1.
Find the center of the circle whose equation is x^2 + y^2 -10x + 12y -10 = 0
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Correct Ans:(5, -6)
Explanation:
For general format ofcircle: ax^{2}+by^{2}+cx+dy+e= 0
The center-radius form of the circle equation is in the format(x–h)^{2}+ (y–k)^{2}=r^{2}, with the center being at the point(h, k) and the radius being "r".
Given eqn is: x^2 + y^2 -10x + 12y -10 = 0
=> x^2 + y^2 -10x + 12y = 10
Group thex-stuff together. And theny-stuff together.
=>(x^2 - 10x) + (y^2 + 12y) = 10
Take thex-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with they-term coefficient.
=> (x^2 - 10x + 25) + (y^2 + 12y + 36) = 10 + 25 + 36
=> (x – 5)^2 + (y + 6)^2 = 71
=> (x – 5)^2 + [y – (-6)]^2 = 71
On comparing it with center-radius form of the circle equation: (x–h)^{2}+ (y–k)^{2}=r^{2}
=> The center is at (h, k) = (5, -6)
2. Find the distance between the points (2,2) and (-1,6)
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Correct Ans:5 units
Explanation:
Given, two points, (x1,y1) = (2,2)
And (x2,y2) = (-1,6)
Distance between the points = sqrt {(x2 – x1)^2 + (y2 – y1)^2}
= sqrt {(-1 – 2)^2 + (6 – 2)^2}
= sqrt {9 + 16}
= sqrt(25)
= 5
Thus, Distance between the points = 5
3. Find the distance between the points (2,3) and (3,4)
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Correct Ans:sqrt(2)
Explanation:
Given two points, (x1,y1) = (2,3)
And (x2,y2) = (3,4)
Distance between the points = sqrt {(x2 – x1)^2 + (y2 – y1)^2}
= sqrt {(3 – 2)^2 + (4 – 3)^2}
= sqrt {1 + 1}
= sqrt(2)
Thus, Distance between the points = sqrt(2)
4. Find the area of the triangle formed by the three points whose coordinates are (2, 3), (4, 5) and (6, 3).
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Correct Ans:4 sq units
Explanation:
Given, three points are
(x1,y1) = (2,3)
(x2,y2) = (4,5)
(x3,y3) = (6,3)
Then, Area of triangle = (1/2) * | (x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)|
= (1/2) * | (2x5 + 4x3 + 6x3) – (4x3 + 6x5 + 2x3)|
= (1/2) * |(10 + 12 + 18) – (12 + 30 + 6)|
= (1/2) * |40 – 48|
= (1/2) * |(-8)|
= (1/2) * 8
= 4
Therefore, Area of triangle = 4 sq.units
5. Find the slope of the line whose equation is 4y +12x - 1 = 0
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Correct Ans:-3
Explanation:
Given, eqn of line is: 4y +12x - 1 = 0
=> 4y = -12x + 1
Dividing 4 on both sides, we get
=> y = -3x + (1/4)
On comparing this eqn with the standard eqn of line: y = mx + c
Here, Slope = m = -3
6. Find the area of the triangle formed by the line 3x+2y=6, in the first coordinate.
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Correct Ans:3 sq units
Explanation:
Given, eqn of line: 3x + 2y = 6
Step 1: To find the points where this line connects with the x and y axis.
x intercept occurs when y = 0
So, subs y = 0 in 3x + 2y = 6
=> 3x + 2(0) = 6
=> 3x = 6
=> x = 2 = base of the triangle.
Now, y intercept occurs when x=0
So, subs x = 0 in 3x + 2y = 6
=> 3(0) + 2y = 6
=> 2y = 6
=> y = 3 = height of the triangle
Step 2: To find the area of the triangle
Area of the triangle = (1/2) * base * height
= (1/2) * 2 * 3
= 3 sq units
Thus, Area of the triangle = 3 sq units.
7. Find the center of the circle whose equation is x^2 + y^2 + 6x - 10y - 120 = 0
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Correct Ans:(-3 , 5)
Explanation:
For general format ofcircle: ax^{2}+by^{2}+cx+dy+e= 0
The center-radius form of the circle equation is in the format(x–h)^{2}+ (y–k)^{2}=r^{2}, with the center being at the point(h, k)and the radius being "r".
