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3 years ago

What is the perimeter of the triangle.

Mathematics

1 answer:

Dmitry [639]3 years ago

8 0

36. Find the hypotenuse by squaring the length of the height and length. It gives you 225...so your side C should be 15. Then add all your sides giving you 36

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Find the area of the triangle.

Vsevolod [243]

Answer:

20 m²

Step-by-step explanation:

Formula (Area of triangle):

$\text{A (Triangle)} = \dfrac{1}{2} \times \text{Base} \times \text{Height}$

From the triangle, we can see that the base of the triangle is "7 + 3" m and the height of the triangle is "4" m.

$\implies \text{A (Triangle)} = \dfrac{1}{2} \times (7 + 3) \times (4)$

Simplify the right-hand-side as needed to evaluate the area.

$\implies \text{A (Triangle)} = \dfrac{1}{2} \times (10) \times (4)$

$\implies \text{A (Triangle)} = \dfrac{40}{2} = 20 \ \text{m}^{2}$

Therefore, the area of the provided triangle is 20 m².

Learn more about this topic: brainly.com/question/15442893

8 0

2 years ago

Find the value of -(-2x+1)=9-14x

Elodia [21]

2x - 1 = 9 - 14x

16x - 1= 9

16x = 10

x = 10/16
x= 5/8 is the answer

7 0

3 years ago

Applying Triangle Classification Theorems

dexar [7]

Answer:

B is the answer on edge

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The sum of two numbers is 39 and the difference is 3. What are the numbers?

frosja888 [35]

Answer: 21 & 18

Step-by-step explanation:

Sum of two numbers is 39

Difference is 3

Let the numbers be x & y

X+y=39........equation 1

X-y=3...........equation 2

X=3+y............equation 3

Substitute equation 3 into equation 1

(3+y)+y=39

3+y+y=39

3+2y=39

2y=39-3

2y=36

Y=36/2

Y=18

Substitute for y in equation 2

X-y=3

X-18=3

X=3+18

X=21

The two numbers are 21 & 18

8 0

3 years ago

Carlos drew a plan for his garden on a coordinate plane. Rose bushes are located at A(–5, 4), B(3, 4), and C(3, –5)

BartSMP [9]

Given:

A(-5,4)

B(3,4)

C(3,-5)

So point D is:

so point D is (-5,-5)

For AB is

Distance between two point is:

$\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}$

so distance between A(-5,4) and B(3,4) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

So AB is 8 unit apart.

For B(3,4) and C(3,-5).

$\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}$

So BC is 9 unit apart.

For fourth bush point is (-5,-5) it left of point C(3,-5) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

so fourth bush is 8 unit left of C.

For fourth bush(-5,-5) below to point A(-5,4)

$\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}$

so fourth bush 9 units below of A.

8 0

1 year ago