A triangle has two sides that measure 10 in. and 8 in. which could be the measure of the third side?

Mathematics

2 answers:

valkas [14]3 years ago

5 0

If the triangle has a right angle, the measure of the third side is 12.8 inches

insens350 [35]3 years ago

5 0

Solutions

To solve this problem we have to use the Pythagorean theorem. You can only use the Pythagorean theorem in Right Triangles. The longest side of the triangle is called the "hypotenuse". C² is the longest side so it is the hypotenuse . To calculate c² we have to do α² + β² = c².

Given

Side (A) = 10 in
Side (B) = 8 in
Side (C) = unknown

The length of the third side must be greater than 0 but less than 18in

10-8 < s < 10+8

2 < s < 18

17 in is the correct answer

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ABC = 25x + 34 CBD= 11x +23 Find ABD CBD and ABC

mote1985 [20]

The measure of <ABD, <DBC and <ABC are 67, 67 and 134 degrees respectively.

<h3>Bisection of angles</h3>

Angles are bisected if they are divided into two equal parts.

If the angle BC bisects <ABC, hence <ABD and <DBC are equal, hence;

2(11x + 23) = <ABC

Given the following parameters

<ABC = 25x + 34

2(11x + 23) = 25x + 34

Expand

22x +46 = 25x + 34

22x-25x = 34 - 46

-3x = -12

x = 4

Determine the measure of the angles

<ABD = 11x + 23 = <DBC
<ABD = 11(4) + 23

<ABD = 44 + 23

<ABD = 67 degrees

<ABC = 2(67)

<ABC = 134 degrees

Hence the measure of <ABD, <DBC and <ABC are 67, 67 and 134 degrees respectively.

Learn more on bisection of angles here: brainly.com/question/25770607

#SPJ1

5 0

1 year ago

In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th

11111nata11111 [884]

Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

Probability that the students passed the second exam = P(S) = 0.72

Probability that the students passed both exams = $P(F \bigcap S)$ = 0.58