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.
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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]
where, S is a semi-perimeter of Tri.. ABC.
Now, we find the Area of Triangle ABC by Heron's Formula ---------
Both Equations (eq. 1 and eq.2) are equal because area of a same triangle ABC.
So,
▪ Eq. [1.] = Eq.[2.]
▪▪ Therefore,
¤¤ AB = [8+y] = [8+7] = 15 cm
and
¤¤ AC = [y+6] = [7+6] = 13 cm
☛ ⛧⛧ Ⓜr. Thakur ⛧⛧
▪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]
where, S is a semi-perimeter of Tri.. ABC.
Now, we find the Area of Triangle ABC by Heron's Formula ---------
Both Equations (eq. 1 and eq.2) are equal because area of a same triangle ABC.
So,
▪ Eq. [1.] = Eq.[2.]
▪▪ Therefore,
¤¤ AB = [8+y] = [8+7] = 15 cm
and
¤¤ AC = [y+6] = [7+6] = 13 cm
☛ ⛧⛧ Ⓜr. Thakur ⛧⛧
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