Q2 of 112 Page 150

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (0, 0), (4, a), (6, 4)


Area (in sq. units): 17

Vertices of triangle A (0, 0), B (4, a) and C (6, 4)


Area of triangle = 17 sq. units


Area of triangle =


x1 = 0, x2 = 4 and x3 = 6


y1 = 0, y2 = a and y3 = 4


⇒


⇒


⇒


⇒ 17 × 2 = 16 – 6a


⇒ 34 = 16 – 6a


⇒ 34 + 6a = 16


⇒ 6a = 16 – 34


⇒ 6a = – 18


⇒


⇒ a = –3


Therefore, the required vertices are (4, –3)


More from this chapter

All 112 →
1

Find the area of the triangle formed by the points

(5, 2), (3, –5) and (–5, –1)

 1

Find the area of the triangle formed by the points

(–4, –5), (4, 5) and (–1, –6)

 2

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, a), (4, 5), (6, –1)


Area (in sq. units): 9

 2

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, –3), (3, a), (–1,5)


Area (in sq. units): 12