The triangles are similar. What is the value of x? Enter your answer in the box. x = two right triangles. the larger triangle has a long leg of 96 units, short leg of 28 units, and the hypotenuse is labeled 6 x plus 28. the smaller triangle has a long leg of 24 units, short leg of 7 units, and hypotenuse of 25 units. Two segments A D and B C intersect at point E to form two triangles A B E and D C E. Side A B is parallel to side D C. A E is labeled 2 x plus 10. E D is labeled x plus 3. A B is 10 units long. D C is 4 units long Two sailboats with sails that are right triangles. The top angle on the sails are marked congruent to each other. The smaller sail has a height of x and a base of 7 inches. The larger sail has a height of 12 inches and a base of 8 inches. two right triangles are shown showing the path the basketball makes from the father to the son. The height from the floor to the fathers hand is 30 inches. The height from the floor to the sons hand is labeled x. The distance from the fathers hand to where the ball hits the floor is 45 inches. The distance from where the ball bounces off the floor to the son is 40 inches.

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23-01-2019
Mathematics

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The triangles are similar. What is the value of x? Enter your answer in the box. x = two right triangles. the larger triangle has a long leg of 96 units, short leg of 28 units, and the hypotenuse is labeled 6 x plus 28. the smaller triangle has a long leg of 24 units, short leg of 7 units, and hypotenuse of 25 units. Two segments A D and B C intersect at point E to form two triangles A B E and D C E. Side A B is parallel to side D C. A E is labeled 2 x plus 10. E D is labeled x plus 3. A B is 10 units long. D C is 4 units long Two sailboats with sails that are right triangles. The top angle on the sails are marked congruent to each other. The smaller sail has a height of x and a base of 7 inches. The larger sail has a height of 12 inches and a base of 8 inches. two right triangles are shown showing the path the basketball makes from the father to the son. The height from the floor to the fathers hand is 30 inches. The height from the floor to the sons hand is labeled x. The distance from the fathers hand to where the ball hits the floor is 45 inches. The distance from where the ball bounces off the floor to the son is 40 inches.

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cryssatemp cryssatemp · Answer 1 · 2019-02-02T02:18:20+01:00

Answers:

These problems can be solved by the Thales’s Theorem, which states:

Two triangles are similar when they have equal angles and proportional sides

In addition, if two triangles are similar, their angles are similar as well. This means that the relation between two sides of the big triangle is equal to the relation between two sides of the small triangle, as follows:

[tex]\frac{B}{A}=\frac{b}{a}[/tex]

Knowing this, lets’s begin with the answers:

1. For this first problem, see the first figure attached. There are shown the two triangles, and we are told both are similar, this means (according the prior explanation above) that we can use the Thale’s Theorem to find [tex]x[/tex].

Therefore, we can establish a relation between two sides of the big triangle which is equal to the relation between two sides of the small triangle; in this case let’s use [tex]B[/tex] and [tex]C[/tex] for the first triangle and [tex]b[/tex] and [tex]c[/tex] for the second:

[tex]\frac{C}{B}=\frac{c}{b}[/tex]

[tex]\frac{6x+28}{96}=\frac{25}{24}[/tex]

Now we find [tex]x[/tex]:

[tex]6x+28=\frac{(25)(96)}{24}[/tex]

[tex]6x+28=100[/tex]

[tex]6x=100-28[/tex]

[tex]x=\frac{72}{6}[/tex]

[tex]x=12[/tex]

Finally, the value of [tex]x[/tex] is 12 units.

2. For this problem, see the second figure attached. In order to find the values of the segments BE and EC, we will use two sides of each triangle (ABE and DCE) according to the Thale’s Theorem:

For the segment BE:

[tex]\frac{BE}{10}=\frac{x+3}{4}[/tex]

[tex]BE=\frac{(x+3)(10)}{4}[/tex]

Simplifying:

[tex]BE=\frac{5}{2}(x+3)[/tex] >>>>>This is the length of the segment BE

For the segment EC:

[tex]\frac{EC}{4}=\frac{2x+10}{10}[/tex]

[tex]EC=\frac{(2x+10)(4)}{10}[/tex]

Simplifying:

[tex]EC=\frac{4}{5}(x+5)[/tex]>>>>>This is the length of the segment EC

3. For this problem, see the third figure attached:

[tex]\frac{12 inches}{8 inches}=\frac{x}{7 inches}[/tex]

Solving for [tex]x[/tex]:

[tex]x=\frac{(12 inches)(7 inches)}{8 inches}[/tex]

Finally:

[tex]x=10.5 inches[/tex]>>>>>This is the height of the smaller sail

4. For this problem, see the fourth figure attached.

[tex]\frac{80 inches}{45 inches}=\frac{x}{40 inches}[/tex]

Solving for [tex]x[/tex]:

[tex]x=\frac{(80 inches)(40 inches)}{45 inches}[/tex]

[tex]x=71.11 inches[/tex] >>>>>This is the height from the floor to the son's hand