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Eva8 [605]
3 years ago
7

Please do Correctly will mark you Brainliest!

Mathematics
2 answers:
allochka39001 [22]3 years ago
5 0

Answer:

\rm\displaystyle \: A _{ \text{ total}}  = 212

Step-by-step explanation:

Prefer the attachment

so we have 3 rectangle

recall that,

\displaystyle \: A _{ \text{ rect}} = l \times w

<h3>rectangle-1:</h3>

the width of the rectangle is 10m

and the length is 12+5=17m

so the area is

\displaystyle \: A _{ \text{ rect - 1}} = 17  \times 10

\displaystyle \: A _{ \text{ rect - 1}} =170 \:  {m}^{2}

<h3>rectangle-2:</h3>

the length is 9 m

the width is 3 m

so the area is

\displaystyle \: A _{ \text{ rect - 2}} = 9 \times3

\displaystyle \: A _{ \text{ rect - 2}} = 27 \:  {m}^{2}

<h3>rectangle-3:</h3>

the length is 5m

the width is 3m

so the area is

\displaystyle \: A _{ \text{ rect - 3}} = 5 \times 3

\displaystyle \: A _{ \text{ rect - 3}} = 15 {m}^{2}

<h3>total area:</h3>

\rm\displaystyle \: A _{ \text{ total}} =170 \:  {m}^{2}  +  {27m}^{2}  +  {15m}^{2}

\rm\displaystyle \: A _{ \text{ total}}  = 212 {m}^{2}

ivolga24 [154]3 years ago
4 0

Answer:

279...................................

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8 0
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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

SP \approx 8.246

PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

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