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yaroslaw [1]
4 years ago
11

Triangle QRS is a right triangle. Complete the similarity statement. ΔSTR ~ Δ

Mathematics
2 answers:
Kitty [74]4 years ago
6 0

Answer:

ΔSTR is similar to ΔRTQ

Step-by-step explanation:

Given QRS is a right angled triangle. we have to find the similarity statement ΔSTR ~ Δ__

Let ∠S=x

In ΔSTR, by angle sum property

∠S+∠STR+∠SRT=180°

⇒ ∠SRT=90°-x

In ΔSRQ, by angle sum property

∠S+∠R+∠Q=180°

⇒ ∠Q=90°-x

In ΔSTR and ΔRTQ

∠SRT=∠Q=90°-x     (proved above)

∠STR=∠RTQ           (each 90°)

RT=RT                      (common)

Hence, by AAS rule ΔSTR≅ΔRTQ

∴ ΔSTR is similar to ΔRTQ

Option 4 is correct.

LenKa [72]4 years ago
5 0

Answer:

RTQ

Step-by-step explanation:

Let ∠S=a, In ΔSTR, using angle sum property, we have

∠S+∠STR+∠SRT=180°

⇒ ∠SRT=90°-a

Again In ΔSRQ, using angle sum property, we have

∠S+∠R+∠Q=180°

⇒ ∠Q=90°-a

Now, In ΔSTR and ΔRTQ

∠SRT=∠Q=90°-a    (proved above)

∠STR=∠RTQ              (each 90°)

 RT=RT                      (common)

Hence, by AAS rule,

ΔSTR≅ΔRTQ

Thus,  ΔSTR is similar to ΔRTQ

Option 4 is correct.

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Step-by-step explanation:  

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Sketch the lines given by x+y =6 and −3x+y = 2 on the same set of axes to solve the system graphically.Then solve the system of
tankabanditka [31]

Answer:

x= 1 and y= 5

The graphical solution is in the attachment.

Step-by-step explanation:

There's a lot of methods for solving a system of equations. For example, the substitution method.

You have to solve one of the equations for one variable (x or y) and replace it in the other equation and solve it for that variable. Therefore you will obtain a solution and then you have to replace that solution in the first equation in order to obtain the second solution.

In this case, solving the first equation for x:

x+y=6

Adding -y both sides:

x + y - y = 6-y

x = 6-y (I)

Replacing it in the second equation:

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Applying the distributive property:

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Dividing by 4 both sides:

y = 20/4 =5

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In order to sketch the given lines, you have to obtain two points for each one and trace the line containing those points. The solution of the system of equations is the point of intersection of the two lines.

<u>For x+y = 6</u>

-The intersection with the x-axis is given by replacing y=0

x + 0 = 6

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P1(6,0)

-The intersection with the y-axsis is given by replacing x=0

0 + y =6

y = 6

P2(0,6)

<u>For -3x + y = 2</u>

-The intersection with the x-axis:

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x = -2/3

P3(-2/3,0)

-The intersection with the y-axis:

-3(0) + y = 2

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The intersection of the two lines is setting the equations equal to each other:

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