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Anuta_ua [19.1K]
3 years ago
11

Are all isosceles triangles similar? Are all right triangle similar? Are all isosceles right triangles similar? Explain.

Mathematics
1 answer:
Alex17521 [72]3 years ago
5 0

9514 1404 393

Answer:

  a) no

  b) no

  c) yes

Step-by-step explanation:

a) Isosceles triangles can have a third angle (the one that is not part of the pair of equal angles) with any value in the range 0 < α < 180°. Triangles with different angle values will not be similar.

Not all isosceles triangles are similar.

__

b) The smallest acute angle of a right triangle can have any measure in the range 0 < α ≤ 45°. Right triangles having different smallest angles will not be similar. The attachment shows dissimilar right triangles.

Not all right triangles are similar.

__

c) If the smallest acute angle of a right triangle is 45°, then both acute angles are 45°, and the right triangle is an isosceles triangle. Each such triangle will be similar to every such triangle.

All isosceles right triangles are similar.

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Answer:

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3 years ago
4x7x5x6x7x6x6x6<br><br><br> yeah
miv72 [106K]

Answer:

1,270,080

Step-by-step explanation:

According to my calculator 4x7x5x6x7x6x6x6 = 1,270,080

7 0
3 years ago
In the coordinate plane, you are given JKL with J(2, 3), K(10,4), and L(6, 10). Using the distance formula, classify JKL
Marrrta [24]

Answer:

Perimeter of JKL = 22.55 unit (Approx.)

Step-by-step explanation:

Given:

Coordinate of J = (2,3)

Coordinate of K = (10,4)

Coordinate of L = (6,10)

Find:

Perimeter of JKL

Computation:

Distance = √(x1 - x2)² + (y1 - y2)²

Distance between JK = √(2 - 10)² + (3 - 4)²

Distance between JK = √64 + 1

Distance between JK = √65

Distance between JK = 8.06 unit

Distance between KL = √(10 - 6)² + (4 - 10)²

Distance between KL = √16 + 36

Distance between KL = √52

Distance between KL = 7.21 unit

Distance between LJ = √(6 - 2)² + (10 - 3)²

Distance between LJ = √4 + 49

Distance between LJ = √53

Distance between LJ = 7.28 unit

Perimeter of JKL = Distance between JK + Distance between KL + Distance between LJ

Perimeter of JKL = 8.06 + 7.21 + 7.28

Perimeter of JKL = 22.55 unit (Approx.)

4 0
2 years ago
If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
3 years ago
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