The area of the yellow region is 112 cm^2. find the value of x
1 answer:
<h3>Answer: x = 7</h3>
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Explanation:
The largest rectangle (composed of the green and yellow sections combined) has area of 11*12 = 132 cm^2.
The yellow region takes up 112 of those 132 sq cm. This must mean the green region takes up 132-112 = 20 cm^2.
The horizontal portion of the green rectangle is 12-x cm. The vertical portion is 11-x cm. We can form the area of the green rectangle as an algebraic expression like so
area = length*width
area = (11-x)*(12-x)
area = 132 - 11x - 12x + x^2 .... apply the FOIL rule
area = x^2 - 23x + 132
Set this equal to the 20 cm^2 we found earlier.
x^2 - 23x + 132 = 20
x^2 - 23x + 132-20 = 0
x^2 - 23x + 112 = 0
We could factor or we could use the quadratic formula. I'll go with the second option.
We'll plug in a = 1, b = -23, c = 112
One of these solutions isn't feasible. Note how if x = 16, then this exceeds both the 11 cm and 12 cm sides. So this x value is not possible.
However, x = 7 is possible.
If x = 7, then the horizontal portion of the green rectangle is 12-x = 12-7 = 5 cm. Also, the vertical portion of the green rectangle would be 11-x = 11-7 = 4 cm. The area then is length*width = 5*4 = 20 cm^2 which matches up with what we got earlier. So the answer is confirmed.
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We need to determine whether is a terminating decimal or a repeating decimal.
Let's solve this question using the long division method
First, let's identify the divisor and dividend. The number to be divided is 11 hence this is the dividend, and it needs to be divided by 128 which is the divisor
Next, since the divisor (128) is greater than the dividend (11) it can not divide 11. Hence, we will introduce a decimal point in quotient, and append a 0 next to 11 and divide 110 by 128. Again, 128 is greater than 110 so we will introduce a 0 in the quotient, and append another 0 next to 110, and will divide 1100 by 128. We will see what multiple of 128 is less than or equal to 1100. That multiple is 8. So we write 8 in the quotient and multiply 128 with 8 and subtract the product (128*8 = 1024) from 1100. The remainder that we get is 76.
Next, we append a 0 to the remainder and divide 760 by 128. Now, we see what multiple of 128 is less than or equal to 760. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 760. Now, the remainder is 120.
Next, we append a 0 to the remainder and divide 1200 by 128. Now, we see what multiple of 128 is less than or equal to 1200. That multiple is 9. So we write 9 next to the quotient and multiply 128 with 9 and subtract the product (1152) from 1200. Now, the remainder is 48.
Next, we append a 0 to the remainder and divide 480 by 128. Now, we see what multiple of 128 is less than or equal to 480. That multiple is 3. So we write 3 next to the quotient and multiply 128 with 3 and subtract the product (384) from 480. Now, the remainder is 96.
Next, we append a 0 to the remainder and divide 960 by 128. Now, we see what multiple of 128 is less than or equal to 960. That multiple is 7. So we write 7 next to the quotient and multiply 128 with 7 and subtract the product (896) from 960. Now, the remainder is 64.
Next, we append a 0 to the remainder and divide 640 by 128. Now, we see what multiple of 128 is less than or equal to 640. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 640. Now, the remainder is 0.
Hence, we have solved the entire problem
Last, we look at the quotient i.e. 0.0859375, which is the solution to the problem. We see that the quotient has a definite number of digits in it, and terminates at 5. Hence, this is a terminating decimal.
A repeating decimal is one in which a particular pattern after the decimal point keeps re-occuring, which is not the case here. Hence, is a terminating decimal.
Please refer to the attached image for visualization