Given eqn is: x^2 + y^2 + 6x - 10y - 120 =0
=> x^2 + y^2 + 6x - 10y = 120
Group thex-stuff together. And theny-stuff together.
=>(x^2 + 6x) + (y^2 – 10y) = 120
Take thex-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with they-term coefficient.
=> (x^2 + 6x + 9) + (y^2 – 10y + 25) = 120 + 9 + 25
=> (x + 3)^2 + (y – 5)^2 = 154
On comparing it with center-radius form of the circle equation: (x–h)^{2}+ (y–k)^{2}=r^{2}
=> The center is at (h, k) = (-3, 5)
8. Which of the following points in the given options does not lie on the circle x^2 + y^2 = 10
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Correct Ans:(2,8)
Explanation:
9. Find the center of the circle whose equation is x^2+y^2-2x-6y=12
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Correct Ans:(1, 3)
Explanation:
For general format ofcircle: ax^{2}+by^{2}+cx+dy+e= 0
The center-radius form of the circle equation is in the format(x–h)^{2}+ (y–k)^{2}=r^{2}, with the center being at the point(h, k)and the radius being "r".
Given eqn is: x^2+y^2-2x-6y=12
Group thex-stuff together. And theny-stuff together.
=>(x^2 – 2x) + (y^2 – 6y) = 12
Take thex-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with they-term coefficient.
=> (x^2 – 2x + 1) + (y^2 – 6y + 9) = 12 + 1 + 9
=> (x – 1)^2 + (y – 3)^2 = 22
On comparing it with center-radius form of the circle equation: (x–h)^{2}+ (y–k)^{2}=r^{2}
=> The center is at (h, k) = (1, 3)
10. In which of the following lines, do these two point (1,3) and (2,6) lies?
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Correct Ans:y = 3x
Explanation:
Given, two points, (x1, y1) = (1,3)
(x2, y2) = (2,6)
Step 1: To find slope of the line:
Slope (m) = (y2 - y1) / (x2 - x1)
= (6-3)/ (2-1)
= 3/1
=> m = 3
Step 2: To find y intercept:
y = mx + c
Here subs, x = 1, y = 3 and m = 3
3 = 3(1) + c
=> 3 = 3 + c
=> C = 3 – 3 = 0
Step 3: Equation of line (in slope- intercept form)
y = mx + c
=> y = 3x ---> which is the correct answer.
11. Find the height of the triangle whose base is 20 with an area of 40 sq. units.
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Correct Ans:4
Explanation:
12. In triangle PQR length of the side QR is less than twice the length of the side PQ by 2 cm. Length of the side PR exceeds the length of the side PQ by 10 cm. The perimeter is 40 cm. The length of the smallest side of the triangle PQR is :
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Correct Ans:8
Explanation:
13. Given a triangle ABC, another triangle is formed by connecting the midpoints of the sides of the triangle ABC, then the area of the new triangle formed will be how many times the area of the triangle ABC
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Correct Ans:One fourth
Explanation:
14. Given triangle ABC, such that AB = AC, then ratio of the angle B to angle C = ?
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Correct Ans:1 : 1
Explanation:
15. Find the area of the triangle whose coordinates are (1, 2) , (3, 4) and (5, 10)
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Correct Ans:4
Explanation:
16. The shortest distance of the point (4,8) form the X-Axis is
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Correct Ans:8
Explanation:
17. A wall is of the form of a trapezium with height 4 m and parallel sides being 3 m and 5 m. What is the cost of painting the wall, if the rate of painting is Rs. 25/- per square metre?
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Correct Ans:400
Explanation:
18. The difference b/w two parallel sides of a trapezium is 8 cm and the perpendicular distance b/w them is 38 cm. Find the lengths of the parallel sides, if the area of the trapezium is 950 cm 2 .
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Correct Ans: 29 and 21
Explanation:
19. Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
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Correct Ans:330
Explanation:
20. Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
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Correct Ans:488
Explanation:
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