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A square piece of paper is folded three times along its diagonal to get an isosceles trianglewhose equal sides are 10 cm. What is the area of the unfolder original piece of paper?
- a)400 sq. cm.
- b)800 sq. cm.
- c)800√2 sq. cm.
- d)1600 sq. cm.
- e)Insufficient data to answer
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A square piece of paper is folded three times along its diagonal to g...
The smaller side of the isosceles triangle = 10
View all questions of this test
Then the side of the square = 20 and the area of the square
= 400 sq. cm.
Option A is the correct answer.
Most Upvoted Answer
A square piece of paper is folded three times along its diagonal to g...
Given:
- A square piece of paper is folded three times along its diagonal.
- The folded paper forms an isosceles triangle with two sides of length 10 cm.
To find:
- The area of the unfolded original piece of paper.
Approach:
1. Draw a diagram to visualize the given information.
2. Use the properties of isosceles triangles to find the dimensions of the unfolded paper.
3. Calculate the area of the unfolded paper using the dimensions obtained.
Solution:
Let's assume the original side length of the square paper is "x".
Folding:
1. When the paper is folded along its diagonal for the first time, it forms a right-angled triangle. The hypotenuse of this triangle is the diagonal of the square paper, which is equal to "x".
2. Folding the paper along its diagonal for the second time results in an isosceles triangle. The equal sides of this triangle are the two equal halves of the hypotenuse formed in the previous step. Therefore, each equal side of the isosceles triangle is x/2.
3. Folding the paper for the third time along its diagonal forms another isosceles triangle. The equal sides of this triangle are each half of the previous sides, i.e., (x/2)/2 = x/4.
Unfolding:
1. The unfolded original piece of paper will form a rectangle.
2. The length of the rectangle is equal to the base of the isosceles triangle, which is x/4.
3. The width of the rectangle is equal to the height of the isosceles triangle, which is obtained using the Pythagorean theorem:
- (x/2)^2 + (x/4)^2 = 10^2
- Simplifying the equation, we get: x^2/4 + x^2/16 = 100
- Combining the terms, we have: 5x^2/16 = 100
- Solving for x, we get: x^2 = 3200
- Taking the square root of both sides, we get: x = √3200 = 40√2/2 = 20√2
Calculating Area:
1. The area of the unfolded paper is equal to the product of its length and width.
2. Area = (x/4) * (x/2) = (20√2/2) * (40√2/2) = 400 sq. cm.
Therefore, the correct answer is option A) 400 sq. cm.
- A square piece of paper is folded three times along its diagonal.
- The folded paper forms an isosceles triangle with two sides of length 10 cm.
To find:
- The area of the unfolded original piece of paper.
Approach:
1. Draw a diagram to visualize the given information.
2. Use the properties of isosceles triangles to find the dimensions of the unfolded paper.
3. Calculate the area of the unfolded paper using the dimensions obtained.
Solution:
Let's assume the original side length of the square paper is "x".
Folding:
1. When the paper is folded along its diagonal for the first time, it forms a right-angled triangle. The hypotenuse of this triangle is the diagonal of the square paper, which is equal to "x".
2. Folding the paper along its diagonal for the second time results in an isosceles triangle. The equal sides of this triangle are the two equal halves of the hypotenuse formed in the previous step. Therefore, each equal side of the isosceles triangle is x/2.
3. Folding the paper for the third time along its diagonal forms another isosceles triangle. The equal sides of this triangle are each half of the previous sides, i.e., (x/2)/2 = x/4.
Unfolding:
1. The unfolded original piece of paper will form a rectangle.
2. The length of the rectangle is equal to the base of the isosceles triangle, which is x/4.
3. The width of the rectangle is equal to the height of the isosceles triangle, which is obtained using the Pythagorean theorem:
- (x/2)^2 + (x/4)^2 = 10^2
- Simplifying the equation, we get: x^2/4 + x^2/16 = 100
- Combining the terms, we have: 5x^2/16 = 100
- Solving for x, we get: x^2 = 3200
- Taking the square root of both sides, we get: x = √3200 = 40√2/2 = 20√2
Calculating Area:
1. The area of the unfolded paper is equal to the product of its length and width.
2. Area = (x/4) * (x/2) = (20√2/2) * (40√2/2) = 400 sq. cm.
Therefore, the correct answer is option A) 400 sq. cm.
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A square piece of paper is folded three times along its diagonal to get an isosceles trianglewhose equal sides are 10 cm. What is the area of the unfolder original piece of paper?a)400 sq. cm.b)800 sq. cm.c)800√2 sq. cm.d)1600 sq. cm.e)Insufficient data to answerCorrect answer is option 'A'. Can you explain this answer?
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A square piece of paper is folded three times along its diagonal to get an isosceles trianglewhose equal sides are 10 cm. What is the area of the unfolder original piece of paper?a)400 sq. cm.b)800 sq. cm.c)800√2 sq. cm.d)1600 sq. cm.e)Insufficient data to answerCorrect answer is option 'A'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A square piece of paper is folded three times along its diagonal to get an isosceles trianglewhose equal sides are 10 cm. What is the area of the unfolder original piece of paper?a)400 sq. cm.b)800 sq. cm.c)800√2 sq. cm.d)1600 sq. cm.e)Insufficient data to answerCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A square piece of paper is folded three times along its diagonal to get an isosceles trianglewhose equal sides are 10 cm. What is the area of the unfolder original piece of paper?a)400 sq. cm.b)800 sq. cm.c)800√2 sq. cm.d)1600 sq. cm.e)Insufficient data to answerCorrect answer is option 'A'. Can you explain this answer?.
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