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Elena L [17]
3 years ago
12

Jake drew this sketch of his triangular rock garden. What is the area of his rock garden?

Mathematics
2 answers:
Oksanka [162]3 years ago
8 0

Answer:

It’s either 30ft or 30ft^2

Charra [1.4K]3 years ago
7 0

Answer:

60

Step-by-step explanation:

To find the area of a triangle you have to multiply base by height base is 12 and height is 5 so you do 12 x 5 = 60! :) hope this helps

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Find the surface area of a pentagon-based pyramid with slant height of 11 m.
Ronch [10]

Answer:

108.75 square meter.

Step-by-step explanation:

<h3>Surface area of pentagon based pyramid:</h3>

   \sf \boxed{\text{\bf Surface area of pentagonal pyramid = $ \bf \dfrac{5}{2}*b*(a + s)$}}

  Here,  b is the base length of the pyramid. b = 3 m

              a is the apothem of the pyramid. a = 3.5 m

              s is the slant height of the pyramid. s = 11 m

     \sf Surface \ area = \dfrac{5}{2}*3*(11 + 3.5)

                         \sf = \dfrac{5}{2}*3*14.5\\\\     = 108.75 \ m^2

5 0
1 year ago
What is the value of y?
coldgirl [10]

<em>Note: You forgot to add the diagram, so I found the missing diagram and attached below, based on which I am solving your query. Hopefully, it would clear your concepts anyways.</em>

<em />

Answer:

Option B is correct, as  \boxed{y=6\sqrt{3}}  

Step-by-step explanation:

From the attached figure,

ΔMTU is a right triangle.

Applying Pythagoras theorem,

TM^2=TU^2+MU^2

\left(6\right)^2=TU^2+\left(3\right)^2

\left(6\right)^2-\left(3\right)^2\:=TU^2

switching sides

TU^2=\left(6\right)^2-\left(3\right)^2\:

TU^2=36-9

TU^2=27

Note that ΔNTU s also right triangle.

\:NT^2\:=\:TU^2\:+\:NU^2

         =27\:+\:9^2           ∵ TU^2=27

         =108

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

NT=\sqrt{108},\:NT=-\sqrt{108}

But length of a side can not be negative, so ignore the negative solution.

so

NT=\sqrt{108}

    y=6\sqrt{3}                ∵ NT = y

Therefore, option B is correct, as  \boxed{y=6\sqrt{3}}  

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3 years ago
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Answer:

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Step-by-step explanation:

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