## TRIANGLE WORKSHEET SOLUTION 2

Triangle Worksheet Solution 2 :

In this section, you will have solutions for the problems on finding area of a triangle when three of its vertices are given.

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

(1/2)(16 – 6 a)  =  17

(1/2)  2 (8 - 3a)  =  17

8 – 3 a  =  17  ==>  -3a  =  17 – 8

-3a  =  9

a  =  9/(-3)

a  =  -3

Hence, the required vertices is (4, -3)

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

Solution :

Area of the triangle  =  9 square units 9  =  (1/2) [(5a - 4 + 6a) – (4a + 30 - a)]

(1/2)[ (11a – 4) – ( 3a+30) ]  =  9

(1/2) [11a – 4 – 3a - 30 ]  =  9

(1/2) [8 a – 34 ]  =  9

8 a – 34  =  18  ==> 8 a  =  18 + 34

8 a  =  52  ==> a  =  58/8  ==>  a  =  13/2

Hence, the required vertex is (13/2 , 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 12  =  (1/2)[ (a² + 15 + 3) – (-9 - a + 5a) ]

(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

## Related topics After having gone through the stuff, we hope that the students would have understood how to find area of a triangle when three of its vertices are given.

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