TRIANGLE WORKSHEET SOLUTION 2

(2) Vertices of the triangle taken in order and their areas are given below. In each of the following find the value of a.

(i) (0, 0), (4, a) and (6, 4) and its area is 17 sq.units

Solution :

Area of the triangle  =  17 square units

(1/2)[(0 + 16 + 0) – (0 + 6a + 0)]  =  17

(1/2)(16 – 6 a)  =  17

8 - 3a  =  17

-3a  =  9

a  =  -3

(ii)  (a, a) , (4 , 5) and ( 6 , -1) and its area is 9 sq.units

Solution :

Area of the triangle  =  9 square units

(1/2)[(5a - 4 + 6a) – (4a + 30 - a)]

(1/2)[(11a – 4) – ( 3a+30)]  =  9

(1/2)[11a – 4 – 3a - 30]  =  9

(1/2)[8a – 34]  =  9

4a – 17  =  9

4a  =  26

a  =  26/4

a  =  13/2

(iii) (a, -3),(3, a) and (-1, 5) and its area is 12 sq.units

Solution :

Area of the triangle = 12 square units

(1/2)[(a2 + 15 + 3) – (-9 - a + 5a)]  =  12

(1/2)[(a2 + 18 ) – (-9 + 4a)]  =  12

[a2 + 18 + 9 - 4a]  =  12 x 2

a2 - 4a + 27  =  24

a2 - 4a + 27 -  24  =  0

a2 - 4a + 3  =  0

(a - 1)(a - 3)  =  0

a  =  1 , 3

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