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### Area of a Triangle 22-25.doc

• 1. 1 (MMP) Problem No. 22 1 2 3 Find the area of the triangle determined by the points (1,1,1), (1,2,1), (1,1,2) P P P Solution
• 2. 2 1 2 3 1 2 2 3 (1,1,1), (1,2,1), (1,1,2) (1 1) (2 1) (1 1) (1 1) (1 2) (2 1) P P P i J k j i J k j k                 P P P P 1 2 2 3 2 3 4 , 0 1 0 0 1 1 ( 1) (1 0) ( 1) (0 0) ( 1) (0 0) Now i j k X i X j X k X i             P P P P 1 2 2 3 1 1 1 1 sq.unit 2 2 (Area is always square while Volume is always cubic) and X i A X      P P P P (MMP) Problem No. 23
• 3. 3 1 2 3 Find the area of the triangle determined by the points (0,0,0), (0,1,2), (2,2,0) P P P Solution 1 2 3 1 2 2 3 (0,0,0), (0,1,2), (2,2,0) (0 0) (1 0) (2 0) 2 (2 0) (2 1) (0 2) 2 2 P P P i J k j k i J k i j k                  P P P P 1 2 2 3 2 3 4 , 0 1 2 2 1 2 ( 1) ( 2 2) ( 1) (0 4) ( 1) (0 2) 4 4 2 Now i j k X i X j X k X i j k                 P P P P
• 4. 4 1 2 2 3 4 4 2 16 16 4 36 6 1 6 3 sq.unit 2 (Area is always square while Volume is always cubic) and X i j k A X             P P P P (MMP) Problem No. 24 1 2 3 Find the area of the triangle determined by the points (1,2,4), (1, 1,3), ( 1, 1,2) P P P    Solution 1 2 3 1 2 2 3 (1,2,4), (1, 1,3), ( 1, 1,2) (1 1) ( 1 2) (3 4) 3 ( 1 1) ( 1 1) (2 3) 2 P P P i J k j k i J k i k                         P P P P
• 5. 5 1 2 2 3 2 3 4 , 0 3 1 2 0 1 ( 1) (3 0) ( 1) (0 2) ( 1) (0 6) 3 2 6 Now i j k X i X j X k X i j k                  P P P P 1 2 2 3 3 2 6 9 4 36 49 7 1 7 7 =3.5 sq.unit 2 2 (Area is always square while Volume is always cubic) and X i j k A X            P P P P (MMP) Problem No. 25 1 2 3 Find the area of the triangle determined by the points (1,0,3), (0,0,6), (2,4,5) P P P Solution
• 6. 6 1 2 3 1 2 2 3 (1,0,3), (0,0,6), (2,4,5) (0 1) (0 0) (6 3) 3 (2 0) (4 0) (5 6) 2 4 P P P i J k i k i J k i j k                   P P P P 1 2 2 3 2 3 4 , 1 0 3 2 4 1 ( 1) (0 12) ( 1) (1 6) ( 1) ( 4 0) 12 5 4 Now i j k X i X j X k X i j k                  P P P P 1 2 2 3 12 5 4 144 25 16 185 13.60 1 13.60 6.80 sq.unit 2 (Area is always square while Volume is always cubic) and X i j k A X             P P P P
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