Class 9 Exam > Class 9 Questions > If O(0, 0), A(4, 0) and B(0, 5) are the verti...
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If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB is
- a)an isosceles triangle
- b)a right triangle
- c)an obtuse triangle
- d)an equilateral triangle
Correct answer is option 'B'. Can you explain this answer?
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If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then &...
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If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then &...
Explanation:
To determine if the triangle DOAB is a right triangle, we need to check if any of its angles is a right angle (90 degrees).
Calculating the Lengths of the Sides:
To begin, we can use the distance formula to calculate the lengths of the sides of the triangle DOAB.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Using the given coordinates:
OA = sqrt((4 - 0)^2 + (0 - 0)^2) = 4
OB = sqrt((0 - 0)^2 + (5 - 0)^2) = 5
AB = sqrt((4 - 0)^2 + (5 - 0)^2) = sqrt(41)
Determining the Angles:
Next, we need to find the angles of the triangle DOAB.
Angle D is opposite to side AB and can be found using the Law of Cosines:
cos(D) = (OA^2 + OB^2 - AB^2) / (2 * OA * OB)
Substituting the values:
cos(D) = (4^2 + 5^2 - sqrt(41)^2) / (2 * 4 * 5)
cos(D) = (16 + 25 - 41) / 40
cos(D) = 0 / 40
cos(D) = 0
Since the cosine of D is 0, angle D is 90 degrees. Therefore, triangle DOAB is a right triangle.
To determine if the triangle DOAB is a right triangle, we need to check if any of its angles is a right angle (90 degrees).
Calculating the Lengths of the Sides:
To begin, we can use the distance formula to calculate the lengths of the sides of the triangle DOAB.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Using the given coordinates:
OA = sqrt((4 - 0)^2 + (0 - 0)^2) = 4
OB = sqrt((0 - 0)^2 + (5 - 0)^2) = 5
AB = sqrt((4 - 0)^2 + (5 - 0)^2) = sqrt(41)
Determining the Angles:
Next, we need to find the angles of the triangle DOAB.
Angle D is opposite to side AB and can be found using the Law of Cosines:
cos(D) = (OA^2 + OB^2 - AB^2) / (2 * OA * OB)
Substituting the values:
cos(D) = (4^2 + 5^2 - sqrt(41)^2) / (2 * 4 * 5)
cos(D) = (16 + 25 - 41) / 40
cos(D) = 0 / 40
cos(D) = 0
Since the cosine of D is 0, angle D is 90 degrees. Therefore, triangle DOAB is a right triangle.
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If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then &...
Because O lies at origin and A lies on x-axis and B lies on y-axis and if we join all the points we will get a right angled triangle. :: hence proved
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If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB isa)an isosceles triangleb)a right trianglec)an obtuse triangled)an equilateral triangleCorrect answer is option 'B'. Can you explain this answer? for Class 9 2025 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB isa)an isosceles triangleb)a right trianglec)an obtuse triangled)an equilateral triangleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 9 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If O(0, 0), A(4, 0) and B(0, 5) are the vertices of a triangle, then ΔOAB isa)an isosceles triangleb)a right trianglec)an obtuse triangled)an equilateral triangleCorrect answer is option 'B'. Can you explain this answer?.
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