Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Geometry – Section Formula

Mathematics Multiple Choice Questions & Answers on “Geometry – Section Formula”.

1. What will be the coordinates of the point which divides the line segment joining the points A(-2, 2) and B(-1, 5) in the ratio 2:5?
a) ((frac {-4}{3}, frac {-20}{9}))
b) ((frac {-4}{3}, frac {20}{9}))
c) ((frac {4}{3}, frac {20}{9}))
d) ((frac {4}{3}, frac {-20}{9}))
Answer: b
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-2, 2) and B(-1, 5) and the ratio is 2:5
∴ x = (frac {2(-1)+5(-2)}{2+7} = frac {-2-10}{9} = frac {-12}{9} = frac {-4}{3})
y = (frac {2(5)+5(2)}{2+7} = frac {10+10}{9} = frac {20}{9} = frac {20}{9})
Hence, the point is ((frac {-4}{3}, frac {20}{9})).

2. What will be the coordinates of the midpoint of the line segment joining the points (-5, 10) and(15, 2)?
a) (-5, -6)
b) (-5, 6)
c) (5, 6)
d) (5, -6)
Answer: c
Clarification: Midpoint lies in the center of the line segment
Hence, it divides the line in the ratio 1:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-5, 10) and B(15, 2) and the ratio is 1:1
∴ x = (frac {1(-5)+1(15)}{2} = frac {-5+15}{2} = frac {10}{2}) = 5
y = (frac {1(2)+1(10)}{2} = frac {2+10}{2} = frac {12}{2}) = 6
Hence, the point is (5,6).

3. In what ratio does the point ((frac {-19}{3}, frac {7}{3})) divide the line segment joining A(3, 7) and B(-11, 0)?
a) 1:2 (externally)
b) 1:2 (internally)
c) 2:1 (externally)
d) 2:1 (internally)
Answer: d
Clarification: Let the ratio in which the point ((frac {-19}{3}, frac {7}{3})) divides the line segment joining the points A(3, 7) and B(-11, 0) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(3, 7) and B(-11, 0) and the ratio is k:1
∴ x = (frac {k(-11)+1(3)}{k+1} = frac {-11k+3}{k+1})
y = (frac {k(0)+1(7)}{k+1} = frac {7}{k+1})
Since, the point is ((frac {-19}{3}, frac {7}{3})).
∴ (frac {-19}{3} = frac {-11k+3}{k+1})
-19(k + 1) = 3(-11k + 3)
-19k – 19 = -33k + 9
-19k + 33k = 19 + 9
14k = 28
k = (frac {28}{14}) = 2
The ratio is 2:1.

4. What will be the value of y, if the ratio in which the point ((frac {3}{4}), y) divides the line segment joining the points A(-1, 4) and B(6, 5)is 1:3?
a) y = (frac {9}{2})
b) y = (frac {5}{2})
c) y = (frac {9}{4})
d) y = (frac {5}{2})
Answer: a
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are (-1, 4)and B(6, 5)in the ratio 1:3
∴ x = (frac {1(6)+3(-1)}{1+3} = frac {6-3}{4} = frac {3}{4})
y = (frac {1(6)+3(4)}{1+3} = frac {6+12}{4} = frac {18}{4})
Therefore y = (frac {9}{2})

5. What will be ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4)?
a) 1:2 (internally)
b) 1:2 (externally)
c) 2:1 (externally)
d) 2:1 (internally)
Answer: b
Clarification: Let the ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4) be k:1.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(0, -1) and B(-3, -4) and the ratio is k:1.
∴ x = (frac {k(-3)+1(0)}{k+1} = frac {-3k}{k+1})
y = (frac {k(-4)+1(-1)}{k+1} = frac {-4k-1}{k+1})
Since, the point ((frac {-3k}{k+1}, frac {-4k-1}{k+1} )) lies on the line 3x+y-11 = 0.
3 ((frac {-3k}{k+1} + frac {-4k-1}{k+1} ))-11 = 0
3(-3k) + (-4k – 1) – 11(k + 1) = 0
-9k – 4k – 1 – 11k – 11 = 0
-24k – 12 = 0
-24k = 12
k = (frac {12}{-24} = frac {-1}{2})
The ratio is 1:2 (externally).

6. In what ratio is the line segment joining the points A(-5, 2) and B(3, 9) divided by the x-axis?
a) 2:5 (internally)
b) 2:5 (externally)
c) 2:9 (externally)
d) 2:9 (internally)
Answer: c
Clarification: Let the ratio in which the x-axis divides the line segment joining the points A(-5, 2) and B(3, 9) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-5, 2) and B(3, 9) and the ratio is k:1
∴ x = (frac {k(3)+1(-5)}{k+1} = frac {3k-5}{k+1})
y = (frac {k(9)+1(2)}{k+1} = frac {9k+2}{k+1})
Since, the point is on x-axis.
Hence, the y-coordinate will be zero.
∴ 0 = (frac {9k+2}{k+1})
0 = 9k+2
k = (frac {-2}{9})
The ratio in which the y-axis cuts the line segment joining the points A(-5, 2) and B(3, 9) will be 2:9 (externally).

