Term 1 class 9 objective question in math in 2021
1.) Can we write 0 in the form of p/q?
a. Yes
b. No
c. Cannot be explained
d. None of the above
Answer: a
Explanation: 0 is a rational number and hence it can be written in the form of p/q.
Example: 0/4 = 0
2.) The three rational numbers between 3 and 4 are:
a. 5/2, 6/2, 7/2
b. 13/4, 14/4, 15/4
c. 12/7, 13/7, 14/7
d.11/4, 12/4, 13/4
Answer: b
Explanation: There are many rational numbers between 3 and 4
To find 3 rational numbers, we need to multiply and divide both the numbers by 3+1 = 4
Hence, 3 x (4/4) = 12/4 and 4 x (4/4) = 16/4
Thus, three rational numbers between 12/4 and 16/4 are 13/4, 14/4 and 15/4.
3.) In between any two numbers, there are:
a. Only one rational number
b. Two rational numbers
c. Infinite rational numbers
d. No rational number
Answer: c
Explanation: Take the reference from question number 2 explained above.
4.) Every rational number is:
a. Whole number
b. Natural number
c. Integer
d. Real number
Answer: d
5.) √9 is __________ number.
a. A rational
b. An irrational
c. Neither rational nor irrational
d. None of the above
Answer: a
Explanation: √9 = 3
Hence, √9 is a rational number.
6.) Which of the following is an irrational number?
a. √16
b. √(12/3)
c. √12
d. √100
Answer: c
Explanation: √12 cannot be simplified to a rational number.
7.) 3√6 + 4√6 is equal to:
a. 6√6
b. 7√6
c. 4√12
d. 7√12
Answer: b
Explanation: 3√6 + 4√6 = (3 + 4)√6 = 7√6
8.) √6 x √27 is equal to:
a. 9√2
b. 3√3
c. 2√2
d. 9√3
Answer: a
Explanation: √6 x √27 = √(6 x 27) = √(2 x 3 x 3 x 3 x 3) = (3 x 3)√2 = 9√2
9.) Which of the following is equal to x3?
a. x6 – x3
b. x6.x3
c. x6/x3
d. (x6)3
Answer: c
Explanation: x6/x3 = x6 – 3 = x3
10.) Which of the following is an irrational number?
a. √23
b. √225
c. 0.3796
d. 7.478478
Answer: a
Explanation: √23 = 4.79583152331…
Since the decimal expansion of the number is non-terminating non-recurring. Hence, it is an irrational number.
11.) Which of the following is an irrational number?
a. 0.14
b.
c.
d. 0.4014001400014…
Answer: d
Explanation: 0.4014001400014…is an irrational number as it is non-terminating and non-repeating.
12.) 2√3+√3 =
a. 6
b. 2√6
c. 3√3
d. 4√6
Answer: c
Explanation: 2√3+√3 = (2+1)√3= 3√3.
13.) Find the value of , if √2 = 1.4142.
a. 5.8282
b. 0.1718
c. 0.4142
d. 2.4142
Answer: c
Explanation:
=
=
= 0.4142
14.) The number obtained on rationalising the denominator of is
a. (√7+2)/3
b. (√7-2)/3
c. (√7+2)/5
d. (√7+2)/45
Answer: a
Explanation:
15.) Which of the following is rational?
a. 4/0
b. 0/4
c. √3
d. π
Answer: b
Explanation: 0/4 is a rational number that is equal to 0. Whereas π and √3 are irrational numbers and 4/0 is meaningless.
16.) The irrational number between 2 and 2.5 is
a. √11
b. √5
c. √22.5
d. √12.5
Answer: b
Explanation: The irrational number between 2 and 2.5 is √5 because the approximate value of √5 is 2. 23606…
17.) The value of √10 times √15 is equal to
a. 5√6
b. √25
c. 10√5
d. √5
Answer: a
Explanation: √10×√15 =(√2.√5)×(√3. √5) = 5√6.
18.) The decimal representation of the rational number is
a. Always terminating
b. Either terminating or repeating
c. Either terminating or non-repeating
d. Neither terminating nor repeating
Answer: b
Explanation: As per the definition of rational number, its decimal representation is either terminating or repeating.
19.) Which of the following is a rational number?
a. 0
b. 2√3
c. 2+√3
d. π
Answer: a
Explanation: 0 is a rational number, and it can be written as 0/1 or 0/2 or 0/3 etc. Whereas 2√3, 2+√3, and π are irrational numbers.
