Math, asked by Anonymous, 11 months ago

a triangle abc is drawn to circumscribe a circle radius 4 cm such that the segment BC and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively. Find the AB and AC.

Answers

Answered by shrihari20
29
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shrihari20: amyone
Answered by MrThakur14Dec2002
23
Solution..........


▪BC is divided by the point of contact D as BD & CD with the length 8cm and 6cm respectively.

▪OD, OE and OF are the radii of the same circle with the length of 4cm each. So,
OD = OE = OF = 4cm.

▪ We Know that , the lengths of tangents from the external point to a circle are equal in length. So,
CD = CE = 6cm

BD = BF = 8cm

Let, the lenght of AE be y cm.
so, AE = AF = y.


==========
Step ---- I
==========


Area of Triangle BOC = 1/2 × base × height
= 1/2 × BC × OD
= 1/2 × 14 × 4
= 28


Thus ,
Area of Tri.. AOC = 1/2 × AC × OE
= 1/2 × [y+6] × 4
= 2y+12.

Similarly
Area Of Tri.. AOB = 1/2 × AB × OF
= 1/2 × [8+y] × 4
= 16+2y.

Now,
¤ Area Of Triangle ABC =

= ar [BOC ] + ar [AOC ] + ar [AOB ]
= 28 + 2y+12 + 16 + 2y
= 56 + 2y ............................. Eq. {1.}



============
Step ----- II
============



a = AB = [8+y]
b = BC = 14
c = AC = [y+6]


 \:  \:  \:  \: s \:  \:  \:  \:  =   \:  \:  \:  \: \frac{a + b + c}{2}

where, S is a semi-perimeter of Tri.. ABC.



s =  \frac{8 + y + 14 + y + 6}{2}  \\  \\ s =  \frac{2y + 28}{2}  \\  \\  s = \frac{2(y + 14)}{2}  \\  \\  s = y + 14 \\  \\

Now, we find the Area of Triangle ABC by Heron's Formula ---------

 =  \sqrt{s(s - a)(s - b)(s - c)}


 =  \sqrt{(y + 14)(y + 14 - 8 - y)(y + 14 - 14)(y + 14 - y - 6)}  \\  \\  =  \sqrt{(y + 14) \times 6 \times y \times 8 }  \\  \\

 \\   = \sqrt{48y(y + 14)}  \\  \\  =  \sqrt{16 \times 3y(y + 14)}  \\  \\


  = 4 \sqrt{3y(y + 14)} ...................eq.(2.)


Both Equations (eq. 1 and eq.2) are equal because area of a same triangle ABC.

So,

▪ Eq. [1.] = Eq.[2.]


56 + 4y = 4 \sqrt{3y(y + 14)}

4(y + 14) = 4 \sqrt{3y(y + 14)}
 \\ squaring \: both \: sides.....


 {(y + 14)}^{2}  = 3y(y + 14)


 \\ 3y =  \frac{ {(y + 14)}^{2} }{(y + 14)}  \\  \\ 3y = y + 14 \\  \\ 3y - y = 14 \\  \\ 2y = 14 \\  \\ y = 7.....

▪▪ Therefore,

¤¤ AB = [8+y] = [8+7] = 15 cm

and

¤¤ AC = [y+6] = [7+6] = 13 cm




 \\ hope \: this \: will \: \: help \: you \: .................. \\  \\  \\  \\  \\  \\  \\  \\ \\ be \: brainly...............






☛ ⛧⛧ Ⓜr. Thakur ⛧⛧
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