7. In what ratio is the line segment joining the points A(2, 4) and B(6, 5) divided by the y-axis?
a) 2:1 (internally)
b) 2:1 (externally)
c) 3:1 (internally)
d) 3:1 (externally)
Answer: d
Clarification: Let the ratio in which the y-axis divides the line segment joining the points A(2, 4) and B(6, 5) be k:1
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(2, 4) and B(6, 5) and the ratio is k:1
∴ x = (frac {k(6)+1(2)}{k+1} = frac {6k+2}{k+1})
y = (frac {k(5)+1(4)}{k+1} = frac {5k+4}{k+1})
Since, the point is on y-axis.
Hence, the x-coordinate will be zero.
∴ 0 = (frac {6k+2}{k+1})
0 = 6k + 2
k = (frac {-6}{2}) = -3
The ratio in which the y-axis cuts the line segment joining the points A(2, 4) and B(6, 5) will be 3:1 (externally).

8. What will be the coordinates of B, if the point C((frac {29}{7}, frac {46}{7} )), divides the line segment joining A (5, 8) and B (a, b) in the ratio 2:5?
a) a = 2, b = 3
b) a = -2, b = 3
c) a = 2, b = -3
d) a = -2, b = -3
Answer: a
Clarification: Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(5, 8)and B(a, b)in the ratio 2:5
∴ x = (frac {2(a)+5(5)}{2+5} = frac {2a+25}{7})
y = (frac {2(b)+5(8)}{2+5} = frac {2b+40}{7})
But the coordinates of C are ((frac {29}{7}, frac {46}{7} ))
Therefore, (frac {2a+25}{7} = frac {29}{7})
a = 2
(frac {2b+40}{7} = frac {46}{7})
b = 3

9. What will be the length of the median through the vertex A, if the coordinates of the vertices of ∆ABC are A(2, 5), B(5, 0), C(-2, 5)?
a) (sqrt {frac {113}{3}}) units
b) (sqrt {frac {13}{2}}) units
c) (sqrt {frac {113}{2}}) units
d) (sqrt {frac {13}{2}}) units
Answer: b
Clarification:

The median through A will bisect the line BC.
Hence, D is the midpoint of BC
Coordinates of D = ((frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {5-2}{2}, frac {0-5}{2} ) =( frac {3}{2}, frac {-5}{2} ))
Distance between A and D = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2-frac {3}{2})^2+ (5+frac {5}{2})^2} )
= ( sqrt {(frac {1}{2})^2+ (frac {15}{2})^2} )
= ( sqrt {frac {1}{4}+ frac {225}{4}} )
= ( sqrt {frac {113}{2}} ) units

10. What will be the coordinates of the fourth vertex S, if P(-1, -1), Q(2, 0), R(2, 3) are the three vertices of a parallelogram?
a) (-5, -12)
b) (5, -12)
c) (5, 12)
d) (-5, 12)
Answer: c
Clarification:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.
∴ O is the mid-point SQ and PR.
Midpoint of PR
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are P(-1, -1) and R(2, 3) and the ratio is 1:1
∴ x = (frac {1(-1)+1(2)}{2} = frac {-1+2}{2} = frac {1}{2})
y = (frac {1(3)+1(-1)}{2} = frac {3-1}{2} = frac {2}{2}) = 1
Hence, the coordinates of O is (5, 6)
Midpoint of QS
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are Q(2, 0) and S(a, b) and the ratio is 1:1
∴ x = (frac {1(a)+1(2)}{2} = frac {a+2}{2})
y = (frac {1(b)+1(0)}{2} = frac {b}{2})
The coordinates of O is (5, 6)
Therefore, (frac {a+2}{2}) = 5
a = 8
(frac {b}{2}) = 6, b = 12
The coordinates of S are (5, 12).

11. What will be the value of a and b, if (-5, a), (-3, -3), (-b, 0) and (-3, 3) are the vertices of the parallelogram?
a) a = 0, b = -1
b) a = -1, b = 1
c) a = 1, b = 1
d) a = 0, b = 1
Answer: d
Clarification:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.
∴ O is the mid-point SQ and PR.
Midpoint of PR
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are P(-5, a) and R(-b, 0) and the ratio is 1:1
∴ x = (frac {1(-b)+1(-5)}{2} = frac {-b-5}{2})
y = (frac {1(0)+1(a)}{2} = frac {a}{2})
Midpoint of QS
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are Q(-3, -3) and S(-3, 3) and the ratio is 1:1
∴ x = (frac {1(-3)+1(-3)}{2} = frac {-6}{2}) = -3
y = (frac {1(3)+1(-3)}{2} = frac {0}{2}) = 0
Therefore, (frac {-b-5}{2}) = -3
b = 1
(frac {a}{2}) = 0
a = 0

12. What will be the centroid of the ∆ABC whose vertices are A(-2, 4), B(0, 0) and C(4, 2)?
a) ((frac {2}{3}), 2)
b) ((frac {2}{3}), 1)
c) ((frac {2}{5}), 2)
d) ((frac {1}{3}), 2)
Answer: a
Clarification: We know, xcentroid = (frac {x_1+x_2+x_3}{3}) and ycentroid = (frac {y_1+y_2+y_3}{3})
xcentroid = (frac {-2+0+4}{3} = frac {2}{3})
ycentroid = (frac {4+0+2}{3}) = 2
The coordinates of the centroid are ((frac {2}{3}), 2).

