ABC is drawn2 circumscribe a circle of radius 4cm such dat the segments BD and DC into which BC is divided D r of lengths 8cm and 6cm,Find AB nd AC.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the notation for the following points: tangency point M lies on side AB and tangency point N lies on side AC.

Since the radius of inscribed circle is orthogonal to each side and it falls in tangency points D,M,N, hence, the segments `BM=BD=8 cm`  and `CN=CD=6 cm` . Thus, the segments `AM=AN` .

`AM = AB-BM = AB-8`

`AN = AC-CN = AC-6`

You need to set equations above equal such that: `AB-8 = AC-6`

You need to use Pythagorean theorem in right triangles AMO and ANO such that:

`AO^2 = AM^2 + MO^2`

`AO^2 = AM^2 + 4^2 =gt AO^2 = AM^2 + 16`

`AO^2 = AN^2 + NO^2 =gt AO^2 = AN^2 + 16`

Notice that equating the relations above yields `AM=AN=a` .

You need to remember the relations between the radius of an incircle and the sides of triangle circumscribing the circle such that: ` r = sqrt(((p-AB)(p-BC)(p-AC))/p)`

`p = (AB+AC+BC)/2`

Substituting all the problem provides in relation above yields:

`4 = sqrt(((p-a-8)(p-14)(p-a-6))/p)`

`p = (a+8+a+6+14)/2 =gt p = (2a+28)/2 =gt p = a+14`

`4 = sqrt(((a+14-a-8)(a+14-14)(a+14-a-6))/(a+14))`

Reducing like terms yields:

`4 = sqrt(((6)(a)(8))/(a+14))`

You need to remove the square root, thus you need to raise to square both sides such that:

`16 = 48a/(a+14) =gt 16a + 16*14 = 48a`

`32a = 224=gt a = 224/32`

`a=7cm `

Hence, the side `AB = 8+7=15 cm ` and the side `AC = 6+7 = 13 cm.`

Hence, evaluating the lengths of sides yields `AB=15 cm ` and `AC = 13 cm.`

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lee88 | Student, Grade 10 | eNotes Newbie

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Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In ABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = x (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area ofΔOCA + Area of ΔOAB

Either x+14 = 0 or x − 7 =0

Therefore, x = −14and 7

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

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