The sides of a triangle are 21, 20 and 13 cm. Find the area of the larger triangle into which the given triangle is divided by the perpendicular upon the longest side from the opposite vertex.

Options

A. `72 cm^(2)`
B. `96 cm^(2)`
C. `168 cm^(2)`
D. `144 cm^(2)`

Answer

To find the area of the larger triangle formed by dropping a perpendicular from the opposite vertex to the longest side of the triangle with sides 21 cm, 20 cm, and 13 cm, we can follow these steps:

### Step 1: Identify the sides of the triangle
The sides of the triangle are given as:
- \( a = 21 \) cm (longest side)
- \( b = 20 \) cm
- \( c = 13 \) cm

### Step 2: Calculate the semi-perimeter (s)
The semi-perimeter \( s \) of the triangle can be calculated using the formula:
\[ s = \frac{a + b + c}{2} \]
Substituting the values:
\[ s = \frac{21 + 20 + 13}{2} = \frac{54}{2} = 27 \text{ cm} \]

### Step 3: Calculate the area of the triangle using Heron's formula
The area \( A \) of the triangle can be calculated using Heron's formula:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{27 \times (27 - 21) \times (27 - 20) \times (27 - 13)} \]
Calculating each term:
- \( s - a = 27 - 21 = 6 \)
- \( s - b = 27 - 20 = 7 \)
- \( s - c = 27 - 13 = 14 \)

Now substituting these values:
\[ A = \sqrt{27 \times 6 \times 7 \times 14} \]

Calculating the product:
\[ 27 \times 6 = 162 \]
\[ 162 \times 7 = 1134 \]
\[ 1134 \times 14 = 15876 \]

Now taking the square root:
\[ A = \sqrt{15876} = 126 \text{ cm}^2 \]

### Step 4: Calculate the height (h) from the vertex opposite to the longest side
Using the area formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the longest side \( a = 21 \) cm. We can rearrange this to find the height:
\[ 126 = \frac{1}{2} \times 21 \times h \]
Multiplying both sides by 2:
\[ 252 = 21h \]
Now, dividing by 21:
\[ h = \frac{252}{21} = 12 \text{ cm} \]

### Step 5: Divide the triangle into two parts
The triangle is divided into two triangles by dropping a perpendicular from the opposite vertex to the longest side. We need to find the lengths of the segments on the base (21 cm) created by the perpendicular.

### Step 6: Use the Pythagorean theorem to find the segments
Let \( AM \) be the segment from vertex A to the foot of the perpendicular M on side BC, and \( MB \) be the remaining segment. We can use the Pythagorean theorem in triangle ACM:
\[ AC^2 = AM^2 + CM^2 \]
Where \( AC = 20 \) cm and \( CM = h = 12 \) cm:
\[ 20^2 = AM^2 + 12^2 \]
Calculating:
\[ 400 = AM^2 + 144 \]
\[ AM^2 = 400 - 144 = 256 \]
\[ AM = \sqrt{256} = 16 \text{ cm} \]

### Step 7: Find the length of MB
Since the total length of the base (BC) is 21 cm:
\[ MB = BC - AM = 21 - 16 = 5 \text{ cm} \]

### Step 8: Calculate the area of the larger triangle ACM
The larger triangle is ACM, with base \( AM = 16 \) cm and height \( h = 12 \) cm:
\[ \text{Area of } \triangle ACM = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 16 \times 12 \]
Calculating:
\[ = \frac{1}{2} \times 192 = 96 \text{ cm}^2 \]

### Final Answer
The area of the larger triangle ACM is \( 96 \text{ cm}^2 \).

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