13. The coordinates of one end of the diameter AB of a circle are A (-2, -3) and the coordinates of diameter are (-2, 0). What will be the coordinates of B?
a) (2, -3)
b) (-2, 3)
c) (2, 3)
d) (-2, -3)
Answer: b
Clarification: We know that the diameter is twice the radius.
Hence, the center is the midpoint of the diameter.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-2, -3) and center is (-2, 0) and the ratio is 1:1
Let the coordinates of other side of the radius be (x, y).
∴ -2 = (frac {1(-2)+1(x)}{2} = frac {-2+x}{2})
-4 = -2 + x
-4 + 2 = x
x = -2
0 = (frac {1(-3)+1(y)}{2} = frac {-3+y}{2})
0 = -3 + y
y = 3
Hence, the point is (-2, 3).

14. The coordinates of the ends of the diameter AB of a circle are A (-4, 7) and B(4, 7). What will be the coordinates of the center of the circle?
a) (0, -8)
b) (0, 8)
c) (0, 7)
d) (0, -7)
Answer: c
Clarification: We know that the diameter is twice the radius.
Hence, the center is the midpoint of the diameter.
Using, section formula x = (frac {mx_2+nx_1}{m+n}) and y = (frac {my_2+ny_1}{m+n})
The points are A(-4, 7) and B(4, 7) and the ratio is 1:1
∴ x = (frac {1(-4)+1(4)}{2} = frac {0}{2}) = 0
y = (frac {1(7)+1(7)}{2} = frac {7+7}{2} = frac {14}{2}) = 7
Hence, the point is (0, 7).

15. The two vertices of ∆ABC are given by A(-3, 0) and B(-8, 5) and its centroid is (-2, 1).What will be the coordinates of the third vertex C?
a) (-5, -2)
b) (5, 2)
c) (-5, 2)
d) (5, -2)
Answer: d
Clarification: The two vertices of triangle are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).
We know, xcentroid = (frac {x_1+x_2+x_3}{3}) and ycentroid = (frac {y_1+y_2+y_3}{3})
Now, xcentroid = (frac {-3-8+x_3}{3})
xcentroid = -2
-2 = (frac {-3 – 8 + x_3}{3})
-6 = -3 – 8 + x3
5 = x3
Now, ycentroid = (frac {0+5+y_3}{3})
ycentroid = 1
1 = (frac {0 + 5 + y_3}{3})
3 = 5 + y3
-2 = y3
The third coordinate is (5, -2).

[CLASS 10] Mathematics MCQs on Geometrical Meaning of Zeros of Polynomial

Mathematics Multiple Choice Questions for Schools on “Geometrical Meaning of Zeros of Polynomial”.

1. The graph of the polynomial 4x2-8x+3 cuts the x-axis at ________ and ________ points.
a) ((frac {3}{4}), 0), ((frac {1}{2}), 0)
b) ((frac {3}{2}), 0), ((frac {1}{2}), 0)
c) ((frac {3}{2}), 0), ((frac {1}{6}), 0)
d) ((frac {7}{2}), 0), ((frac {3}{2}), 0)

Answer: b
Clarification: The graph of the polynomial cuts the x-axis. Only the zeros of the polynomial cut the x-axis.
4x2-8x+3=0
4x2-6x-2x+3=0
2x(2x-3)-1(2x-3)=0
(2x-3)(2x-1)=0
x=(frac {3}{2}, frac {1}{2})
Hence, the graph of the polynomial cuts the x-axis at ((frac {3}{2}), 0) and ((frac {1}{2}), 0).

2. The graph of the polynomial 2x2-8x+5 cuts the y-axis at __________
a) (6, 0)
b) (0, 7)
c) (0, 5)
d) (8, 9)

Answer: c
Clarification: The graph of the polynomial 2x2-8x+5 cuts the y-axis.
Hence, the value of x will be 0.
y(0)=2(0)2-8(0)+5
y=5
The graph cuts the y-axis at (0,5)

3. How many points will the graph of x2+2x+1 will cut the x-axis?
a) 3
b) 1
c) 2
d) 0

Answer: d
Clarification: The graph of x2+2x+1 does not cut the x-axis, because it has imaginary roots.
x2+2x+1=0
x2+x+x+1=0
x(x+1)+(x+1)=0
(x+1)(x+1)=0
x=-1, -1

4. The graph of the quadratic polynomial -x2+x+90 will open upwards.
a) False
b) True

Answer: a
Clarification: The graph of the polynomial will have a downward opening since, a
The graph for the same can be observed here,
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5. If the graph of a polynomial cuts the x-axis at 3 points, then the polynomial is ______
a) Linear
b) Quadratic
c) Cubic
d) Biquadratic

Answer: c
Clarification: Since, the graph of the polynomial cuts the x-axis at 3 points, hence, it will be a cubic polynomial. A polynomial is said to be linear, quadratic, cubic or biquadratic according to the degree of the polynomial.

