What is the speed of a snake that moves 20 m in 5 seconds ?
1 answer:
The speed of the snake is 8.95 miles per hour.
Step-by-step explanation:
Step 1:
The snake travels at a speed of 20 meters in 5 seconds.
We will need to determine the distance it covered in a single second. To do that we divide the distance covered in 5 seconds by 5.
The distance covered in 1 second
So the snake is moving at a distance of 4 meters per second.
Step 2:
To convert the speed of the snake from meters per second to miles per hour, we multiply the speed in meters per second by 2.23694.
So to attain 4 meters per second in miles per hour, we multiply it by 2.23694.
4 meters per second miles per hour.
Rounding this off, we get the speed of the snake as 8.95 miles per hour.
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Answer:
Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.
To prove: Point A,D,P are collinear.
Proof:
→Case 1. When vertices A and P are opposite side of Base BC.
In Δ ABD and Δ ACD
AB= AC [Given]
AD is common.
BD=DC [median of a triangle divides the side in two equal parts]
Δ ABD ≅Δ ACD [SSS]
∠1=∠2 [CPCT].........................(1)
Similarly, Δ PBD ≅ Δ PCD [By SSS]
∠ 3 = ∠4 [CPCT].................(2)
But, ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]
2 ∠2 + 2∠ 4=360° [using (1) and (2)]
∠2 + ∠ 4=180°
But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.
So, point A, D, P lies on a line.
CASE 2.
When ΔABC and ΔPBC lie on same side of Base BC.
In ΔPBD and ΔPCD
PB=PC[given]
PD is common.
BD =DC [Median of a triangle divides the side in two equal parts]
ΔPBD ≅ ΔPCD [SSS]
∠PDB=∠PDC [CPCT]
Similarly, By proving ΔADB≅ΔADC we will get, ∠ADB=∠ADC[CPCT]
As PD and AD are medians to same base BC of ΔPBC and ΔABC.
∴ P,A,D lie on a line i.e they are collinear.
