Worksheet on Finding the Centroid of a Triangle

Students who are interested to learn how to find the centroid of a triangle can refer to this page. You can find different types of questions in the Worksheet on Finding the Centroid of a Triangle. Before finding the centroid of the triangle recall the formula for finding the centroid of a triangle. You can solve the problems easily only if you learn the formulas related to the centroid of a triangle. Gain immense practice using this Free Printable Centroid of a Triangle Worksheet with Answers and gain adequate skills regarding the concept.

See More:

  • Medians and Altitudes of a Triangle
  • Properties of Triangle

Centroid of a Triangle Worksheet PDF

Example 1.
Calculate the coordinates of the centroid of the triangle ABC is A(6,2) B(0,2) C(1,4)

Solution:

We know that
Centroid formula =( x1 + x2 + x3/3, y1 + y2 + y3/3)
(6 + 0 + 1 /3 , 2 + 2 + 4 /3)
(7/3, 8/3)

Example 2.
A(3,x), B(4,3) and C( y,2) are the vertices of the triangle ABC whose centroid is the origin and calculate the values of x and y.

Solution:

Given centroid of the triangle ABC is the origin
We know that the coordinates of the point dividing (x1, y1) and (x2,y2) in the ratio m:n is given by
(m1x1 + m2xx1/m1 + m2 , m1y2 + m2y1/m1 + m2)
(m1, m2) = (0,0)
(0,0) = (3 + 4 + y/3, x + 3 + 2 /3)
(0,0) = (7 + y/3, x + 5/3)
0 = 7 + y/3 and 0 = x + 5/3
y = -7 and x = -5

Example 3.
Find the centroid of the triangle whose vertices are (1,2) (3,4) and (5,6).

Solution:

Let the vertices of the triangle be A(1,2) (3,4) and (5,6).
The centroid of triangle (G) = (x1 + x2 + x3/3, y1 + y2 + y3 /3)
= (1 + 3 + 5/3 , 2 + 4 + 6/3)
= (9/3 ,2/3)
=( 3,4)

Example 4.
If the centroid of the triangle is at (4,2) and two of its vertices are (3,2) and (5,2) then find the third vertex of the triangle.

Solution:

Let the vertices of a triangle be A(3,2) B(5,2) and c(x3,y3)
The centroid of the triangle is (4,2)
Centroid formula = (x1 + x2 + x3/3 , y1 + y2 + y3/3)
(4,2) = (3 + 5 + x3/3 , 2 + 2 + y3/3)
(4,2) = (8 + x3/3, 4 + y3/3)
8 + x3/3 = 4
8 + x3 = 12
x3 = 12 – 8
x3 = 4.
4 + y3/3 = 2
4 + y3 = 6
y3 = 6-4
y3 = 2
The third vertex is (4,2)

Example 5.
Find the length of the median through A of a triangle whose vertices are A(1,3) B(1,1) and C(5,1)

Solution:

AD is the median of the ∆ABC.
D is the mid point of BC
Midpoint of a line = (x1 + x2/2, y1 + y2/2)
Midpoint of BC =( 1 + 5/2 , 1 + ½)
= (6/2, 2/2) = (3,1)
Length of the median AD = √(x2 – x1)² + (y3 – y1)²
= √(3 + 1)² + (0 + 3)²
= √4² + 3²
= √16 + 9
√25 = 5
The length of the median AD is 5 units.

Example 6.
The vertices of a triangle are (1,2) (h,3) and (4,k) if the centroid of the triangle is at the point (5,1) then find the value of √(h + k)² + (h + 3k)²

Solution:

Let the vertices A(1,2) B(h,3) and c(4,k).
Centroid of a ∆ABC = (x1 + x2 + x3/3, y1 + y2 + y3/3).
(5,1) = (1 + h + 4/3, 2 + 3 + k/3)
(5 + h/3, 5 + k/3)
5 + h/3 = 5
5 + h = 15
h = 15 – 5
h = 10
5 + k/3 = 1
5 + k = 3
k = 3 – 5
k = -2.
The value of √(h + k)² + (h + 3k)²
= √(10 – 2)² + (10+ 3(-2))²
= √8² + (10 – 6)²
= √8² + 4²
= √64 + 16
= √80
= 8.9

Example 7.
The A(h,6) B(2,3) and C(6,k) are the coordinates of vertices of a triangle whose centroid is G(1,5) find h and k.

Solution:

Given that
A(x1,y2) = A(h,6)
B(x2,y2) = B(2,3)
C(x3,y3) = C(6,k)
Centroid G(x,y) = G(1,5)
G is the centroid.
By centroid formula
x = x1 + x2 + x3/3
1 = h + 2 + 6/3
3 = h + 2+ 6
3 = h + 8
h = 3 – 8
h = -5.
y = y1 + y2 + y3/3
5 = 6 + 3 + k/3
15 = 9 + k
k = 15 – 9
k = 6
(h,k) = (-5, 6)

Example 8.
Find the coordinates of the point of trisection of the line segment AB with A(2,7) and B(4,8).