6. What will be the nature of the zeros of a quadratic polynomial if it cuts the x-axis at two different points?
a) Real
b) Distinct
c) Real, Distinct
d) Complex

Answer: c
Clarification: The zeros of the quadratic polynomial cut the x-axis at two different points.
∴ b2 – 4ac ≥ 0
Hence, the nature of the zeros will be real and distinct.

7. The graph of a quadratic polynomial cuts the x-axis at only one point. Hence, the zeros of the quadratic polynomial are equal and real.
a) True
b) False

Answer: a
Clarification: If the graph meets x-axis at one point only, then the quadratic polynomial has coincident zeros. Also, the discriminant of the quadratic polynomial is zero, therefore roots will be real.

8. A real number is called zeros of the polynomial p(x) if _________
a) p(α)=4
b) p(α)=1
c) p(α)≠0
d) p(α)=0

Answer: d
Clarification: A number is called zero of polynomial when it satisfies the equation of the polynomial.

9. If a < 0, then the graph of ax2+bx+c, has a downward opening.
a) True
b) False

Answer: a
Clarification: The leading coefficient of the polynomial is less than zero, hence, it has downward opening. For example, the graph of -x2 is
” alt=”” width=”521″ height=”385″ data-src=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9.png” data-srcset=”2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9.png 521w, 2020/12/mathematics-questions-answers-geometrical-meaning-zeros-polynomial-q9-300×222.png 300w” data-sizes=”(max-width: 521px) 100vw, 521px” />

10. A polynomial is said to be linear, quadratic, cubic or biquadratic according to the degree of the polynomial.
a) False
b) True

Answer: b
Clarification: The degree of the polynomial is the highest of the degree of the polynomial. Hence, a polynomial with highest degree one is linear, two as quadratic and so on.

11. Which of the following is a polynomial?
a) x2+2x+5
b) √x+2x+4
c) x(frac {2}{3})+10x
d) 5x+(frac {5}{x})

Answer: a
Clarification: An expression in the form of (x)=a0+a1x+a2x2+…+anxn, where an≠0, is called a polynomial where a1, a2 … an are real numbers and each power of x is a non-negative integer.
In case of √x+2x+4 , the power of √x is not an integer. Similarly for x(frac {2}{3})+10x, (frac {2}{3}) is a fraction.
Now, 5x+(frac {5}{x}) in this case the power of x is a negative integer. Hence it is not a polynomial.

12. The biquadratic polynomial from the following is ______
a) (x2+3)(x2-3)
b) x2-7
c) x7+x6+x5
d) 5x-3

Answer: a
Clarification: A biquadratic polynomial has highest power 4.
Hence, the polynomial with the highest power as 4 is x4-9 or (x2+3)(x2-3).

13. Which of the following is not a polynomial?
a) x2+5x+10
b) √x+2x+4
c) x10+10x
d) 5x+4

Answer: b
Clarification: An expression in the form of (x)=a0+a1x+a2x2+…+anxn, where an≠0, is called a polynomial where a1, a2 … an are real numbers and each power of x is a non-negative integer.
In case of √x+2x+4, the power of x is not an integer.
Therefore it is not a polynomial.

14. If the zeros of a polynomial are 3 and -5, then they cut the x-axis at ____ and _____ points.
a) (8, 0) and (-4, 0)
b) (3, -3) and (-5, 5)
c) (-3, 0) and (5, 0)
d) (3, 0) and (-5, 0)

Answer: d
Clarification: Since, the zeros of the polynomial are 3 and -5.
Therefore, x = 3 and x = -5 and they cut the x-axis so the y-coordinate will be zero.
Hence, the points it cuts the x-axis will be (3, 0) and (-5, 0).

15. If the graph of the quadratic polynomial is completely above or below the x-axis, then the nature of roots of the polynomial is _____
a) Real and Distinct
b) Distinct
c) Real
d) Complex

Answer: d
Clarification: Since, the graph is completely above or below the x-axis, hence, it has no real roots. If a polynomial has real roots only then it cuts the x-axis. If it lies above or below, the roots are complex in nature.

[CLASS 10] Mathematics MCQs on Tangent to a Circle

Mathematics Multiple Choice Questions & Answers on “Tangent to a Circle”.

1. Identify the parts of a circle.
a) Sides
b) Chord
c) Corners
d) Faces
Answer: b
Clarification: Chord is one of the parts of a circle. A line segment that joins any two points on a circle is called a chord. The other parts of a circle are arc, segment, secant, radius, diameter, etc.

2. Choose the correct statement.
a) The center of the circle belongs to the circle
b) The angle in a semi-circle is a complete angle
c) A circle can have only two parallel tangents
d) Radius equals to twice of the diameter
Answer: c
Clarification: A circle can have only two parallel tangents. These two parallel tangents touch either end of the diameter. Hence, ‘A circle can have only two parallel tangents’ is the only correct statement.

3. The angle in a semi-circle is _____
a) 90°
b) 180°
c) 0°
d) 360°
Answer: b
Clarification: The angle in a semi-circle is 180° whereas the angle in a circle is a complete angle that is 360° and an angle is said to be 0° when the circle is in the form of a point.

