Answer:
The area of the composite figure = 32 + 6 + 6 = 44 units²
Explanation:
* To find the area of this composite figure lets divide it into 3
familiar figures trapezium and 2 triangles
* The vertices of the trapezium are:
(3 , 2) , (3 , -1) , (-5 , -3) , (-5 , 2)
∵ The vertices of the parallel bases are:
Small base ⇒ (3 , 2) and (3 , -1) its length = 2 - -1 = 3 units ⇒
vertical distance (same x-coordinates)
Large base ⇒ (-5 , -3) and (-5 , 2) its length = 2 - -3 = 5 units ⇒
vertical distance (same x-coordinates)
The vertices of the its height are:
(3 , 2) and (-5 , 2) its length = 3 - -5 = 8 ⇒ horizontal distance
(same y-coordinates)
∵ Area trapezium = 1/2 (base1 + base2) × height
∴ Its area = 1/2 × (3 + 5) × 8 = 32 units² ⇒ (1)
* The vertices of the 2 triangles are:
First Δ: (3 , 2) , (-1 , 2) , (1 , 5)
The length of its base = 3 - -1 = 4 ⇒ horizontal distance
(same y-coordinates)
The length of its height = 3 (at x = 1 the ⊥ distance from the
vertex to the base = 5 - 2 = 3)
∵ The area of the Δ = 1/2 (base)(height)
∴ Its area = 1/2 × 4 × 3 = 6 units² ⇒ (2)
Second Δ: (-1 , 2) , (-5 , 2) , (-3 , 5)
The length of its base = -1 - -5 = 4 ⇒ horizontal distance
(same y-coordinates)
The length of its height = 3 (at x = -3 the ⊥ distance from the
vertex to the base = 5 - 2 = 3)
∵ The area of the Δ = 1/2 (base)(height)
∴ Its area = 1/2 × 4 × 3 = 6 units² ⇒ (3)
Add (1) , (2) and (3)
∴ The area = 32 + 6 + 6 = 44 units²