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iren2701 [21]
4 years ago
5

Which triangles could not be similar to triangle ABC ? triangle a b c where angle b is a right angle. side a b is 15 cm. side b

c is 8 cm. side a c is 17 cm. Select each correct answer. Right triangle G H I with right angle at H. G H is 77 centimeters. G I is 85 centimeters. H I is 36 centimeters. Right triangle J K L with right angle at K. J K is 60 centimeters. J L is 68 centimeters. K L is 32 centimeters. Right triangle M N O with right angle at N. M N is 35 centimeters. M O is 37 centimeters. N O is 12 centimeters. Right triangle D E F with right angle at E. D E is 22.5 centimeters. D F is 25.5 centimeters. E F is 12 centimeters.

Mathematics
1 answer:
STALIN [3.7K]4 years ago
3 0

Answer:

option (b) and (d) are correct.

 Right triangle J K L with right angle at K. J K is 60 centimeters. J L is 68 centimeters. K L is 32 centimeters.

and Right triangle D E F with right angle at E. D E is 22.5 centimeters. D F is 25.5 centimeters. E F is 12 centimeters.

Step-by-step explanation:

Given : triangle a b c where angle b is a right angle. side a b is 15 cm. side b c is 8 cm. side a c is 17 cm.

We have to select triangle from given options that are similar to ΔABC,

Two triangles are said to be  similar when their corresponding sides are in same ratios  and their corresponding angles are equal.

Consider : Right triangle J K L with right angle at K. J K is 60 centimeters. J L is 68 centimeters. K L is 32 centimeters.

FINDING CORRESPONDING RATIOS,

\frac{AB}{JK}=\frac{15}{60}=\frac{1}{4}

\frac{BC}{KL}=\frac{8}{32}=\frac{1}{4}

\frac{AC}{JL}=\frac{17}{68}=\frac{1}{4}

Since, ratios are equal. Thus, triangles are similar.

ΔABC ≅ ΔJKL

Consider : Right triangle D E F with right angle at E. D E is 22.5 centimeters. D F is 25.5 centimeters. E F is 12 centimeters.

FINDING CORRESPONDING RATIOS,

\frac{AB}{DE}=\frac{15}{22.5}=\frac{2}{3}

\frac{BC}{EF}=\frac{8}{12}=\frac{2}{3}

\frac{AC}{DF}=\frac{17}{25.5}=\frac{2}{3}

Since, ratios are equal. Thus, triangles are similar.

ΔABC ≅ ΔDEF

Thus, option (b) and (d) are correct.


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5 0
3 years ago
In order to qualify for finals in racing event,a race car must achieve an average speed of 250km/hr on a track with a total leng
Kazeer [188]

Answer:

The minimum average speed needed in the second half is 270 km/hr

Step-by-step explanation

We can divide the track in two parts. For the first half of the track the average speed the car achieved was 230 km/hr and we need to make sure that the average speed of the full track is 250 km/hr. Then, we can calculate the average speed of the two parts of the track and force this to be equal to 250 km/hr. In equation, defining v_2 as the average speed of the second half:

\frac{230 + v_2}{2}=250

Solving for x

x=250*2-230\\x=500-230\\x=270

Therefore, achieving a speed of 270 km/hr in the second half would be enough to achieve an average speed of 250 on the track.  

6 0
3 years ago
Question 6 of 25
Rom4ik [11]

Answer:

I pretty sure the answer is c

3 0
3 years ago
Solve for x= 7x + 15 = 38
777dan777 [17]
In order to solve for x in the equation <span>7x + 15 = 38, you must first subtract 15 from both sides.
</span><span>7x + 15 = 38
</span><span>7x + 15 - 15 = 38 - 15
7x = 23
Then you must divide so there is only 1x = a number. This can be done by dividing both sides by 7.
7x = 23
7x / 7 = 23 / 7
x ≈ </span>3.286
So the solution for x in the equation 7x + 15 = 38 is x ≈ 3.286.
Hope this helps!
7 0
4 years ago
How do I answer this—
Katen [24]

Answer:

angle 2 = 87 degrees

" /_ 2=87° "

4 0
3 years ago
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