4. Length of the tangent is _____
a) (sqrt {d^2+c^2})
b) (sqrt {r^2-d^2})
c) (sqrt {d^2-c^2})
d) (sqrt {d^2-r^2})
Answer: d
Clarification: Length of the tangent (l) = (sqrt {d^2-r^2})
Where, l = Length of the tangent.
d = Distance between center of the circle and the external point of the circle.
r = Radius of the circle.

5. The line intersecting a circle at exactly one point is called as _____
a) arc
b) secant
c) chord
d) tangent
Answer: d
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and an infinite number of tangents are drawn to a circle whereas secant is a line that intersects two distinct points on a circle.

6. What is the name of the point which is common to circle and the tangent?
a) Intersection point
b) Tangential point
c) Point of contact
d) Point of touch
Answer: c
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and the point which is common for both circle and tangent is called the point of contact.

7. The radius is always _____ to the tangent.
a) equal
b) perpendicular
c) twice
d) parallel
Answer: b
Clarification: Radius is measured from the center of the circle to any point on the circle and it is always perpendicular to the tangent drawn at their common point of contact.

8. Find the length of the tangent if d = 5 cm and r = 4 cm.
a) 3 cm
b) 4 cm
c) 5 cm
d) 9 cm
Answer: a
Clarification: The length of the tangent = (sqrt {d^2-r^2})
= (sqrt {5^2-4^2})
= √9
= 3

9. Find the radius of a circle if 8 m is the length of the tangent, 11 m is the distance between the center of the circle the external point.
a) 7 m
b) 5 m
c) √57 m
d) √58 m
Answer: c
Clarification: Length of the tangent = (sqrt {d^2-r^2})
8 = (sqrt {11^2-r^2})
r2 = 112 – 82
r = (sqrt {11^2-8^2})
r = (sqrt {121-64})
r = √57

10. Identify the tangents in the circle.

a) Only A
b) A and B
c) Only C
d) A and C
Answer: c
Clarification: The line only C is the tangent of the given circle because line C is the only line that touches the given circle at only a single point and the lines A, B are the secants of the circle because they touch the circle in two distinct points.

11. Identify the secants in the circle.

a) Only A
b) A and B
c) Only C
d) A and C
Answer: b
Clarification: The lines A and B are the secants of the circle because they touch the circle in two distinct points whereas line C is the only tangent of the circle because it intersects the circle at only a single point.

12. Identify the type of tangents.

a) Parallel tangents
b) Concurrent tangents
c) Perpendicular tangents
d) Equal tangents
Answer: a
Clarification: The type of tangents l and m are parallel tangents because they both are parallel to each other and a circle can have only two parallel tangents.

13. Identify the type of tangents.

a) Parallel tangents
b) Concurrent tangents
c) Perpendicular tangents
d) Equal tangents
Answer: c
Clarification: The type of tangents A and B are perpendicular tangents because they both are perpendicular to each other. The formation of 90° between both the tangents makes them a pair of perpendicular tangents.

14. Find the length of the tangent.

a) 10 cm
b) 11 cm
c) √197 cm
d) √199 cm
Answer: d
Clarification: The length of the tangent = (sqrt {d^2-r^2})
= (sqrt {12^2-5^2})
= (sqrt {144-25})
= √199

15. Find the area of the circle if 8 cm is the length of the tangent, 11 cm is the distance between the center of the circle the external point.
a) 100 cm
b) 110 cm
c) 197.14 cm
d) 179.14 cm
Answer: d
Clarification: Length of the tangent = (sqrt {d^2-r^2})
8 = (sqrt {11^2-r^2})
r2 = 112 – 82
r = (sqrt {11^2-8^2})
r = √57
Area of the circle = πr2
= (frac {22}{7})(√57)2
= 179.14

[CLASS 10] Mathematics MCQs on Geometry – Distance Formula

Mathematics Multiple Choice Questions & Answers on “Geometry – Distance Formula”.

1. The distance between the points (5, 7) and (8, -5) is ________
a) √153
b) √154
c) √13
d) √53
Answer: a
Clarification: Using distance formula,
Distance between (5, 7) and (8, -5) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(8-5)^2 + (-5-7)^2})
= ( sqrt {(3)^2 + (-12)^2} )
= ( sqrt {9 + 144})
= √153

2. The distance of the point (9, -12) from origin will be ___________
a) 13
b) 15
c) 14
d) 17
Answer: b
Clarification: Distance between (9, -12) and (0, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0-9)^2 + (0 + 12)^2} )
= ( sqrt {(9)^2 + (-12)^2} )
= ( sqrt {81 + 144})
= √225 = 15

3. What will be the value of x, if the distance between the points (5, 11) and (2, x) is 10?
a) -11 + √91, -11 – √91
b) 11 + √91, 11 – √91
c) 11 + √91, 11 + √91
d) -11 + √91, 11 – √91
Answer: b
Clarification: Distance between (5, 11) and (2, x) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2-5)^2 + (x-11)^2} )
= ( sqrt {x^2-22x + 121 + (-3)^2} )
= ( sqrt {x^2-22x + 121 + 9} )
= ( sqrt {x^2-22x + 130} )
The distance between (5, 11) and (2, x) is 10
∴ ( sqrt {x^2-22x + 130} ) = 10
Squaring on both sides we get,
x2 – 22x + 130 = 100
x2 – 22x + 130 – 100 = 0
x2 – 22x + 30 = 0
x = 11 + √91, 11 – √91

