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seropon [69]
3 years ago
12

PLEASE HELP ILL GIVE BRAINLIEST THIS IS GEOMETRY

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

the first option x = 6sqrt2 y =6

Step-by-step explanation:

if this is special right triangles than this is the 45 45 90 one soo the 45 angles are 6 and the 90 (hypotenuse) will be x (6) with a square root of 2

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If a perimeter of a room is 78 feet, what is the area of the floor of the room?
Andrei [34K]
The perimeter doesn't tell you the area, and neither of them
can be calculated by knowing the other one.

Check this out:

Dimensions   Perimeter   Area      .

1  by  38              78          38 square feet
2  by  37              78          74 square feet         
3  by  36              78         108 square feet         
5  by  34              78         170 square feet         
10  by  29            78         290 square feet         
15  by  24            78         360 square feet         
17  by  22            78         374 square feet         
19  by  20            78         380 square feet         
19.5 by 19.5        78          380.25 square feet

Nine same perimeters, nine different areas !

Here's one more ... a long skinny one:

1 inch by 38ft 11in     Perimeter = 78 ft.    Area = 3.24 square feet.    
6 0
4 years ago
Fie triunghiul ABC isoscel cu AB=AC=3 cm daca mediatoarea laturi AC intersectat cu latura BC in M si perimetrul thriunghiului AM
NemiM [27]

Answer:

MC = 4.5cm

Step-by-step explanation:

Question:

Let the isosceles triangle ABC with AB = AC = 3 cm. if the mediator of the sides AC intersects with the side BC in M ​​and the perimeter of the triangle AMC = 12 cm. Calculate MC.

Solution:

Find attached the diagram used in solving the question.

Given:

∆ABC is an isosceles triangle (two sides and angles are equal)

AB = BC = 3cm

Perimeter of ∆AMC = 12cm

From the diagram, M cuts AC at the the middle.

AD = CD = AC/2 = 3/2

Perimeter of Right angled ∆AMD = AM + AD + MD

= 3/2 + AM +MD

Perimeter of Right angled ∆CMD =CM + CD + MD

= 3/2 + CM +MD

Right angled ∆AMD = Right angled ∆CMD

CM = AM

Therefore ∆AMC is an isosceles triangle

CM = AM (two sides of an isosceles triangle are equal)

Let CM = AM = x

Perimeter of ∆AMC = AM + CM + AC

12 = x + x + 3

12 = 2x + 3

2x = 12-3

2x = 9

x = 9/2 = 4.5

CM = AM = 4.5cm

MC = CM = 4.5cm

3 0
3 years ago
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