Question

find the area of the kite.
skip
? m²

Explanation:

Step1: Find length of first diagonal

The vertical diagonal is composed of two segments: 6 m and 6 m. So its length is \( 6 + 6 = 12 \) m.

Step2: Find length of second diagonal

The horizontal diagonal is composed of 4 m and 12 m? Wait, no, looking at the diagram, the horizontal segments: wait, actually, the horizontal diagonal parts: wait, no, the kite's diagonals: one diagonal is split into 4 m and 12 m? Wait, no, let's re - examine. Wait, the vertical diagonal: from top to bottom, the two parts are 6 m and 6 m, so total vertical diagonal \( d_1=6 + 6=12 \) m. The horizontal diagonal: the left part is 4 m and the right part is 12 m? Wait, no, the horizontal diagonal length is \( 4+12 = 16 \) m? Wait, no, maybe I misread. Wait, the kite can be divided into four triangles. Alternatively, the formula for the area of a kite is \( A=\frac{1}{2}\times d_1\times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

Wait, looking at the diagram again: the vertical diagonal has two segments of 6 m each, so \( d_1 = 6+6 = 12 \) m. The horizontal diagonal: the left segment is 4 m and the right segment is 12 m? Wait, no, that can't be. Wait, maybe the horizontal diagonal is \( 4 + 12=16 \) m? Wait, no, let's calculate the area by summing the areas of the four triangles.

First triangle (top - left): area \( A_1=\frac{1}{2}\times4\times6 \)
Second triangle (top - right): area \( A_2=\frac{1}{2}\times12\times6 \)
Third triangle (bottom - left): area \( A_3=\frac{1}{2}\times4\times6 \)
Fourth triangle (bottom - right): area \( A_4=\frac{1}{2}\times12\times6 \)

Wait, but maybe a better way: the kite is made up of two pairs of congruent triangles. The two upper triangles: one with base 12 m and height 6 m, and the other with base 4 m and height 6 m? No, wait, no. Wait, the vertical diagonal is 12 m (6 + 6), and the horizontal diagonal is \( 4+12 = 16 \) m? Wait, no, let's use the formula for the area of a kite. The area of a kite is the sum of the areas of the four triangles.

Alternatively, the two diagonals: let's see, the vertical diagonal length: from the top vertex to the bottom vertex, the distance is \( 6 + 6=12 \) m. The horizontal diagonal: from the left vertex to the right vertex, the distance is \( 4 + 12 = 16 \) m? Wait, no, maybe I made a mistake. Wait, looking at the diagram, the horizontal segments: the left part is 4 m, the right part is 12 m. So the total horizontal diagonal \( d_2=4 + 12 = 16 \) m? No, wait, no, the kite's diagonals are perpendicular. So the area of a kite is \( \frac{1}{2}\times d_1\times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the two diagonals.

Wait, let's calculate the lengths of the diagonals correctly. The vertical diagonal: the two parts are 6 m and 6 m, so \( d_1=6 + 6 = 12 \) m. The horizontal diagonal: the two parts are 4 m and 12 m? Wait, no, the horizontal diagonal is split into 4 m and 12 m? So \( d_2=4 + 12=16 \) m? Then the area would be \( \frac{1}{2}\times12\times16=96 \)? Wait, no, that doesn't match. Wait, maybe the horizontal diagonal is 12 + 4 = 16? Wait, no, let's calculate the area by adding the areas of the four triangles.

First triangle (top - left): base = 4 m, height = 6 m. Area \( A_1=\frac{1}{2}\times4\times6 = 12 \) \( m^2 \)
Second triangle (top - right): base = 12 m, height = 6 m. Area \( A_2=\frac{1}{2}\times12\times6=36 \) \( m^2 \)
Third triangle (bottom - left): base = 4 m, height = 6 m. Area \( A_3=\frac{1}{2}\times4\times6 = 12 \) \( m^2 \)
Fourth triangle (bottom - right): base = 12 m, height = 6 m. Area \( A_4=\frac{1}{2}\times12\times6 = 36 \) \( m^2 \)

Now, sum all these areas: \( A=A_1 + A_2+A_3 + A_4=12 + 36+12 + 36=96 \) \( m^2 \)

Wait, but another way: the kite can be considered as two congruent triangles on the top and two congruent triangles on the bottom. The top two triangles: their combined base is \( 4 + 12 = 16 \) m? No, wait, the top two triangles: one with base 4 and height 6, the other with base 12 and height 6. Similarly for the bottom. So total area is \( 2\times(\frac{1}{2}\times4\times6+\frac{1}{2}\times12\times6)=2\times(12 + 36)=2\times48 = 96 \) \( m^2 \)

Or, using the formula for the area of a kite: \( A=\frac{1}{2}\times d_1\times d_2 \), where \( d_1 \) is the length of one diagonal and \( d_2 \) is the length of the other diagonal. Here, \( d_1=6 + 6 = 12 \) m (vertical diagonal), \( d_2=4 + 12 = 16 \) m (horizontal diagonal). Then \( A=\frac{1}{2}\times12\times16 = 96 \) \( m^2 \)

Answer:

\( 96 \)