4. What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?
a) (10, 0)
b) (13, 0)
c) (11, 0)
d) (12, 0)
Answer: c
Clarification: Let the point on x-axis be (x, 0)
Distance between (9, 8) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-9)^2 + (0-8)^2} )
= ( sqrt {x^2-18x + 81 + (-8)^2} )
= ( sqrt {x^2-18x + 81 + 64} )
= ( sqrt {x^2-18x + 145} )
Distance between (3, 2) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-3)^2 + (0-2)^2} )
= ( sqrt {x^2-6x + 9 + (-2)^2} )
= ( sqrt {x^2-6x + 9 + 4} )
= ( sqrt {x^2-6x + 13} )
Since, the point ( x, 0) is equidistant to (3, 2) and (9, 8)
The distances will be equal
∴ ( sqrt {x^2-18x + 145} = sqrt {x^2-6x + 13} )
Squaring on both sides we get,
x2 – 18x + 145 = x2 – 6x + 13
-18x + 145 = -6x + 13
-18x + 6x = -145 + 13
-12x = -132
x = ( frac {132}{12} ) = 11
The point is (11, 0)

5. What will be the point of y-axis which will be equidistant from the points (-1, 0) and (3, 9)?
a) (5, ( frac {89}{18} ))
b) (1, ( frac {89}{18} ))
c) (16, ( frac {89}{18} ))
d) (0, ( frac {89}{18} ))
Answer: d
Clarification: Let the point on y-axis be (0, y)
Distance between (-1, 0) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0 + 1)^2 + (y-0)^2} )
= ( sqrt {y^2 + (1)^2} )
= ( sqrt {y^2 + 1} )
Distance between (3, 9) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0-3)^2 + (y-9)^2} )
= ( sqrt {y^2-18y + 81 + (-3)^2} )
= ( sqrt {y^2-18y + 81 + 9} )
= ( sqrt {y^2-18y + 90} )
Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ ( sqrt {y^2 + 1} = sqrt {y^2-18y + 90} )
Squaring on both sides we get,
y2 + 1 = y2 – 18y + 90
1 – 90 = -18y
-89 = -18y
y = ( frac {89}{18} )
The point is (0, ( frac {89}{18} ))

6. If the point P(a, b) is equidistant from the points (3, 1) and (2, 0) then ____________
a) a + b = -3
b) a – b = -3
c) a + b = 3
d) a – b = 3
Answer: a
Clarification: The point is (a, b)
Distance between (3, 1) and (a, b) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(a-3)^2 + (b-1)^2} )
= ( sqrt {a^2-6a + 9 + b^2-2b + 1} )
= ( sqrt {a^2-6a + 10 + b^2-2b} )
Distance between (2, 0) and (a, b) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(a-2)^2 + (b-0)^2} )
= ( sqrt {a^2-4a + 4 + b^2 } )
Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ ( sqrt {a^2-6a + 10 + b^2-2b} = sqrt {a^2-4a + 4 + b^2 } )
Squaring on both sides we get,
a2 – 6a + 10 + b2 – 2b = a2 – 4a + 4 + b2
-6a + 10 – 2b = -4a + 4
-2a – 6 = 2b
-a – b = 3
a + b = -3

7. The point on y-axis which is at a distance 5 unit from the point (-5, 7) is ___________
a) (7, 0)
b) (0, 7)
c) (1, 7)
d) (7, 7)
Answer: b
Clarification: Let the point on y-axis be (0, y)
Distance between (-5, 7) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0 + 5)^2 + (y-7)^2} )
= ( sqrt {y^2-14y + 49 + 25} )
= ( sqrt {y^2-14y + 74} )
The distance between (-5, 7) and (0, y) is 5
∴ ( sqrt {y^2-14y + 74} ) = 5
Squaring on both sides, we get,
y2 – 14y + 74 = 25
y2 – 14y + 49 = 0
y = 7, 7
Hence, the point is (0, 7)

8. The point on x-axis which is at a distance 12 unit from the point (4, 6) is ___________
a) (-4 + ( sqrt {11i} ), 0), (-4 – ( sqrt {11i} ), 0)
b) (-4 – ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
c) (4 – ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
d) (4 + ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
Answer: d
Clarification: Let the point on x-axis be (x, 0)
Distance between (4, 6) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-4)^2 + (0-6)^2} )
= ( sqrt {x^2-8x + 16 + 36} )
= ( sqrt {x^2-8x + 52} )
The distance between (4, 6) and (x, 0) is 12
∴ ( sqrt {x^2-8x + 52} ) = 12
Squaring on both sides, we get,
x2 – 8x + 52 = 25
x2 – 8x + 27 = 0
x = 4 + ( sqrt {11i} ), 4 – ( sqrt {11i} )

9. If A(0, 3), B(5, 0) and C(-5, 0) are the vertices of ∆ABC, then the triangle is __________
a) Right-angled
b) Isosceles
c) Scalene
d) Equilateral
Answer: b
Clarification: Distance between (0, 3) and (5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(5-0)^2 + (0-3)^2} )
= ( sqrt {5^2 + -3^2 } )
= ( sqrt {25 + 9} )
= √34
Distance between (5, 0) and (-5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-5-5)^2 + (0-0)^2} )
= ( sqrt {-10^2} )
= 10
Distance between (0, 3) and (-5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-5-0)^2 + (0-3)^2} )
= ( sqrt {-5^2 + (-3)^2} )
= ( sqrt {25 + 9} )
= √34
Since, the two sides of the triangle are equal.
Hence, the triangle will be isosceles triangle.