20.) Which of the following is an irrational number?
a. √(4/9)
b. √12/√3
c. √7
d. √81
Answer: c
Explanation: √7 is an irrational number. Because other options given are simplified into a rational number.
21) The name of the horizontal line in the cartesian plane which determines the position of a point is called:
a. Origin
b. X-axis
c. Y-axis
d. Quadrants
Answer: b
22) The name of the vertical line in the cartesian plane which determines the position of a point is called:
a. Origin
b. X-axis
c. Y-axis
d. Quadrants
Answer: c
23) The section formed by horizontal and vertical lines determining the position of the point in a cartesian plane is called:
a. Origin
b. X-axis
c. Y-axis
d. Quadrants
Answer: d
24) The point of intersection of horizontal and vertical lines determining the position of a point in a cartesian plane is called:
a. Origin
b. X-axis
c. Y-axis
d. Quadrants
Answer: a
25) If the coordinates of a point are (0, -4), then it lies in:
a. X-axis
b. Y-axis
c. At origin
d. Between x-axis and y-axis
Answer: b
Explanation: Since, x=0 and y=-4. Hence, the point will lie in the negative y-axis 4 units far from the origin.
26) If the coordinates of a point are (3, 0), then it lies in:
a. X-axis
b. Y-axis
c. At origin
d. Between x-axis and y-axis
Answer: a
Explanation: Since, x = 3 and y = 0, therefore, the point will lie at the positive x-axis 3 units far from the origin.
27) If the coordinates of a point are (-3, 4), then it lies in:
a. First quadrant
b. Second quadrant
c. Third quadrant
d. Fourth quadrant
Answer: b
Explanation: Since x = -3 and y = 4, then if we plot the point in a plane, it lies in the second quadrant.
28) If the coordinates of a point are (-3, -4), then it lies in:
a. First quadrant
b. Second quadrant
c. Third quadrant
d. Fourth quadrant
Answer: c
Explanation: Since, x = -3 and y = -4, then if we plot the point in a plane, it lies in the third quadrant.
29) Points (1, 2), (-2, -3), (2, -3);
a. First quadrant
b. Do not lie in the same quadrant
c. Third quadrant
d. Fourth quadrant
Answer: b
30) If x coordinate of a point is zero, then the point lies on:
a. First quadrant
b. Second quadrant
c. X-axis
d. Y-axis
Answer: d
31) Signs of the abscissa and ordinate of a point in the second quadrant are respectively
a. +, +
b. +, –
c. -, +
d. -, –
Answer: c
Explanation: The signs of abscissa (x-value) and ordinate(y-value) in the second quadrant are – and + respectively.
32) The point (-10, 0) lies in
a. Third quadrant
b. Fourth quadrant
c. On the negative direction of the x-axis
d. On the negative direction of the y-axis
Answer: c
Explanation: The lies (-10, 0) lies in the negative direction of the x-axis.
33) A quadrant in which both x and y values are negative is
a. First quadrant
b. Second quadrant
c. Third quadrant
d. Fourth quadrant
Answer: c
Explanation: In the third quadrant, both the abscissa and ordinate values are negative. Example (-2, -2), which lies in the third quadrant.
34) Abscissa of all the points on the x-axis is
a. 0
b. 1
c. 2
d. Any number
Answer: d
Explanation: Abscissa of all the points on the x-axis can be any number. The coordinates of any point on the x-axis is (x, 0), where x can take any value.
35) Ordinate of all points on the x-axis is
a. -1
b. 0
c. 1
d. Any number
Answer: b
Explanation: The ordinate of all points on the x-axis is 0. We know that the coordinates of any point on the x-axis is (x, 0). Here, the abscissa can take any value and the ordinate is zero.
36) Abscissa of a point is positive in
a. I quadrant
b. I and II quadrants
c. II quadrant only
d. I and IV quadrants
Answer: d
Explanation: In a coordinate plane, x can take positive values in the first and fourth quadrants. For example, (2, 2) and (2, -4) lies on the first and fourth quadrants, respectively.
37) Points (1, -1), (2, -2), (4, -5), (-3, -4)
a. lie in II quadrant
b. lie in III quadrant
c. lie in IV quadrant
d. Does not lie in the same quadrant
Answer: d
Explanation: The point (1, -1), (2, -2) and (4, -5) lies in the fourth quadrant, where (-3, -4) lies in the third quadrant.
