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crimeas [40]
2 years ago
6

A sports teacher had 1025 sweets. After distributing these equally amongst all the participants in a sports meet, he had just 1

sweet left with him. Which of these could be the number of sweets given away to each participant?
Mathematics
1 answer:
Cerrena [4.2K]2 years ago
5 0

Step-by-step explanation:

Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).

As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):

1) 512 => 2 partecipants

2) 256 => 4 partecipants

3) 128 => 8 partecipants

4) 64 => 16 partecipants

5) 32 => 32 partecipants

6) 16 => 64 partecipants

7) 8 => 128 partecipants

9) 4 => 256 partecipants

10) 2 => 512 partecipants

The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.

Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.

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Using the given information, length of QR = 16.6 cm

The area of the trapezium is 304.5 cm²

<h3>Calculating area of a trapezium</h3>

From the question we are to calculate the length of QR and the area of trapezium PQRS

In the given diagram,

Let the midpoint of PS be T

Then, we can write that

|OP|² = |OT|² + |PT|²  (<em>Pythagorean theorem</em>)

|OP| = radius = 13 cm

|PT| = 1/2 × PS = 1/2 × 24 cm = 12 cm

∴ 13² = |OT|² + 12²

169 = |OT|² + 144

|OT|² = 169 -144

|OT|² = 25

|OT| = √25

|OT| = 5 cm

Also, let the midpoint of QR be U

Then, we can write that

|OQ|² = |OU|² + |QU|² (<em>Pythagorean theorem</em>)

|OQ| = radius = 13 cm

From the given information,

The distance of QR from O is twice the distance of PS from O

∴ |OU| = 2 × |OT|= 2 × 5 cm = 10cm

Thus,

13² = ||² + 12²

169 = 10² + |QU|²

|QU|² = 169 -100

|QU|² = 69

|QU| = √69

|QU| = 8.3 cm

Now,

Length of QR = 2 × |QU| = 2 × 8.3 cm

Length of QR = 16.6 cm

b)

Area of the trapezium = 1/2(|QR| + |PS|) × |TU|

Area of the trapezium = 1/2(16.6 + 24) × 15

NOTE: |TU| = |OT| + |OU|

Area of the trapezium = 1/2(40.6) × 15

Area of the trapezium = 20.3 × 15

Area of the trapezium = 304.5 cm²

Hence, the area of the trapezium is 304.5 cm²

Learn more on Calculating area of trapezium here: brainly.com/question/3435635

#SPJ1

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