10. The area of the triangle if A (-1, -1), B(-1, 3) and C (2, -1) are the vertices of the triangle is ____________
a) 8 units
b) 4 units
c) 6 units
d) 5 units
Answer: c
Clarification: Distance between A (-1, -1) and B (-1, 3) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-1 + 1)^2 + (3 + 1)^2} )
= ( sqrt {4^2} )
= √16
= 4
Distance between B (-1, 3) and C(2, -1) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2 + 1)^2 + (-1-3)^2} )
= ( sqrt {3^2 + (-4)^2} )
= ( sqrt {9 + 16} )
= 5
Distance between A (-1, -1) and C (2, -1) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2 + 1)^2 + (-1 + 1)^2} )
= ( sqrt {3^2 + 0^2} )
= 3
Now, AC2 + AB2 = 42 + 32 = 16 + 9 = 25
BC2 = 52 = 25
Hence, it is a right-angled triangle, right-angled at A.
Area of triangle = ( frac {1}{2}) × base × height = ( frac {1}{2}) × 4 × 3 = 6 units

[CLASS 10] Mathematics MCQs on Irrational and Rational Numbers

Mathematics Multiple Choice Questions & Answers on “Irrational and Rational Numbers”.

1. If x is a number whose simplest form is (frac {p}{q}), where p and q are integers and q≠0, then x is a terminating decimal only when q is of the form _________
a) 3m×5n
b) 2m×6n
c) 2m×5n
d) 7m×5n
Answer: b
Clarification: Let’s, take a number where q is of the form 2m×5n, say 250×510and p can be any integer
(frac {p}{2^{50}times 5^{10}} = frac {p times 5^{40}}{10^{50}})
The number (frac {p times 5^{40}}{10^{50}}) will terminate after 50 decimal places.
Hence, if q is of the form 2m×5n, it will terminate after some decimal places.

2. If x is a number whose simplest form is (frac {p}{q}), where p and q are integers and q ≠ 0, then x is a non-terminating repeating decimal only when q is not of the form ________
a) 2m×2n
b) 5m×5n
c) 2m×5n
d) 3m×4n
Answer: c
Clarification: Let’s, take a number where q is of the form 2m×5n, say 250×510and p can be any integer
(frac {p}{2^{50}times 5^{10}} = frac {p times 5^{40}}{10^{50}})
The number (frac {p times 5^{40}}{10^{50}}) will terminate after 50 decimal places.
Hence, if q is of the form 2m×5n, it will terminate after some decimal places.

3. Which of the following rational is non-terminating repeating decimal?
a) 0.25
b) (frac {4}{5})
c) (frac {4}{55})
d) (frac {2}{5})
Answer: c
Clarification: The value of (frac {4}{55}) is 0.07272727272…., which is non-terminating repeating decimal.
The other numbers terminate after few places of decimal.

4. The terminating rational number from the following numbers is _________
a) (frac {4}{9})
b) (frac {4}{3})
c) 0.146
d) (frac {4}{5})
Answer: d
Clarification: The value of (frac {4}{5}) is 0.8, which is terminating decimal.

5. The simplest form of the rational number 0.196 is ________
a) (frac {1}{6})
b) (frac {3}{6})
c) (frac {13}{66})
d) (frac {2}{5})
Answer: c
Clarification:
10x = 1.969696…..(1)
1000x = 196.9696…(2)
Subtracting (1) from (2)
We get,
990x=195
x = (frac {195}{990} = frac {13}{66})

6. The numbers of the form (frac {p}{q}) are integers, and q≠0 are called irrational number.
a) True
b) False
Answer: b
Clarification:
Irrational numbers cannot be written in the form of (frac {p}{q}).
For example, ∛4 cannot be written in a fraction form as it has non-terminating and non-repeating decimals.

7. After how many places of decimal, will the decimal expansion of the rational number (frac {57}{2^4 5^6}) terminate?
a) 4
b) 6
c) 7
d) 8
Answer: b
Clarification:
We have,
(frac {57}{2^4 5^6} = frac {57 times 2^2}{2^6 5^6} = frac {228}{10^6}) = 0.000228
The number (frac {57}{2^4 5^6}) will terminate after 6 decimal places.

8. From the following numbers, which number is not a rational number?
a) π
b) (frac {22}{7})
c) (frac {3}{4})
d) 0.666666…..
Answer: a
Clarification: A rational number has terminating or non-terminating but repeating decimals.
In case of π, it has a non-terminating as well as non-repeating decimal.
The other three numbers have terminating or non-terminating but repeating decimal, therefore, they are rational numbers.
Hence, it is an irrational number.

9. An irrational number has ________
a) Non-terminating decimal
b) Non-repeating decimal
c) Non-terminating and non-repeating decimal
d) Terminating decimal
Answer: c
Clarification: An irrational number has both non-terminating as well as non-repeating decimals.
For example, the number 1.353353335… has non-terminating as well as non-repeating decimals.