38) Abscissa of all the points on the y-axis is
a. 0
b. 1
c. -1
d. Any number
Answer: a
Explanation: The abscissa of all the points on the y-axis is 0. We know that the coordinates of any point on the y-axis is (0, y). Here, the ordinate can take any value and the abscissa is zero.
39) Ordinate of all the points on the y-axis is
a. 0
b. 1
c. -1
d. Any number
Answer: d
Explanation: The ordinate of all the points on the y-axis can be any number. The coordinates of any point on the y-axis is (0, y). Here, abscissa can take only the value of 0 and the ordinate can take any value.
40) The point which lies on the y-axis at a distance of 5 units in the negative direction of the y-axis is
a. (5, 0)
b. (0, 5)
c. (-5, 0)
d. (0, -5)
Answer: d
Explanation: The coordinates of any point on the y-axis is (0, y).
Hence, abscissa should be 0.
41) The linear equation 3x-11y=10 has:
a. Unique solution
b. Two solutions
c. Infinitely many solutions
d.No solutions
Answer: c
Explanation: 3x-11y=10
y=(3x-10)/11
Now for infinite values of x, y will also have infinite solutions.
42) 3x+10 = 0 will has:
a. Unique solution
b. Two solutions
c. Infinitely many solutions
d.No solutions
Answer: a
Explanation: 3x+10 = 0
x = -10/3.
Hence, only one solution is possible.
43) The solution of equation x-2y = 4 is:
a. (0,2)
b. (2,0)
c. (4,0)
d. (1,1)
Answer: c
Explanation: Putting x=4 and y = 0, on the L.H.S. of the given equation, we get;
4-2(0) = 4 – 0 = 4
Which is equal to R.H.S.
44) Find the value of k, if x = 1, y = 2 is a solution of the equation 2x + 3y = k.
a. 5
b. 6
c. 7
d. 8
Answer: d
Explanation: 2x + 3y = k
k=2(1)+3(2) = 2+6 = 8
45) Point (3, 4) lies on the graph of the equation 3y = kx + 7. The value of k is:
a. 4/3
b. 5/3
c. 3
d. 7/3
Answer: b
Explanation: 3y = kx + 7
Here, x = 3 and y = 4
Hence,
(3×4) = (kx3) + 7
12 = 3k+7
3k = 12–7
3k = 5
k = 5/3
46) The graph of linear equation x+2y = 2, cuts the y-axis at:
a. (2,0)
b. (0,2)
c. (0,1)
d. (1,1)
Answer: c
Explanation: x+2y = 2
y = (2-x)/2
If x=0, then;
y=(2-0)/2 = 2/2 = 1
Hence, x+2y=2 cuts the y-axis at (0,1).
47) Any point on line x = y is of the form:
a. (k, -k)
b. (0, k)
c. (k, 0)
d. (k, k)
Answer: d
48) The graph of x = 3 is a line:
a. Parallel to the x-axis at a distance of 3 units from the origin
b. Parallel to the y-axis at a distance of 3 units from the origin
c. Makes an intercept 3 on the x-axis
d. Makes an intercept 3 on the y-axis
Answer: b
49) In equation, y = mx+c, m is:
a. Intercept
b. Slope
c. Solution of the equation
d. None of the above
Answer: b
50) If x and y are both positive solutions of equation ax+by+c=0, always lie in the:
a. First quadrant
b. Second quadrant
c. Third quadrant
d. Fourth quadrant
Answer: a
51) A linear equation in two variables is of the form ax + by + c = 0, where
(a) a = 0, c = 0
(b) a ≠ 0, b = 0
(c) a = 0, b ≠ 0
(d) a ≠ 0, b ≠ 0
Answer: d
Explanation: A linear equation in two variables is of the form ax + by + c = 0, where a ≠ 0, b ≠ 0. If the values of “a” and “b” are equal to 0, the equation becomes c =0. Hence, the values of a and b should not be equal to 0.
52) Any point on the x-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)
Answer: c
Explanation: Any point on the x-axis is of the form (x, 0). On the x-axis, x can take any values, whereas y should be equal to 0.
53) Any point on the y-axis is of the form
(a) (y, y)
(b) (0, y)
(c) (x, y)
(d) (x, 0)
Answer: b
Explanation: Any point on the y-axis is of the form (0, y). On the y-axis, y can take any values and x should be equal to 0.
54) The linear equation 2x – 5y = 7 has
(a) No solution
(b) unique solution
(c) Two solutions
(d) Infinitely many solutions
Explanation: The linear equation 2x-5y has infinitely many solutions. Because, the equation 2x-5y = 7 is a single equation, that involves two variables. Hence, for different values of x, we will get different values of y and vice-versa.
