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3 years ago

PLEASE HELP ME IT’S DUE TODAY

Mathematics

2 answers:

Alexxx [7]3 years ago

6 0

Answer:

No 4: the triangle are similar by congruency SAS(side angle side)

No 5: this is also similar by SAS

No 6: this is similar by AAS ( angle angle side)

No 7: this is similar by RHS ( right angle hypotenuse one side )

No 8: this pair of triangles are not similar

No 9: this is similar by SAS

No 10: this pair of triangles are similar by SAS ( side side side)

No 11: this triangle is similar by SSS

No 12: this triangle is similar by SSS

Mama L [17]3 years ago

3 0

Answer:

ummmm

Step-by-step explanation:

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Alekssandra [29.7K]

The area of a triangle is (base*height)/2
Therefore: 12*10=120/2=60. Since there are three triangles then you multiply 60 by three. 60*3=180
Since the triangle in the middle is equilateral the height is 10. (10*10)/2=50
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4 years ago

Please help me with questions 26 and 27

grandymaker [24]

26) Answer is choice A: y = -3
This is the horizontal line that cuts the figure into two congruent pieces.
--------------------------------------------------------------

27) Answer is choice A: AB = 24 and DE = 12
CA/DF = 12/6 = 2
CB/FE = 20/10 = 2
The ratio of the larger to smaller corresponding sides is 2:1 or just 2

So AB needs to be twice as long as DE
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4 years ago

What is the area of ABC?

Salsk061 [2.6K]

Answer:

Area of triangle ABC = 10

Option C is correct.

Step-by-step explanation:

We need to find area of triangle ABC

We are given A(2,1) B(4,7) and C(6,3)

The formula used to find $Area \ of \ triangle=\frac{1}{2}base\times height$ ABC is:

In the given triangle base= AC and height = BC

First we need to find the distance between A and C and B and C

The formula used to find distance between AC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=2, y_1=1, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-2)^2+(3-1)^2}\\Distance=\sqrt{(4)^2+(2)^2}\\Distance=\sqrt{16+4}\\Distance=\sqrt{20}$

find distance between BC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=4, y_1=7, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-4)^2+(3-7)^2}\\Distance=\sqrt{(2)^2+(4)^2}\\Distance=\sqrt{4+16}\\Distance=\sqrt{20}$

So, We have height = $\sqrt{20}$ and base = $\sqrt{20}$

Finding area of triangle

$Area \ of \ triangle=\frac{1}{2}base\times height\\Area \ of \ triangle=\frac{1}{2}\sqrt{20} \times \sqrt{20} \\Area \ of \ triangle=\frac{1}{2}\times 20\\Area \ of \ triangle=10$