In this page triangle worksheet solution5 we are going to see solution for each problems of the topic area of triangle worksheet.

(5) If the three points (h,0) (a,b) and (0,k) lie on a straight line, then using the area of the triangle formula show that (a/h) + (b/k) = 1,where h,k≠ 0

**Solution:**

Let A(h,0) B (a,b) and C (0,k) are the three points

Since the three points A (h,0) B (a,b) and C (0,k) lie on a straight line we can say that the three points are collinear

So ,Area of triangle ABC = 0

(1/2) [ (h b + a k + 0) – (0 + 0 + kh) ] = 0

[ h b + a k – k h ] = 0 x 2

h b + a k - k h = 0

h b + a k = k h

Divided by (k h) on both sides we get

(h b)/(k h) + (a k)/(k h) = (k h)/(k h)

(b/k) + (a/h) = 1

(a/h) + (b/k) = 1

(7) Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are (0,-1) (2,1) and (0,3). Find the ratio of this area to the area of the given triangle.

**Solution:**

Let A (0,-1) B (2,1) and C (0,3) are the vertices of the triangle . D,E and F are the midpoints of the side AB,BC and CA respectively.

Midpoint of AB = (x1+x2)/2 , (y1+y2)/2

= (0 + 2)/2 , (-1 + 1)/2

= 2/2 , 0/2

= D (1,0)

Midpoint of BC = (x1+x2)/2 , (y1+y2)/2

= (2 + 0)/2 , (1 + 3)/2

= 2/2 , 4/2

= E (1,2)

Midpoint of CA = (x1+x2)/2 , (y1+y2)/2

= (0 + 0)/2 , (3 + (-1))/2

= 0/2 , 2/2

= F (0,1)

= (1/2) [(0 + 6 + 0) – (-2 + 0 + 0)]

= (1/2) [6+2]

= (1/2) [8]

= 4 square units

= (1/2) [(2 + 1 + 0) – (0 + 0 + 1)]

= (1/2) [3-1]

= (1/2) [2]

= 1 square units

Area of triangle ABC: Area of triangle DEF

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triangle worksheet solution5 triangle worksheet solution5

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