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zheka24 [161]
3 years ago
6

Given: PSTK is a trapezoid, m∠P=90° SK =13, PK = 12, ST=8 Find: The area of PSTK.

Mathematics
1 answer:
guajiro [1.7K]3 years ago
5 0
<h3>Given</h3>

trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8

<h3>Find</h3>

the area of PSTK

<h3>Solution</h3>

It helps to draw a diagram.

∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as

... KS² = PS² + KP²

... 13² = PS² + 12²

... PS = √(169 -144) = 5

This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is

... A = (1/2)(b1 +b2)h

... A = (1/2)(12 +8)·5

... A = 50

The area of trapezoid PSTK is 50 square units.

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Step-by-step explanation:

Given data

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3 years ago
The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of
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the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches

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Step-by-step explanation:

-2v+8(1+2v)=-90

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