রেখা বিভাজন ও অনুপাত

Find the perimeter of the triangle formed by the points (3,5),(4,8)(3, 5), (4, 8) and (5,6)(5, 6).

হানি নাটস

To find the perimeter of the triangle formed by the points A(3,5)A(3,5), B(4,8)B(4,8) and C(5,6)C(5,6), we must find the distance between these three points that can be determined as follows:

AB=(43)2+(85)2=12+32=1+9=10AB=\sqrt { { (4-3) }^{ 2 }+{ (8-5) }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 1+9 } =\sqrt { 10 }

BC=(54)2+(68)2=12+22=1+4=5BC=\sqrt { { (5-4) }^{ 2 }+{ (6-8) }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 1+4 } =\sqrt { 5 }

AC=(53)2+(65)2=22+12=4+1=5AC=\sqrt { { (5-3) }^{ 2 }+{ (6-5) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 4+1 } =\sqrt { 5 }

Now the perimeter is AB+BC+ACAB+BC+AC that is

AB+BC+AC=10+5+5=10+25=5(2+2)AB+BC+AC=\sqrt { 10 } +\sqrt { 5 } +\sqrt { 5 } =\sqrt { 10 } +2\sqrt { 5 } =\sqrt { 5 } (2+\sqrt { 2 } )

Hence the perimeter is 5(2+2)\sqrt { 5 } (2+\sqrt { 2 } ).

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