In this page triangle worksheet solution2 we are going to see solution for each problems of the topic area of triangle worksheet.

(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) x 2 (8 - 3a) = 17

8 – 3 a = 17

-3a = 17 – 8

-3a = 9

a = 9/(-3)

a = -3

Therefore 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

Therefore 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

triangle worksheet solution2 triangle worksheet solution2

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