Solution:

Let P(x1, y1) and Q(x2, y2) divide the line AB into 3 equal parts.
AP = PQ = QB
AP/PB = AP/PQ + QB = AP/AP + AP = QB/2AP = ½
So, P divides AB in the ratio 1:2.
x1 = 1×4 + 2×2/2 + 1 = 0
y1 = 1×8 + 2×7/2 + 1 = 2
(x1,y1) = (0,2)
Also
AQ/QB = AP+PB/QB. QB + QB/QB = 2/1
x2 = 2×4 + 1×2/2 + 1 = 8 +⅔ = 10/3
y2 = 2×8 + 1×7/2 + 1 = 16 + 7/3 = 23/3
(x2, y2) = (10/3, 23/3)

Example 9.
The coordinates of the centroid of the triangle PQR are (2,5) if a = (6,5) and R = (11,8) calculate the coordinates of the vertex D.

Solution:

Let G be the centroid of DPQR
Whose coordinates are (2,5)
And let (x,y) be the coordinates of vertex P. Coordinates of G are
G(2,5) = G(x + 6 + 11/3, y + 5 + 8/3)
2 = x + 5/3
6 = x + 5
6 – 5 = x
x = 1
5 = y + 13/3
15 = y + 13
15 – 3 = y
y = 2
Coordinates of vertex P are (1,2).

Example 10.
A(5,x), B(4,3) and C(y,2) are the vertices of the triangle ABC whose centroid is the origin and calculate the values of x and y.

Solution:

Given the centroid of the triangle WBC is the origin.
(0,0) = (5+4+y/3, x+3+2/3)
(0,0) = (9 + y/3 , x + 5/3)
9 + y/3 = 0
9 + y = 0
y = -9.
x+5/3 = 0
x + 5 = 0
x = -5
(x,y) = (-9,-5)

Problems on Distance Formula

Problems on Distance Formula are available on this page. Distance between two points in coordinate geometry can be calculated using the distance formula. The distance formula can be found using the Pythagoras theorem. Learn how to find the distance between the points with the help of the distance formula.

Refer Similar Posts:

  • Distance and Section Formulae
  • Distance Formula

Distance Formula Problems with Solutions PDF

Example 1.
If the distance between the points (6, – 2) and (2, a) is 5, find the values of a.
Solution:
We know that,
The distance between (x1, y1) and (x2, y2)
is √(x1−x2)² + (y1−y2)²
Here, the distance = 5,
x1 = 6, x2 = 2, y1 = -2 and y2 = a
Therefore, 5 = √(6−2)² + (−2−a)²
25 = √(4)² + (-2-a)²
Squaring on both sides
25 = 16 + (2 + a)²
(2 + a)² = 25 – 16
(2 + a)² = 9
Taking square root, 2 + a = ±3
a = -2 +or- 3
a = (1, -5)

Example 2.
Find the value of a, if the distance between the points P(3, 6) and Q(3, a) is 10 units.
Solution:
Let the given points be:
P(3, 6) = (x1, y1)
Q(3, a) = (x2, y2)
Using distance formula,
Distance between the points P(3, 6) and Q(3, a) is:
[(3 – 3)² + (a – 6)²] = 10 units
Squaring on both sides of the equation,
(0)² + (a – 6)² = 100
(a – 6)² = 100
Taking root on both the sides, we get;
a – 6 = ±10
Case I: Considering +10,
a – 6 = 10,
a = 10 + 6 = 16
Case II: Considering -10
a – 6 = -10
a = -10 + 6
a = – 4
a = (16, -4)

Example 3.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:
Let P(x, y) be the point which is equidistant from the points A(6, 1) and B(3, 5).
Given,
AP = BP
AP² = BP²
(x – 6)² + (y – 1)² = (x – 3)² + (y – 5)² (by distance formula)
x² – 12x + 36 + y² – 2y + 1 = x² – 6x + 9 + y² – 10y + 25
-12x + 37 – 2y + 6x + 10y – 34 = 0
-6x + 8y = – 3
This is the required relation between x and y

Example 4.
Find a point on the y-axis which is equidistant from points A(6, 5) and B(– 4, 3).
Solution:
We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B. Then:
AP = BP
AP2 = BP2
(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2
36 + 25 + y2– 10y = 16 + 9 + y2 – 6y
61 – 10y = 25 – 6y
10y – 6y = 61 – 25
4y = 36
y = 9
So, the required point is (0, 9).
Verification:
AP = √[(6 – 0)2 + (5 – 9)2]
= √(36+16)
= √52
BP = √[(-4-0)2+(3-9)2]
=√(16+36)
=√52
Hence, we conclude that the point (0, 9) is equidistant from the given two points.

