Question

$\color{red}\underline{\underline{\huge\bf Question :-}}$
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The radius of the Incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle.
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Topic : Circles
Class : 10th
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Thankieww <3 ​

Answer 1

$\huge\mathfrak\pink{Answer}$

BF=BD=6cm

[length of tangent from external point are equal]

CE=CD=8cm

[length of tangent from external point are equal]

Let AF=AE=x

[length of tangent from external point are equal ]

Now,

AreaofΔABC=AreaofΔAOB+AreaofΔBOC+AreaofΔCOA

= 1/2×4(6+x)+ 21×4(14)+ 21×4(8+x)

= 1/2×4(28+2x)

=4(14+x)

Also, area of ΔABC by Heron's formula

S= 14+6+x+8+x/2=14+x

Area of ΔABC= root(14+x)(8)(6)x

So, 4(14+x)= root48x(14+x)

16(14+x)^2 =48x(14+x)

14+x=3x

2x=14⇒x=7

So, AB=6+x=6+7=13cm

AC=8+x=8+7=15cm

Sum = 13+15=28cm

Answer 2

Step-by-step explanation:

Given:

A circle with centre O

Let triangle AB circumscribe the circle

with OD = radius = 4 cm

CD = 6 cm

BD = 8 cm

To find:

AB AC and OB

Let AC, AB intersect circle at E & F respectively

Solution

lengths of tangents drawn from external point are equal

Hence,

CE = CD = 6 cm

BF = BD = 8 cm

AE = AF = x

Now, our three sides are

CB=CD+DB=6+8=14 cm

AC=AE+EC=(x+6)cm

AB=AF+FC=(x+8)cm

Now, we find Area of ∆ ABC using Herons formula Area of triangle

$\pmb{ \sqrt {(s(s - a)(s - b)(s - c)}}$

Here,

AC=a=(x+6)cm

AB=b=(x+8)cm

BC=c=6+8=14 cm

$\bf \: s \: = \frac{a + b + c}{2} \\ \bf \: = \frac{(x + 6) + (x + 8) + 14}{2} \\ \bf \: = \frac{2x + 28}{2} \\ \bf \: = \frac{2(x + 14)}{2} = x + 14$

$\pmb{Area \: \Delta \: ABC = \sqrt{(s(s - a) (s - b) (s - c)}}$

Putting values

$\bf=\sqrt {(x+14)(x+14)-(x+6)(x+14)-(x+8) \: (x+14-14)} \\ \bf \: = \sqrt{ (x+14)(x+14-x-6)(x+14-x-8)(x+14-14)}$

$\bf= \sqrt{ (x+14)(8)(6)(x)} \\ \bf \: = \sqrt{ (x+14)(48)(x)} \\ \bf \: = \sqrt{ (x+14)(48x)} \\ \bf \: = \sqrt {x(48x)+14(48x) } \\ \bf= \sqrt {48x^ 2 +672x}$

$\pmb{ \: Area \: \Delta \: A B C = \sqrt{(48x ^ 2 + 672x)}}$

Join OE & OF

Here,

OD = OF = OE = radius =4 cm

Also,

we know that tangent is perpendicular to the radius

So,

$\pmb{ \: OD \perp BC , OF \perp AB \: and \: OE \perp AC}$

Also,

$\bf \: Area \: of \Delta \: A B C = Are \: triangle \: AOC+Area \Delta \: A O B + \: Area \: \Delta B O C$

We find Area ∆ AOC , Area ∆ AOB & Area ∆ BOC

Area ∆ AOC

$\bf= \frac{1}{2} \times Base \times Height \\ \bf = \frac{1}{2} \times OE \times AC \\ \bf = \frac{1}{2} \times 4 \times (x+6) \\ \bf =2(x+6)\\ \bf =2x+12$

Area ∆ AOB

$\bf= \frac{1}{2} \times Base \times Height \\ \bf = \frac{1}{2} \times OF \times AB \\ \bf= \frac{1}{2} \times 4 \times (x+8) \\ \bf=2(x+8) \\ \bf=2x+16$

Area ∆BOC

$\bf= \frac{1}{2} \times Base \times Height \\ \bf = \frac{1}{2} \times OD \times BC \\ \bf = \frac{1}{2} \times 4 \times 14 \\ \bf =2 \times 14 \\ \bf=28$

Now,

Area ∆ AOC , Area ∆ AOB & Area ∆ BOC

Putting values

$\bf \sqrt{(48x ^ 2 + 672x) = (2x + 12) + (2x + 16) + 28)} \: \\ \bf \sqrt{(48x ^ 2 + 672x) = 4x + 56 }$

Squaring both sides

$\bf \sqrt{(48x ^ 2 + 672x)^ 2 = (4x + 56) ^ 2} \\ \bf48x ^ 2 + 672x = (4x) ^ 2 + 56 ^ 2 + 2 \times 4x \times 56 \\\bf 48x ^ 2 + 672x = 16x ^ 2 + 3136 + 448x \\\bf 48x ^ 2 - 16x ^ 2 + 672x - 448x - 3136 = 0 \\\bf 32x ^ 2 + 224x - 3136 = 0 \\ \bf32(x ^ 2 + 7x - 98) = 0 \\ \bf x ^ 2 + 7x - 98 = 0 \\\bf x ^ 2 + 14x - 7x - 98 = 0$

$\bf \: x(x + 14) - 7(x + 14) = 0 \\ \bf (x - 7)(x + 14) = 0 \\ \bf \: So, x = 78x = - 14$

So,

x = 7 x = -14

Since x cannot be negative,

So,x=7

Now,

$\bf \: AB=x+8=7+8=15 cm \\\bf AC=x+6=7+6=13 cm$

$\large\bf{\red{\mathfrak{Hope\:It\:Helps}}}$