABC) \\ \sf\rightarrow \: \frac{1}{2} \times 8 \times r+ \frac{1}{2} \times 6 \times r + \frac{1}{2} \times 10 \times r = \frac{1}{2} \times 6 \times 8 \\ \sf\rightarrow \: 4r + 3r + 5r = 24 \\ \sf\rightarrow \: 12r = 24 \\ \sf\rightarrow \: r = \frac{24}{12} \\ \sf\rightarrow \: 2 \: cm$

Therefore, 2 cm is the radius.