Find the area of the kite QRST. HELP PLEASE!!!
A. 90m^2
B. 108m^2
C. 216 m^2
D. 135 m^2
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ANSWER
D. 135 m^2
EXPLANATION
The area of a kite is half the product of the diagonals.
The first diagonal is
9m+6m=15m
Note that the vertical diagonal is the axis of symmetry of the kite.
This diagonal bisect the kite
Therefore the second diagonal is
2(9m)=18m
The area of the kite
[tex] = \frac{1}{2} \times 18 \times 15[/tex]
[tex] = 9 \times 15[/tex]
[tex] = 135 {m}^{2} [/tex]
Answer:
[tex]A_{T}=135 m^{2[/tex]
Step-by-step explanation:
Hello
To solve this problem we can divide the kite into simpler geometric figures to find its area, we have four triangles
the area of a triangle =(b*h)/2
b is the base and his the heigth
Step 1
triangle 1 (LEFT UP)
[tex]b=9\ m\\\ h= 9\ m\\A=\frac{9 m* 9m}{2}\\\\A_{1} =40.5\ m^{2}[/tex]
Step 2
triangle 2 (RIGHT UP)
[tex]b=9\ m\\\ h= 9\ m\\A=\frac{9 m* 9m}{2}\\\\A_{2} =40.5\ m^{2}[/tex]
Step 3
triangle 3 (LEFT DOWN)
[tex]b=9\ m\\\ h= 6\ m\\A=\frac{9 m* 6m}{2}\\\\A_{3} =27\ m^{2}[/tex]
Step 4
triangle 4 (RIGHT DOWN)
[tex]b=9\ m\\\ h= 6\ m\\A=\frac{9 m* 6m}{2}\\\\A_{4} =27\ m^{2}[/tex]
Step 5
Total Area
the total area is
[tex]A_{T}=A_{1}+A_{2}+A_{3}+A_{4} \\A_{T}=(40.5+40.5+27+27)m^{2} \\A_{T}=135 m^{2}[/tex]
Have a great day