Question

Find the area of the triangle formed by the following points
(i) (2, 0), (1, 2), (1, 6)
(ii) (3, 1), (5, 0), (1, 2)
(iii) (−1.5, 3), (6, 2), (−3, 4)
What do you observe?

Answer 1

if vertices of triangles are $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ then, area of triangle =$\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

(i) (2,0), (1,2) , (1,6)
area of triangle = 1/2 |2(2 - 6) + 1(6 - 0) + 1(0 - 2)|
= 1/2|2 × -4 + 1 × 6 + 1 × -2 |
= 1/2|-8 + 6 - 2|
= 1/2 |-4|
= 1/2 × 4
= 2 sq unit

(ii) (3,1) , (5,0) , (1,2)
area of triangle = 1/2|3(0 - 2) + 5(2 - 1) + 1(2 - 1)|
= 1/2| 3 × -2 + 5 × 1 + 1 × 1 |
= 1/2 | -6 + 5 + 1 |
= 1/2 | -6 + 6|
= 1/2 × 0 = 0
e.g., area of triangle = 0
means givens points are collinear.

(iii) (-1.5, 3) , (6,2) and (-3,4)
area of triangle= 1/2|-1.5(2 - 4) + 6(4 - 3) -3(3 - 2)|
= 1/2| -1.5 × -2 + 6 × 1 - 3 × 1 |
= 1/2 | 3 + 6 - 3 |
= 1/2 | 6 |
= 1/2 × 6 = 3 sq unit

Answer 2

Answer:

Step-by-step explanation:

Let the lengths and breadth of rectangle be l & b respectively.

Now l = b+5, perimeter = 2(l+b) = 58 cm

2(b +b +5) = 58

2b + 5 = 29

2b = 24

b = 12 and length = b +5 = 17