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Irina-Kira [14]
2 years ago
7

help please! the pair of figures to the right are similar. what is the ratio of the perimeters and the ratio of the areas?

Mathematics
1 answer:
jasenka [17]2 years ago
4 0

Answer:

Perimeters = 3 in : 7 in

Areas = 9 in² : 49 in²

Step-by-step explanation:

    <em>If the figures are </em><em>squares</em><em>, and the measurements are the length of that sides (as you have said), then we will do the following. </em>

[The left square]

P = 4a

P = 4(15)

P = 60 in

-

A = a²

A = 15²

A = 225 in²

[The right square]

P = 4a

P = 4(35)

P = 140 in

-

A = a²

A = (35)²

A = 1,225 in²

[Simplifying ratios]

Perimeters = 60 in : 140 in

Perimeters = 6 in : 14 in

Perimeters = 3 in : 7 in

-

Areas = 225 in² : 1,225 in²

Areas = 45 in² : 245 in²

Areas = 9 in² : 49 in²

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1) 0.48 3     2) 7/8     3) 7 1/8
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X, Y and Z form the vertices of a triangle. XY = 12.4m, XZ = 10.4m and YZ = 8.7m. Find the angle ∠ YXZ rounded to 1 DP.
sweet-ann [11.9K]

Answer:

43.8°

Step-by-step explanation:

Applying,

Cosine rule,

From the diagram attached,

x² = y²+z²-2yxcos∅.................... Equation 1

where ∅ = ∠YXZ

Given: x = 8.7 m, y = 10.4 m, z = 12.4 m

Substitute these values  into equation 1

8.7² = 10.4²+12.4²-[2×10.4×12.4cos∅]

75.69 = (108.16+153.76)-(257.92cos∅)

75.69 = 261.92-257.92cos∅

collect like terms

257.92cos∅ = 261.92-75.69

257.92cos∅ = 186.23

Divide both sides by the coefficient of cos∅

cos∅ = 186.23/257.92

cos∅ = 0.722

Find the cos⁻¹ of both side.

∅ = cos⁻¹(0.7220)

∅ = 43.78°

∅ = 43.8°

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X + (x + 11) =79 solve x
posledela

Answer:

x=34

Step-by-step explanation:

x + (x + 11) =79

Combine like terms

2x+11 = 79

Subtract 11 from each side

2x+11-11 = 79-11

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Divide by 2

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In the figure shown, ABC is a right triangle with side lengths a, b, and c, and CD is an altitude to side AB. The side lengths o
ycow [4]

Answer:

A

Step-by-step explanation:

In the figure shown, ABC is a right triangle with side lengths a, b, and c, and CD is an altitude to side AB. This altitude divides the triangle into two right triangles ADC and BDC. In these triangles,

  • \angle CBD\cong \angle ACD\cong \angle CBA
  • \angle DCB\cong \angle DAC\cong \angle BAC

So,

\triangle ABC\sim \triangle CBD\sim \triangle ACD

1. From the similarity \triangle ABC\sim \triangle CBD, you have

\dfrac{AB}{BC}=\dfrac{BC}{BD}\\ \\\dfrac{c}{a}=\dfrac{a}{s}

2. From the similarity \triangle ABC\sim \triangle ACD, you have

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