Plz help due 2morrow 14 and 15
1 answer:
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The complete question in the attached figure
Part a)
we know that
ABC is a right triangle
∠ACB=45°
AC=hypotenuse------> 6√2 cm
sin 45=AB/AC-----> AB=AC*sin 45----> AB=6√2*√2/2----> AB=6 cm
the answer part a) is
AB=6 cm
Part b)
we know that
volume of the pyramid=(1/3)*Area of the base*height
area of the base=50 cm²
height=6 cm
so
volume of the pyramid=(1/3)*50*6----> 100 cm³
the answer part b) is
100 cm³
Part a)
we know that
ABC is a right triangle
∠ACB=45°
AC=hypotenuse------> 6√2 cm
sin 45=AB/AC-----> AB=AC*sin 45----> AB=6√2*√2/2----> AB=6 cm
the answer part a) is
AB=6 cm
Part b)
we know that
volume of the pyramid=(1/3)*Area of the base*height
area of the base=50 cm²
height=6 cm
so
volume of the pyramid=(1/3)*50*6----> 100 cm³
the answer part b) is
100 cm³