55) The linear equation 3x – y = x – 1 has
(a) No solution
(b) unique solution
(c) Two solutions
(d) Infinitely many solutions
Answer: d
Explanation: The linear equation 3x – y = x – 1 has infinitely many solutions.
On simplification, the given equation becomes 2x-y= -1, which is a single equation with two variables. Thus, 3x – y = x – 1 has infinitely many solutions.
56) The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
(a) (2, 0)
(b) (0, 2)
(c) (3, 0)
(d) (0, 3)
Answer: b
Explanation:
Given that the graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point. Let the point be “P”. Hence, the x -coordinate of point P is 0.
Now, substitute x= 0 in the given equation,
2(0) + 3y = 6
3y = 6
y=2
Hence, the cooridnate point is (0, 2).
57) The equation 2x + 5y = 7 has a unique solution, if x, y are:
(a) Rational numbers
(b) Real numbers
(c) Natural numbers
(d) Positive real numbers
Answer: c
Explanation: The equation 2x + 5y = 7 has a unique solution, if x, y are natural numbers.
In natural numbers, there exists only one pair (1, 1) which satisfies the given equation. But for rational numbers, real numbers, positive real numbers, there exist many solution pairs to satisfy the equation.
58) The point of the form (a, a) always lies on:
(a) On the line x + y = 0
(b) On the line y = x
(c) x-axis
(d) y-axis
Answer: b
Explanation: The point of the form (a, a) always lies on the line y = x. If the point having the same x and y values, it should lie on the same line.
59) If we multiply or divide both sides of a linear equation with the same non-zero number, then the solution of the linear equation:
(a) Remains the same
(b) Changes
(c) Changes in case of multiplication only
(d) Changes in case of division only
Answer: a
Explanation: If we multiply or divide both sides of a linear equation with the same non-zero number, then the solution of the linear equation remains the same.
60) If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is
(a) 2
(b) 4
(c) 5
(d) 6
Answer: b
Explanation:
Substitute x=2 and y=0 in the given equation, we get
2(2) + 3(0) = k
k = 4+0
k = 4.
Hence, the value of k is 4.
61) A line joining two endpoints is called:
a. Line segment
b. A ray
c. Parallel lines
d. Intersecting lines
Answer: a
62) An acute angle is:
a. More than 90 degrees
b. Less than 90 degrees
c. Equal to 90 degrees
d. Equal to 180 degrees
Answer: b
63) A reflex angle is:
a. More than 90 degrees
b. Equal to 90 degrees
c. More than 180 degrees
d. Equal to 180 degrees
Answer: c
64) A straight angle is equal to:
a. 0°
b. 90°
c. 180°
d. 360°
Answer: c
65) Two angles whose sum is equal to 180° are called:
a. Vertically opposite angles
b. Complementary angles
c. Adjacent angles
d. Supplementary angles
Answer: d
66) Intersecting lines cut each other at:
a. One point
b. Two points
c. Three points
d. Null
Answer: a
Explanation: Two lines always intersect each other at one point.
67) Two parallel lines intersect at:
a. One point
b. Two points
c. Three points
d. Null
Answer: d
Explanation: If two lines are parallel to each other, they don’t intersect each other.
68) If two lines intersect each other, then the vertically opposite angles are:
a. Equal
b. Unequal
c. Cannot be determined
d. None of the above
Answer: a
71) An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is
(a) 37 ½°
(b) 72 ½°
(c) 75°
(d) 52 ½°
Answer: d
Explanation:
The exterior angle of triangle = 105°
Let the interior angles be “x”.
By using, exterior angle theorem, Exterior angle = Sum of interior opposite angles
Therefore, 105° = x+x
2x = 105°
x = 52 ½°
72) If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
(a) 50°
(b) 65°
(c) 145°
(d) 155°
Answer: d
Explanation:
Assume a triangle ABC, such that ∠BAC=130°
Also, the bisectors of ∠B and ∠C meet at O.
