Question

what is the total area of this shape
type your answer in the boxes

Explanation:

Step1: Identify the shapes

The figure consists of a rectangle and a triangle. The rectangle has length \( 6 \, \text{cm} \) and width \( 5 \, \text{cm} \). The triangle has a base equal to the width of the rectangle (\( 5 \, \text{cm} \)) and height \( 14 - 6 = 8 \, \text{cm} \).

Step2: Calculate area of rectangle

The area of a rectangle is \( \text{length} \times \text{width} \). So, \( A_{\text{rectangle}} = 6 \times 5 = 30 \, \text{cm}^2 \).

Step3: Calculate area of triangle

The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). Here, base \( = 5 \, \text{cm} \), height \( = 8 \, \text{cm} \). So, \( A_{\text{triangle}} = \frac{1}{2} \times 5 \times 8 = 20 \, \text{cm}^2 \).

Step4: Total area

Add the areas of the rectangle and triangle: \( A_{\text{total}} = 30 + 20 = 50 \, \text{cm}^2 \)? Wait, no, wait. Wait, maybe I misread the dimensions. Wait, the vertical side: the rectangle's height? Wait, maybe the rectangle is \( 5 \, \text{cm} \) (width) and \( 6 \, \text{cm} \) (height), and the triangle's height is \( 14 - 6 = 8 \, \text{cm} \), base \( 5 \, \text{cm} \). Wait, no, maybe the horizontal side: the rectangle is \( 6 \, \text{cm} \) (length) and \( 5 \, \text{cm} \) (width), and the triangle is attached to the bottom. Wait, maybe I made a mistake. Wait, let's re-examine. The figure: the top part is a rectangle with length \( 6 \, \text{cm} \) (vertical) and width \( 5 \, \text{cm} \) (horizontal). Then the bottom part is a triangle with base \( 5 \, \text{cm} \) and height \( 14 - 6 = 8 \, \text{cm} \) (vertical). Wait, no, maybe the horizontal length: the total vertical length is \( 14 \, \text{cm} \), the rectangle is \( 6 \, \text{cm} \) tall, so the triangle is \( 14 - 6 = 8 \, \text{cm} \) tall. The base of the triangle is equal to the width of the rectangle, which is \( 5 \, \text{cm} \). So rectangle area: \( 6 \times 5 = 30 \). Triangle area: \( 0.5 \times 5 \times 8 = 20 \). Total area: \( 30 + 20 = 50 \)? Wait, that seems low. Wait, maybe the rectangle is \( 5 \, \text{cm} \) (width) and \( 6 \, \text{cm} \) (height), and the triangle is \( 5 \, \text{cm} \) base and \( 8 \, \text{cm} \) height. Wait, but maybe the horizontal side is \( 5 \, \text{cm} \), vertical: rectangle is \( 6 \, \text{cm} \), triangle is \( 8 \, \text{cm} \). Wait, maybe I messed up the dimensions. Wait, let's check again. The figure: the top rectangle has width \( 5 \, \text{cm} \) (horizontal) and height \( 6 \, \text{cm} \) (vertical). The triangle is below the rectangle, with base \( 5 \, \text{cm} \) (same as rectangle's width) and height \( 14 - 6 = 8 \, \text{cm} \) (vertical). So area of rectangle: \( 5 \times 6 = 30 \). Area of triangle: \( 0.5 \times 5 \times 8 = 20 \). Total: \( 30 + 20 = 50 \, \text{cm}^2 \). Wait, but maybe the rectangle is \( 6 \, \text{cm} \) (length) and \( 5 \, \text{cm} \) (width), and the triangle is \( 5 \, \text{cm} \) base and \( 8 \, \text{cm} \) height. Yes, that makes sense. So total area is \( 30 + 20 = 50 \, \text{cm}^2 \)? Wait, no, wait, maybe the vertical side is \( 14 \, \text{cm} \), the rectangle is \( 6 \, \text{cm} \) tall, so the triangle is \( 14 - 6 = 8 \, \text{cm} \) tall. The base of the triangle is equal to the width of the rectangle, which is \( 5 \, \text{cm} \). So rectangle area: \( 5 \times 6 = 30 \). Triangle area: \( 0.5 \times 5 \times 8 = 20 \). Total: \( 30 + 20 = 50 \, \text{cm}^2 \).

Answer:

\( 50 \, \text{cm}^2 \)