Question

three circles with centres A,B and C touch externally each other.The radius of circle center A is x cm, radius of circle with center B has radius 6cm and with centre C has radius 20cm if Angle BAC =90degree,frame an equation in X and solve for it.


the answer from the answer sheet is 4cm.

please if you answer it correctly I will mark ‼️ you as brainlist
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Answer 1

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$\underline{\textsf{\textbf{Question\::}}}$

Three circles with centres A , B and C touch externally each other. Radius of circle with center A is x cm, radius of circle with center B has radius 6 cm and with centre C has radius 20 cm , if Angle BAC = 90°, frame an equation in x and solve for it.

$\underline{\textsf{\textbf{Given\::}}}$

• Radius of circle with center A = x cm
• Radius of circle with center B = 6 cm
• Radius of circle with center C = 20 cm
• Measure of angle BAC = 90°

$\underline{\textsf{\textbf{To\:find\::}}}$

• Radius of circle with center A , i.e , we had to find value of x ?

$\underline{\textsf{\textbf{Solution\::}}}$

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★ Finding the length of AB :

$:\implies\:\sf Length\:of\:AB\:=\:Radius_{(circle\:with\:center\:A)}\:+\:Radius_{(circle\:with\:center\:B)}$

$:\implies\:\bf{Length\:of\:AB\:=\:\red{(x + 6)\:cm}}$

★ Finding the length of BC :

$:\implies\:\sf Length\:of\:BC\:=\:Radius_{(circle\:with\:center\:B)}\:+\:Radius_{(circle\:with\:center\:C)}$

$:\implies\:\sf Length\:of\:BC = 6 + 20$

$:\implies\:\bf{Length\:of\:BC\:=\:\red{26\:cm}}$

★ Finding the length of AC :

$:\implies\:\sf Length\:of\:AC\:=\:Radius_{(circle\:with\:center\:A)}\:+\:Radius_{(circle\:with\:center\:C)}$

$:\implies\:\bf{Length\:of\:AC\:=\:\red{(x + 20)\:cm}}$

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Now,

★ Using pythagoras on triangle ABC :

$\dashrightarrow\:\sf BC^{2} = AB^{2} + AC^{2}$

★ Values that we have :

• BC = 26 cm
• AB = (x + 6) cm
• AC = (x + 20) cm

★ Putting all values :

$\dashrightarrow\:\sf (26)^{2} = (x + 6)^{2} + (x + 20)^{2}$

$\tiny \dag \: {\underline{\sf{Using\: identity\: \red{[(a + b)^2 = a^2 + b^2 + 2ab]}\:on\: \red{(x + 6)^2}\: and\: \red{(x + 20)^2}\::}}}$

$\dashrightarrow\:\sf 676 = [x^2 + (6)^2 + 2(x)(6)] + [x^2 + (20)^2 + 2(x)(20)]$

$\dashrightarrow\:\sf 676 = x^2 + 36 + 12x + x^2 + 400 + 40x$

$\dashrightarrow\:\sf 676 = x^2 + x^2 + 36 + 400 + 40x + 12x$

$\dashrightarrow\:\sf 676 = 2x^2 + 436 + 52x$

$\dashrightarrow\:\sf 0 = 2x^2 + 436 + 52x - 676$

$\dashrightarrow\:\sf 0 = 2x^2 + 52x + 436 - 676$

$\dashrightarrow\:\sf 2x^2 + 52x - 240 = 0$

$\dashrightarrow\:\sf 2(x^2 + 26x - 120) = 0$

$\dashrightarrow\:\sf x^2 + 26x - 120 = \dfrac{0}{2}$

$\dashrightarrow\:\sf x^2 + 26x - 120 = 0$

• Quadratic equation formed !!

★ Solving it by middle splitting method :

$\dashrightarrow\:\sf x^2 + 30x - 4x - 120 = 0$

$\dashrightarrow\:\sf x(x + 30) - 4(x + 30) = 0$

$\dashrightarrow\:\sf (x - 4)\:( x + 30) = 0$

$\dashrightarrow\:\sf x - 4 = 0\:,\:x + 30 = 0$

$\dashrightarrow\:\bf{ x = \red{4}\:,\:x = \red{-30}}$

• Radius can't be negative !!

∴ We will reject x = -30 !!

$\small\underline{\boxed{\sf{\therefore\:Radius\:of\:circle\:with\:center\:A\:=\:x\:=\:\rm\purple{4\:cm}}}}$

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Answer 2

$\huge\tt\red{Q}\tt\pink{U}\tt\blue{E}\tt\green{S}\tt\purple{T}\tt\orange{I}\tt\orange{O}\huge\tt\red{N}$

Three circles with centres A , B and C touch externally each other. Radius of circle with center A is x cm, radius of circle with center B has radius 6 cm and with centre C has radius 20 cm , if Angle BAC = 90°, frame an equation in x and solve for it.

$\huge\tt\red{❥}\tt\pink{A}\tt\blue{N}\tt\green{S}\tt\purple{W}\tt\orange{E}\tt\orange{R}\huge\tt\purple{᭄}$

Answer is Attached

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