What is cos P? Right triangle PQR with a short leg measuring 5, a long leg measuring 12, and the hypotenuse as 13
1 answer:
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Easy
the easiest way is to imagine that you have a rectangle with a corner cut off
the area of the figure=area of rectangle-area cut off
area that was cut off was a triangle
area of rectangle=length times width
area of triangle=1/2base times height
so we draw imaginary lines like in attachment
find area of rectangle
legnth=9 inch
width=9 inch
area=9 times 9=81 in^2
triangle
the top part is ?+5=9, so the base of triangle=4
the side is ?+4=9, so the base is 5
area=1/2 times 4 times 5=10 in^2
area=rectangle-triangle
area=81-10=71 in^2
the easiest way is to imagine that you have a rectangle with a corner cut off
the area of the figure=area of rectangle-area cut off
area that was cut off was a triangle
area of rectangle=length times width
area of triangle=1/2base times height
so we draw imaginary lines like in attachment
find area of rectangle
legnth=9 inch
width=9 inch
area=9 times 9=81 in^2
triangle
the top part is ?+5=9, so the base of triangle=4
the side is ?+4=9, so the base is 5
area=1/2 times 4 times 5=10 in^2
area=rectangle-triangle
area=81-10=71 in^2