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ivann1987 [24]
3 years ago
11

a ladder 13m long is placed on the ground in such a way that it touches the top of a vertical wall 12m high. find the distance o

f the foot of the ladder from the bottom of the wall​
Mathematics
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

by applying Pythagoras theorem;

the ladder(13m) acts as hypotenuse

the wall (12m) acts as height

the distance between the the two is the base

therefore

a² + b² = c²

b²= c²- a²

b²= 169 - 144

b² = 25

√b² = √25

b= 5m

the distance is 5m

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Eric's father work 5 hours each afternoon

Step-by-step explanation:

Let x be the no. of hours Eric's father work each afternoon

He works for hours in morning each day = 3

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He works for total hours in afternoon in 5 days=5x

We are given that he works a total of 40 hours each 5-day workweek.

ATQ

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If you travel 40 miles at 45 miles per hour, how many minutes will that take you?
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Read 2 more answers
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
2 years ago
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