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2 years ago

What is the perimeter of abc

Mathematics

1 answer:

Ede4ka [16]2 years ago

5 0

perimeter ≈ 18.9 units ( to 1 dec. place )

Calculate the lengths of the 3 sides using the distance formula

d = √( (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = A(1, 1) and (x₂, y₂ ) = B(9, 2 )

d = √( (9 - 1 )² + (2 - 1 )² ) = √(64 + 1 ) = √65

repeat with (x₁, y₁ ) = B(9, 2) and (x₂, y₂ ) = C(4,5 )

d = √( (4 - 9 )² + (5 - 2 )² ) = √(25 + 9 ) = √34

let (x₁, y₁ ) = A(1, 1 ) and (x₂, y₂ ) = C(4, 5 )

d = √( (4 - 1 )² + (5 - 1 )² ) = √(9 + 16 = √25 = 5

perimeter = √65 + √34 + 5 ≈ 18.9 units

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Leni [432]

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Step-by-step explanation:

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Read 2 more answers

If triangle ABC is similar to triangle DEC, find the values of x and y.

Mumz [18]

The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal, it is obtained by Pythagoras theorem.

Step-by-step explanation:

The given are,

From ΔABC,

AB= 6

BC= 10

AC = x

From ΔDEC,

CD= 28

DE= 21

CE = y

Step:1

Pythagoras theorem from ΔABC,

$BC^{2}=AB^{2} + AC^{2}$ ...............(1)

Substitute the values,

$10^{2}$ = $6^{2}$ + $AC^{2}$

100 = 36 + $AC^{2}$

$AC^{2}$ = 100 - 36

= 64

AC = $\sqrt{64}$

AC = 8

AC = x = 8

Step:2

Pythagoras theorem for ΔDEC,

$CE^{2} = CD^{2} + DE^{2}$ ................(2)

From the values,

$CE^{2}$ = $28^{2}$ + $21^{2}$

$CE^{2}$ = 784 + 441

= 1225

CE = $\sqrt{1225}$

CE = 35

CE = y = 35

Result:

The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal.

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