To find: ∠BOC
In a triangle △ABC,
∠BAC+∠ABC+∠ACB=180°
By using the angle sum property of the triangle,
130°+∠ABC+∠ACB=180°
∠ABC+∠ACB=50°
½ (∠ABC+∠ACB)=25°
Since OB and OC bisect ∠ABC and ∠ACB
∠OBC+ ∠OCB=25°
Now, consider △OBC,
∠OBC+ ∠OCB+∠BOC=180°
25°+∠BOC=180°
∠BOC=155°
73) If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is
(a) 54°
(b) 108°
(c) 120°
(d) 136°
Answer: b
74) If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) a right triangle
(b) an isosceles triangle
(c) an equilateral triangle
(d) an obtuse triangle
Answer: a
Explanation: If one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle. We know that the sum of interior angles of a triangle is equal to 180°. In the right triangle, one angle should be equal to 90°, and the remaining two angles are acute angles, and their sum is equal to 90°.
75) The angles of a triangle are in the ratio 5 : 3: 7. The triangle is
(a) a right triangle
(b) an acute-angled triangle
(c) an obtuse-angled triangle
(d) an isosceles triangle
Answer: b
Explanation: If the angles are in the ratio of 5:3:7, then a triangle is an acute angle triangle.
We know that the sum of the interior angles of a triangle is 180°
Therefore, 5x+3x+7x = 180°
15x = 180°
x = 180°/15 = 12°
Thus, 5x = 5(12°) =60°
3x = 3(12°) =36°
7x = 7(12°) =84°
Since all the angles are less than 90°, the triangle is an acute angle triangle.
76) Angles of a triangle are in the ratio 2: 4 : 3. The smallest angle of the triangle is
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Answer: b
Explanation:
We know that the sum of the interior angles of a triangle is 180°
Given that, the angles of a triangle are in the ratio of 2:4:3
Hence, 2x+4x+3x = 180°
9x = 180°
x= 20°
Therefore,
2x = 2(20) = 40°
4x = 4(20) = 80°
3x = 3(20) = 60°
Hence, the angles are 40°, 80° and 60°.
Therefore, the smallest angle of a triangle is 40°.
77) In triangle ABC, if AB=BC and ∠B = 70°, ∠A will be:
a. 70°
b. 110°
c. 55°
d. 130°
Answer: c
Explanation: Given,
AB = BC
Hence, ∠A=∠C
And ∠B = 70°
By angle sum property of triangle we know:
∠A+∠B+∠C = 180°
2∠A+∠B=180°
2∠A = 180-∠B = 180-70 = 110°
∠A = 55°
78) For two triangles, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle. Then the congruency rule is:
a. SSS
b. ASA
c. SAS
d. None of the above
Answer: b
79) A triangle in which two sides are equal is called:
a. Scalene triangle
b. Equilateral triangle
c. Isosceles triangle
d. None of the above
Answer: c
80) The angles opposite to equal sides of a triangle are:
a. Equal
b. Unequal
c. supplementary angles
d. Complementary angles
Answer: a
81) If E and F are the midpoints of equal sides AB and AC of a triangle ABC. Then:
a. BF=AC
b. BF=AF
c. CE=AB
d. BF = CE
Answer: d
Explanation: AB and AC are equal sides.
AB = AC (Given)
∠A = ∠A (Common angle)
AE = AF (Halves of equal sides)
∆ ABF ≅ ∆ ACE (By SAS rule)
Hence, BF = CE (CPCT)
82) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively. Then:
a. BE>CF
b. BE<CF
c. BE=CF
d. None of the above
Answer: c
Explanation:
∠A = ∠A (common arm)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC
Hence, BE = CF (by CPCT)
83) If ABC and DBC are two isosceles triangles on the same base BC. Then:
a. ∠ABD = ∠ACD
b. ∠ABD > ∠ACD
c. ∠ABD < ∠ACD
d. None of the above
Answer: a
Explanation: AD = AD (Common arm)
AB = AC (Sides of isosceles triangle)
BD = CD (Sides of isosceles triangle)
So, ΔABD ≅ ΔACD.
∴ ∠ABD = ∠ACD (By CPCT)
84) If ABC is an equilateral triangle, then each angle equals to:
a. 90°
B.180°
c. 120°
d. 60°
Answer: d
Explanation: Equilateral triangle has all its sides equal and each angle measures 60°.
AB= BC = AC (All sides are equal)
Hence, ∠A = ∠B = ∠C (Opposite angles of equal sides)
Also, we know that,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
85) If AD is an altitude of an isosceles triangle ABC in which AB = AC. Then:
a. BD=CD
b. BD>CD
c. BD<CD
d. None of the above
Answer: a
Explanation: In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)
∴ ΔABD ≅ ΔACD (By RHS congruence condition)
BD = CD (By CPCT)
86) In a right triangle, the longest side is:
a. Perpendicular
b. Hypotenuse
c. Base
d. None of the above
Answer: b
Explanation: In triangle ABC, right-angled at B.
