Question

this item has two parts: part a and part b. use the figure to answer part a and part b.
(figure of a right triangle with legs 2√3 and 2√3, hypotenuse 2√6)
2
part a: (a) the area of triangle? is (1/2)(2√3)(2√3). what is the value of this expression?
a 6
b 2√3
c 18
d 12
(b) part b: the perimeter of the triangle? is given by the expression (2√3) + 2√3 + 2√6. what is the simplest form of the expression?
a 6√12
b 6√6
c 4√3 + 2√6
d 4√9 + 2√6

Explanation:

Step1: Solve Part (a)

The formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. Here, base and height are both $2\sqrt{3}$. So we substitute into the formula:
$\frac{1}{2} \times (2\sqrt{3}) \times (2\sqrt{3})$
First, multiply the coefficients and the square roots separately. $(2\sqrt{3}) \times (2\sqrt{3}) = 2\times2\times\sqrt{3}\times\sqrt{3} = 4\times3 = 12$ (since $\sqrt{a}\times\sqrt{a}=a$). Then multiply by $\frac{1}{2}$: $\frac{1}{2} \times 12 = 6$.

Step2: Solve Part (b)

The perimeter of a triangle is the sum of its three sides. The sides are $2\sqrt{3}$, $2\sqrt{3}$, and $2\sqrt{6}$. So we add them together:
$(2\sqrt{3}) + (2\sqrt{3}) + (2\sqrt{6})$
Combine like terms: $2\sqrt{3} + 2\sqrt{3} = 4\sqrt{3}$. So the perimeter is $4\sqrt{3} + 2\sqrt{6}$.

Answer:

(a) A. 6
(b) C. $4\sqrt{3} + 2\sqrt{6}$