Find the perimeter of the triangle whose vertices are the following specified points in the plane. (5,6), (4,1) and (-4,2)

23 units

Step-by-step explanation:

Let the vertices of the triangle be A(5,6), B(4,1) and C(-4,2)

Perimeter of the triangle = AB + AC + BC

AB = √(5-4)²+(6-1)²

AB = √1²+5²

AB = √1+25

AB = √26

For BC where B(4,1) and C(-4,2)

BC = √(4-+4)²+(1-2)²

BC = √8²+1²

BC = √64+1

BC = √65

For AC where A(5,6) and C(-4,2)

AC = √(5+4)²+(6-2)²

AC = √9²+4²

AC = √81+16

AC = √97

Perimeter = √97 + √26 + √65

Perimeter = 9.85+5.09+8.06

Perimeter = 23

Hence the perimeter of the triangle is approx 23 units

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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