∠B = 90
By angle sum property, we know:
∠A + ∠B + ∠C = 180
Hence, ∠A + ∠C = 90
So, ∠B is the largest angle.
Therefore, the side (hypotenuse) opposite to the largest angle will be the longest one.
87) Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Answer: c
Explanation:
SSA is not a criterion for the congruence of triangles. Whereas SAS, ASA and SSS are the criteria for the congruence of triangles.
88) In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) Isosceles and congruent
(b) Isosceles but not congruent
(c) Congruent but not isosceles
(d) Neither congruent nor isosceles
Answer: b
Explanation:
Consider two triangles, ABC and PQR. If the sides AB = AC and ∠C = ∠P and ∠B = ∠Q, then the two triangles are said to be isosceles, but they are not congruent.
89) In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 2 cm
(b) 2.5 cm
(c) 4 cm
(d) 5 cm
Answer: c
Explanation:
Given that, in a triangle PQR, ∠R = ∠P.
Since, ∠R = ∠P, the sides opposite to the equal angles are also equal.
Hence, the length of PQ is 4 cm.
90) If AB = QR, BC = PR and CA = PQ, then
(a) ∆ PQR ≅ ∆ BCA
(b) ∆ BAC ≅ ∆ RPQ
(c) ∆ CBA ≅ ∆ PRQ
(d) ∆ ABC ≅ ∆ PQR
Answer: c
Explanation:
Consider two triangles ABC and PQR.
Given that, AB = QR, BC = PR and CA = PQ.
By using Side-Side-Side (SSS rule),
We can say, ∆ CBA ≅ ∆ PRQ.
91) If ∆ ABC ≅ ∆ PQR, then which of the following is not true?
(a) AC = PR
(b) BC = PQ
(c) QR = BC
(d) AB = PQ
Answer: b
Explanation:
Given that, ∆ ABC ≅ ∆ PQR
Hence, AB = PQ
BC = QR
AC =PR
Thus, BC = PQ is not true, if ∆ ABC ≅ ∆ PQR.
92) In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 100°
Answer: b
93) Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(a) 3.4 cm
(b) 3.6 cm
(c) 3.8 cm
(d) 4.1 cm
Answer: a
Explanation: If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of the third side of the triangle cannot be 3.4 cm. Because the difference between the two sides of a triangle should be less than the third side.
94) In ∆ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Answer: b
Explanation:
Given that, in a triangle ABC, AB = AC and ∠B = 50°.
Since the given triangle is an isosceles triangle, the angles opposite to the equal sides are also equal. Hence, ∠C = 50°.
95) In ∆ PQR, if ∠R > ∠Q, then
(a) QR < PR
(b) PQ < PR
(c) PQ > PR
(d) QR > PR
Answer: c
Explanation: In a triangle PQR, if ∠R > ∠Q, then PQ > PR, because the side opposite to the greater angle is longer.
95) It is given that ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true?
(a) DF = 5 cm, ∠F = 60°
(b) DF = 5 cm, ∠E = 60°
(c) DE = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40
Answer: b
Explanation:
Given that, ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°.
96) Area of the triangle is equal to:
a. Base x Height
b. 2(Base x Height)
c. ½(Base x Height)
d. ½ (Base + Height)
Answer: c
97) If the perimeter of an equilateral triangle is 180 cm. Then its area will be:
a. 900 cm2
b. 900√3 cm2
c. 300√3 cm2
d. 600√3 cm2
Answer: b
Explanation: Given, Perimeter = 180 cm
3a = 180 (Equilateral triangle)
a = 60 cm
Semi-perimeter = 180/2 = 90cm
Now as per Heron’s formula,
Hence, if we put the values here, we get:
A = 900√3
98) The sides of a triangle are 122 m, 22 m and 120 m respectively. The area of the triangle is:
a. 1320 sq.m
b. 1300 sq.m
c. 1400 sq.m
d. 1420 sq.m
Answer: a
Explanation: Given,
a = 122 m
b = 22 m
c = 120 m
Semi-perimeter, s = (122+22+120)/2 = 132 m
Using heron’s formula:
Put the values of s, a, b and c, to get the answer equal to 1320 sq.m.