10. Which of the following numbers is not an irrational number?
a) π
b) (frac {22}{7})
c) 1.5353353335….
d) 2.7878878887….
Answer: b
Clarification:
An irrational number is expressible in the decimal form as non-terminating and non-repeating decimals.
From the given options,
π, 1.5353353335…, 2.7878878887… are non-terminating and non-repeating decimal.
Whereas, (frac {22}{7}) is non-terminating but repeating decimal.

11. The product of (frac {33}{2}) and (frac {5}{4}) is an irrational number.
a) True
b) False
Answer: b
Clarification:
(frac {33}{2} times frac {5}{4} = frac {165}{8})
(frac {165}{8}) is a rational number

12. The product of a rational and an irrational number is rational number.
a) True
b) False
Answer: b
Clarification: Take a rational and an irrational number, say 2 and 3√3
Product of 2 × 3√3 = 6√3.
6√3 is an irrational number
Hence, the product of a rational and an irrational number is a irrational number.

13. The product of two irrational numbers is an irrational number.
a) True
b) False
Answer: b
Clarification: Consider an irrational number, say √10
√10 × √10=10
10 is a rational number. Hence, the product of two irrational numbers is not always irrational.

14. The sum of two rational numbers is a rational number.
a) False
b) True
Answer: b
Clarification: Consider two rational numbers, say (frac {8}{9}, frac {3}{5})
Sum of these number = (frac {8}{9} + frac {3}{5} = frac {67}{45}), which is rational number.
Hence, the sum of two rational numbers is a rational number.

15. The sum of two irrational numbers is a rational number.
a) False
b) True
Answer: a
Clarification: Consider two irrational numbers, say, √2 and √5
Sum of these number = √2 + √3 = 3.14626… which is an irrational number.
Hence, the sum of two irrational numbers is an irrational number.

[CLASS 10] Mathematics MCQs on Number of Tangents from a Point on the Circle

Mathematics Online Quiz for Class 10 on “Number of Tangents from a Point on the Circle”.

1. How many tangents can a circle have?
a) Zero
b) Infinity
c) Estimated on the value of the radius
d) Fixed for every kind of circle
Answer: b
Clarification: A circle is a set of points on a plane that is equidistant from a fixed point and an infinite number of tangents can be drawn for any given circle.

2. Find the number of tangents that can be drawn for the given figure.

a) Estimated by using the Pythagorean theorem
b) Estimated on the value of the angle
c) Cannot be drawn
d) Estimated on the value of the diameter
Answer: c
Clarification: Tangent is part of a circle. It cannot be drawn for any other figure other than a circle. Hence, tangents cannot be drawn to the given figure which is a right – angled triangle.

3. How many tangents can be drawn at one point on a circle?
a) Only one
b) Three
c) Zero
d) Two
Answer: a
Clarification: A circle is a set of points on a plane that is equidistant from a fixed point and we can draw only one tangent at one point on any circle given.

4. A tangent touches a circle at a single point.
a) False
b) True
Answer: b
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and an infinite number of tangents are drawn to a circle whereas secant is a line that intersects two distinct points on a circle.

5. Number of tangents passing through a circle is _____
a) 2
b) 3
c) 1
d) 0
Answer: d
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and a tangent to a circle doesn’t pass through the circle.

6. What happens to the length of the chord when the chord comes closer to the center?
a) Decreases
b) Becomes an arc
c) Increases
d) Becomes a segment
Answer: c
Clarification: The length of the chord of a circle increases when it comes closer and closer to the center of the circle. Hence, the longest chord becomes the diameter.

7. Find the area of the sector if the radius is 6 cm and with an angle of 60°.
a) 18.35 cm
b) 18.85 cm
c) 18.00 cm
d) 18.05 cm
Answer: b
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {60^{circ }}{360^{circ }}) x (frac {22}{7}) × 62
= 18.85 cm

8. The area of the sector is _____
a) (frac {x^{circ }}{360^{circ }}) × πr2
b) (frac {x^{circ }}{360^{circ }}) – πr2
c) (frac {x^{circ }}{360^{circ }}) + πr3
d) (frac {x^{circ }}{360^{circ }}) × πr3
Answer: a
Clarification: The area of the sector is (frac {x^{circ }}{360^{circ }}) × πr2
Where x° is the degree measure of the angle at the center and r is the radius of the circle.

9. Find the radius of a circle if 2 m is the length of the tangent, 6 m is the distance between the center of the circle the external point.
a) 7 m
b) 5 m
c) √32 m
d) √38 m
Answer: c
Clarification: Length of the tangent = (sqrt {d^2 – r^2} )
2 = (sqrt {6^2 – r^2} )
r2 = 62 – 22
r = (sqrt {6^2 – 2^2} )
r = (sqrt {36 – 4} )
r = √32 m

10. Line C is secant to the circle.

a) False
b) True
Answer: a
Clarification: Tangent is a line that touches the circle at a single point. Hence, line C is a tangent, not a secant. Line A and line B are secants of the circle because they’re touching the circle at two distinct points.

To practice Mathematics Online Quiz for Class 10,