The problem has three parts: (a) Express the area $L$ of the triangle in Figure I in terms of $y$. (b) Given that the area of the triangle is $48 \text{ cm}^2$, find the value of $y$. (c) Figure II shows a polygon formed by combining six identical triangles from Figure I. Calculate the area of the polygon and state its name.

1. Problem Description

The problem has three parts:

(a) Express the area $L$ of the triangle in Figure I in terms of $y$.

(b) Given that the area of the triangle is $48 \text{ cm}^2$, find the value of $y$.

(c) Figure II shows a polygon formed by combining six identical triangles from Figure I. Calculate the area of the polygon and state its name.

2. Solution Steps

(a) The area $L$ of a triangle is given by the formula:

$L = \frac{1}{2} \times \text{base} \times \text{height}$

From Figure I, the base of the triangle is $(y+4)$ cm and the height is $y$ cm. Therefore, the area $L$ is given by:

$L = \frac{1}{2} \times (y+4) \times y$

$L = \frac{1}{2}y(y+4)$

$L = \frac{1}{2}(y^2+4y)$

$L = \frac{y^2}{2}+2y$

(b) Given that the area of the triangle is $48 \text{ cm}^2$, we have:

$\frac{1}{2}y(y+4) = 48$

$y(y+4) = 96$

$y^2 + 4y = 96$

$y^2 + 4y - 96 = 0$

We can factor the quadratic equation as:

$(y-8)(y+12) = 0$

So, $y = 8$ or $y = -12$.

Since $y$ represents the height of the triangle, it must be a positive value. Therefore, $y = 8$.

(c) The polygon in Figure II is formed by six identical triangles from Figure I. The area of each triangle is given as $48 \text{ cm}^2$.

Therefore, the area of the polygon is:

Area = $6 \times \text{Area of one triangle}$

Area = $6 \times 48$

Area = $288 \text{ cm}^2$.

The polygon has six sides, thus it is a hexagon.

3. Final Answer

(a) $L = \frac{y^2}{2} + 2y$

(b) $y = 8$

(c) Area of the polygon = $288 \text{ cm}^2$. The name of the polygon is a hexagon.