99) The area of triangle with given two sides 18cm and 10cm, respectively and perimeter equal to 42 cm is:
a. 20√11 cm2
b. 19√11 cm2
c. 22√11 cm2
d. 21√11 cm2
Answer: d
Explanation: Perimeter = 42
a+b+c=42
18+10+c=42
c=42-28=14 cm
Semiperimeter, s = 42/2 = 21cm
Using Heron’s formula:
Put the values of s, a, b and c, to get the answer equal to 21√11 cm2
100) The sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540cm. The area is:
a. 1000 sq.cm.
b. 5000 sq.cm.
c. 9000 sq.cm.
d. 8000 sq.cm.
Answer: c
Explanation: The ratio of the sides is 12: 17: 25
Perimeter = 540 cm
Let the sides of the triangle be 12x, 17x and 25x.
Hence,
12x+17x+25x = 540 cm
54x = 540 cm
x = 10
Therefore,
a = 12x=12 x 10 = 120
b = 17x = 17 x 10 = 170
c = 25x = 25 x 10 = 250
Semi-perimeter, s = 540/2 = 270 cm
Putting the values of s, a, b and c in the Heron’s formula, we will get the area equal to 9000 sq.cm.
101) The equal sides of the isosceles triangle are 12 cm, and the perimeter is 30 cm. The area of this triangle is:
a. 9√15 sq.cm
b. 6√15 sq.cm
c. 3√15 sq.cm
d. √15, sq.cm.
Answer: a
Explanation: Given,
Perimeter = 30cm
Semiperimeter, s = 30/2 = 15cm
a = b = 12cm
c=?
a+b+c = 30
12 +12+c=30
c=30-24 = 6cm
By putting the values of s, a, b and c in Heron’s formula, we can get the value of area.
102) The area of an equilateral triangle having side length equal to √3/4cm is:
a. 2/27 sq.cm
b. 2/15 sq.cm
c. 3√3/64 sq.cm
d. 3/14 sq.cm
Answer: c
Explanation: Here, a = b = c = √3/4
Find the semi-perimeter of the triangle and use Heron’s formula to find the answer.
103) The sides of a parallelogram are 100 m each and length of the longest diagonal is 160m. The area of a parallelogram is:
a. 9600 sq.m
b. 9000 sq.m
c. 9200 sq.m
d. 8800 sq.m
Answer: a
Explanation: The diagonal divides the parallelogram into two equivalent triangles. Hence, its area will be equal to the sum of the area of the two triangles.
Hence, we can determine the area of the two triangles using Heron’s formula.
104) The sides of a triangle are in the ratio of 3: 5: 7 and its perimeter is 300 cm. Its area will be:
a. 1000√3 sq.cm
b. 1500√3 sq.cm
c. 1700√3 sq.cm
d. 1900√3 sq.cm
Answer: b
105) The base of a right triangle is 8 cm and the hypotenuse is 10 cm. Its area will be
(a) 24 cm2
(b) 40 cm2
(c) 48 cm2
(d) 80 cm2
Answer: a
Explanation:
Given: Base = 8 cm and Hypotenuse = 10 cm
Hence, height = √[(102 – 82) = √36 = 6 cm
Therefore, area = (½)×b×h = (½)×8×6 = 24cm2.
106) The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is
(a) Rs 2.00
(b) Rs 2.16
(c) Rs 2.48
(d) Rs 3.00
Answer: b
Explanation:
Given: a = 6cm, b= 8cm, c = 10 cm.
s = (6+8+10)/2 = 12
Hence, by using Heron’s formula, we can write:
A = √[12(12-6)(12-8)(12-10)]= √[(12)(6)(4)(2)]= √576 = 24cm2
Therefore, the cost of painting at a rate of 9 paise per cm2 = 24×9 paise = Rs. 2.16
107) An isosceles right triangle has area of 8 cm2. The length of its hypotenuse is
(a) √32 cm
(b) √16 cm
(c) √48 cm
(d) √24 cm
Answer: a
Explanation:
Given that area of isosceles triangle = 8 cm2.
As the given triangle is isosceles triangle, let base = height = h
Hence,
(½)×h×h = 8
(½)h2 =8
h2=16
h= 4 cm
Since it is isosceles right triangle, Hypotenuse2 = Base2+Height2
Hypotenuse2= 42+42
Hypotenuse2 = 32
Hypotenuse = √32 cm
108) The area of an isosceles triangle having a base 2 cm and the length of one of the equal sides 4 cm, is
(a) √15 cm2
(b) √(15/2) cm2
(c) 2√15 cm2
(d) 4√15 cm2
Answer: a
Explanation:
Given that a = 2 cm, b= c = 4 cm
s = (2+4+4)/2 = 5
By using Heron’s formula, we get:
A =√[5(5-2)(5-4)(5-4)] = √[(5)(3)(1)(1)] = √15 cm2.
