Question

a triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC which BC is divided by the point of contact the are of length 8 cm and 6 cm find the sides ab and ac

Answer 1

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Answer 2

Step-by-step explanation:

Given: Radius of Circle is $4cm$, and BC is divided into two lengths, $8cm$ and $6cm$

To find: Length of AB and AC.

Please find the figure attached.

For calculation of ab and ac,

Let $AF=AE=x$

i. ∴ Area of ΔABC, $BC=14$, $AB=8 + x,$ and $AC=6+x$

⇒ $Area (S)= \frac{a+b+c}{2}$

⇒ $\frac{14+8+x+6+x}{2}$

⇒ $\frac{28+2x}{2}$

⇒ $S=14+x$

$Area= \sqrt{S(S-a)(S-b)(S-c)}$

⇒ $Area= \sqrt{14+x(14+x-14)(1414+x-8x)(14+x-6x)}$

⇒ $Area= \sqrt{(14+x)(x)(6)(8)}$

⇒$\sqrt{(48x)(14+x)}$

ii. We know that Area Δ$ABC=$ Area Δ$COB+$ Area Δ$AOB+$ Area Δ$AOC$

⇒ $\frac{1}{2}$ × $14$ × $4 + \frac{1}{2}$ × $(8+x)$ × $4+\frac{1}{2}$ × $(6+x)$ × $4$

⇒ $28+16+2x+12+2x$

⇒ $56+4x$

∴$\sqrt{(48x)(14+x)} = 56+4x$

⇒ $(48x) (14+x)= (56+4x)^{2}$

⇒ $672x+48x^{2} = 3136+448x+16x^{2}$

⇒ $48x^{2}-16x^{2} +672x-448x=3136$

⇒ $32x^{2} +224x=3136$

⇒ $32(x^{2} +7x-98) = 0$

⇒ $x^{2} + 7x- 98=0$

⇒ $x^{2} +14x-7x-98=0$

⇒ $x(x+14)-7(x+14) = 0$

⇒ $(x-7)(x+14)=0$

So, $x=7$ and $x=-14$

Since, $x$ cannot be negative, $x=7$

iii. Calculation of AC,

⇒$x+6 = 7+6$

⇒$AC=13cm$

iv. Calculation of AB,

⇒ $x+8=7+8$

⇒ $15cm$

The value of AC is $13cm$ and AB is $15 cm$.