Find the area of the
ABC
whose perimeter is
96cm
and
90o  = B
and
30o  = ACB .
(Assume
3 1.732 =
)

We are tasked with finding the area of the triangle \( \Delta ABC \) where:

- The perimeter is 96 cm.
- \( \angle B = 90^\circ \).
- \( \angle ACB = 30^\circ \).
- The approximation \( \sqrt{3} \approx 1.732 \).

### Step-by-Step Solution:

1. **Label the sides**:  
   Since \( \angle B = 90^\circ \), triangle \( \Delta ABC \) is a right-angled triangle. Using conventional labeling, let's denote:
   - Side \( AC = c \) (the hypotenuse).
   - Side \( AB = a \) and \( BC = b \) as the two other sides of the triangle.

2. **Apply angle relationships**:  
   Since \( \angle ACB = 30^\circ \), we can use trigonometric ratios in the right triangle to express the sides.

   - Using \( \sin(30^\circ) = \frac{1}{2} \), we know:
     \[
     \sin(30^\circ) = \frac{AB}{AC} = \frac{a}{c} \quad \Rightarrow \quad a = \frac{c}{2}.
     \]
   
   - Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we also know:
     \[
     \cos(30^\circ) = \frac{BC}{AC} = \frac{b}{c} \quad \Rightarrow \quad b = \frac{c \sqrt{3}}{2}.
     \]

3. **Perimeter condition**:  
   The perimeter is given as 96 cm. Thus, we can write:
   \[
   a + b + c = 96.
   \]
   Substituting \( a = \frac{c}{2} \) and \( b = \frac{c \sqrt{3}}{2} \) into the equation, we get:
   \[
   \frac{c}{2} + \frac{c \sqrt{3}}{2} + c = 96.
   \]
   Simplifying the left-hand side:
   \[
   \frac{c}{2} + \frac{1.732c}{2} + c = 96,
   \]
   \[
   \frac{c (1 + 1.732)}{2} + c = 96,
   \]
   \[
   \frac{c \times 2.732}{2} + c = 96.
   \]
   Multiply both sides by 2:
   \[
   c(2.732 + 2) = 192,
   \]
   \[
   c \times 4.732 = 192.
   \]
   Solving for \( c \):
   \[
   c = \frac{192}{4.732} \approx 40.58 \, \text{cm}.
   \]

4. **Calculate side lengths**:
   - \( a = \frac{c}{2} = \frac{40.58}{2} = 20.29 \, \text{cm} \).
   - \( b = \frac{c \sqrt{3}}{2} = \frac{40.58 \times 1.732}{2} \approx 35.12 \, \text{cm} \).

5. **Area of the triangle**:  
   The area \( A \) of a right triangle is given by:
   \[
   A = \frac{1}{2} \times a \times b.
   \]
   Substituting the values of \( a \) and \( b \):
   \[
   A = \frac{1}{2} \times 20.29 \times 35.12 \approx \frac{1}{2} \times 713.54 = 356.77 \, \text{cm}^2.
   \]

Thus, the area of \( \Delta ABC \) is approximately \( \boxed{356.77 \, \text{cm}^2} \).

---

Do you want more details or have any questions?  
Here are 5 related questions to expand on this topic:

1. How do we use the sine and cosine functions in right-angled triangles?
2. What is the relationship between angles and side lengths in a 30-60-90 triangle?
3. How would the solution change if the perimeter were different?
4. Can the area of the triangle be found using Heron’s formula instead?
5. What are other methods to find the area of a triangle when given angles and sides?

**Tip:** In any right triangle, knowing two angles allows you to determine all side lengths using trigonometric ratios like sine and cosine.

Find the area of the triangle ABC whose perimeter is 96 cm, where angle B is 90° and angle ACB is 30° (Assume √3 ≈ 1.732).

Learn how to calculate the area of a right triangle with a 96 cm perimeter, a 90° angle at B, and a 30° angle at ACB. This problem uses trigonometric ratios and perimeter relationships to solve for side lengths and area. Suitable for students in Grades 9-12.

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