109) The perimeter of an equilateral triangle is 60 m. The area is
(a) 10√3 m2
(b) 15√3 m2
(c) 20√3 m2
(d) 100√3 m2
Answer: d
Explanation:
Given: Perimeter of an equilateral triangle = 60m
3a = 60 m (As the perimeter of an equilateral triangle is 3a units)
a = 20 cm.
We know that area of equilateral triangle = (√3/4)a2 square units
A = (√3/4)202
A = (√3/4)(400) = 100√3 cm2.
110) The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(a) 16√5 cm
(b) 10√5 cm
(c) 24√5 cm
(d) 28 cm
Answer: c
Explanation:
Given: a =35 cm, b=54cm, c =61cm
s = (35+54+61)/2 = 75 cm.
Hence, by using Heron’s formula, A = √[75(75-35)(75-54)(75-61)] = √(882000) = 420√5 cm2
The area of triangle with longest altitude “h” is given as”
(½)×a×h = 420√5
(½)×35×h = 420√5
h= (840√5)/35 = 24√5 cm.
111) The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(a) 1322 cm2
(b) 1311 cm2
(c) 1344 cm2
(d) 1392 cm2
Answer: c
Explanation:
Since, all the sides of a triangle are given, we can find the area of a triangle using Heron’s formula.
Let a = 56 cm, b= 60 cm, c = 52 cm
s = (56+60+52)/2 = 84 cm.
Area of triangle using Heron’s formula, A = √[s(s-a)(s-b)(s-c)] square units
A = √[84(84-56)(84-60)(84-52)] = √(1806336) =1344 cm2.
112) If the area of an equilateral triangle is 16√3 cm2, then the perimeter of the triangle is
(a) 48 cm
(b) 24 cm
(c) 12 cm
(d) 36 cm
Answer: b
Explanation:
Given: Area of equilateral triangle = 16√3 cm2
(√3/4)a2 = 16√3
a2 = [(16√3)(4)]/√3
a2 = 64
a = 8cm
Therefore, perimeter = 3(8) = 24 cm.
113) The area of an equilateral triangle with side 2√3 cm is
(a) 5.196 cm2
(b) 0.866 cm2
(c) 3.496 cm2
(d) 1.732 cm2
Answer: a
Explanation:
Given: Side = 2√3 cm
We know that, area of equilateral triangle = (√3/4)a2 square units
A = (√3/4)(2√3)2 = (√3/4)(12) = 3√3 = 3(1.732) = 5.196 cm2 .
114) The length of each side of an equilateral triangle having an area of 9√3 cm2 is
(a) 8 cm
(b) 36 cm
(c) 4 cm
(d) 6 cm
Answer: d
Explanation:
Given: Area of equilateral triangle = 9√3 cm2
Hence, (√3/4)a2 = 9√3
a2 = [(9√3)(4)]/√3
a2 = 36
a = 6cm.
115) The ratio of the sum of observations and the total number of observations is called:
a. Mean
b. Median
c. Mode
d. Central tendency
Answer: a
116) The mean of x+2, x+3, x+4 and x-2 is:
a. (x+7)/4
b. (2x+7)/4
c. (3x+7)/4
d. (4x+7)/4
Answer: d
Explanation: Mean = (x+2+x+3+x+4+x-2)/4 = (4x+7)/4
117) The median of the data: 4, 6, 8, 9, 11 is
a. 6
b. 8
c. 9
d. 11
Answer: b
118) The median of the data: 155 160 145 149 150 147 152 144 148 is
a. 149
b. 150
c. 147
d. 144
Answer: a
Explanation: First arrange the data in ascending order.
144 145 147 148 149 150 152 155 160
Since, the number of observations here is odd, therefore,
Median = (n+1)/2 th = (9+1)/2 = 10/2 = 5th number = 149
119) The median of the data: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 is:
a. 10
b. 24
c. 12
d. 8
Answer: c
Explanation: Arrange the given data in ascending order:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Since, the number of observations givere here is even, hence,
Median will be average of two middle terms.
n/2th = 16/2 = 8th term
(n/2 +1)th = (16/2 + 1)th = 9th term
Therefore,
Median = (10+14)/2 = 12