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2 years ago

Can someone help me find the length of b of the unknown side in this triangle? Thanks! :)

Mathematics

2 answers:

creativ13 [48]2 years ago

7 0

Answer: 4

Step-by-step explanation:

Using pythagoras theorem =>

b= $\sqrt{c^2-a^2$

b= $\sqrt{5^2-3^2}$

b= $\sqrt{25-9}$

b= $\sqrt{16}$

b=4

givi [52]2 years ago

6 0

Answer:

Step-by-step explanation:

Use the Pythagorean Theorem:

$a^2+b^2=c^2$

The measure of the hypotenuse is 5. The measure of the given leg is 3.

$3^2+b^2=5^2\\\rightarrow 3^2 = 9; 5^2=25\\9+b^2=25\\9-9+b^2=25-9 \leftarrow \text {Subtraction Property of Equality} \\b^2=16\\\sqrt{b^2}=\sqrt{16} \leftarrow \text {Square Root Property of Equality}\\\boxed {b=4}$

The unknown side of the triangle should be four units.

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Step-by-step explanation:

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2 years ago

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2 years ago

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3 years ago

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has

Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0), Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.

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3 years ago