Example 5.
Find the distance between the points (1, 2) and (1, 5).
Solution:
The formula for the distance D between two points (a,b) and (c,d) is given by
D = √(c – a)² + (d – b)²
Apply the formula given above to find distance D between the points (1, 2) and(1, 5) as follows.
D = √(1 – 1)² + (5 – 2)²
√(0)² + (3)²
√0 + 9
√9 = 3

Example 6.
Find a relationship between x and y so that the distance between the points (3, 4) is equal to 12.
Solution:
We apply the distance formula
12 = √(3 – x)² + (4 – y)²
Square both sides
144 = (3 – x)² + (4 – y)²
The above relationship between x and y is the equation of the circle.
144 = 9 – 6x + x² + 16 – 8y + y²
144 = x² + y² – 6x – 8y + 25
x² + y² – 6x – 8y = 144 – 25
x² + y² – 6x – 8y = 119

Example 7.
Find the distance between the points (2,4) and (1,3)
Solution:
The given two points are (x1, y1) = (2,4) and (x2, y2) = (1,3).
Using the distance formula
√(x2 – x1)² + (y2 – y1)²
= √(1 -2)² + (3-4)²
=√(-1)² + (-1)²
= √1 + 1
= √2
Therefore the distance between the points is √2.

Example 8.
Find the distance between the points (3,6) to the line 3x – 4y = 5
Solution:
The given point is (x1,y1) = (3,6)
The given line can be written has 3x – 4y – 5 = 0
Comparing this with ax + by + c = 0
We get a = 3 and b= -4, c = -5
Using the distance formula to find the distance from a point to aline
d = ax1 + by1 + c/√a² + b²
d = 3(3) – 4(6) -5/√3² + (-4)²
d = 9-24-5/√9 + 16
d = -20/√25
d = -20/5
The distance from the given point to the given line -20/5 units

Example 9.
Find the distance point (3,6) and the midpoint of the line segment joining (1,2) and (1,4).
Solution:
We first find the coordinates of the midpoint M of the segment joining (1,2) and (1,4).
We know that
Midpoint formula = [a1 + a2/2 , b1 + b2/2]
M = [1 + 1/2, 2 + 4/2]
M = [2/2, 6/2]
M = [1, 3]
We now use the distance formula to find the distance between the points (3,6) and (1,3)
D = √(a2 – a1)² + (b2 – b1)²
D = √(1 – 2)² + (6 – 4)²
D = √ (-1)² + (– 2)²
D = √1 + 4
D = √5
Therefore the distance is √5

Example 10.
The town network is mapped on a coordinate grid with the origin being at city hall. Rishi's house is located at point (2,6) and Lora's house is located at (1,4). How far is it from Rishi's house to lora's house?
Solution:
Using the distance formula
D = √(x2 – x1)² + (y2 – y1)²
D =√ (2 – 3)² + (5 – 4)²
D = √(-1)² + (1)²
D = √1 + 1
D = √2

Distance Properties in Some Geometrical Figures

Hey Guys!!! Would you like to know deeply about the properties of geometrical figures? If our guess is correct, then you are in the right place. This is the place where you can get the solutions for all your doubts. So, refer to our page to practice the problems on the distance and section formulae. Let us discuss in detail Distance Properties in some Geometrical Figures here.

Distance Properties in Some Geometrical Figures

1. A triangle is said to be an equilateral triangle if and only if AB = BC = CA.
2. A triangle is said to be an isosceles triangle if AB = AC or AB = BC or AC = BC.
3. A triangle is said to be a right-angled triangle if
AB = BC + CA or BC = CA + AB or AC = AB + BC
4. The distance from any point from the origin to the center is known as the radius of the circle.
5. A quadrilateral ABCD is said to be rhombus only if AB = BC = CD = DA
6. A quadrilateral ABCD is said to be a parallelogram if the opposite sides are equal and also if AB = CD and AD = BC.
7. A quadrilateral ABCD is a parallelogram but not a rectangle when the opposite sides are equal and diagonals are not equal.
8. A quadrilateral ABCD is a rectangle if the opposite sides and diagonals of the rectangles are equal.
9. A quadrilateral ABCD is a square only if the opposite sides and diagonals of the rectangles are equal. AB = BC= CD= DA.
10. A quadrilateral is a rhombus but not square if its sides are equal but the diagonals are not equal.

See More:

  • Distance and Section Formulae
  • Distance Formula

FAQs on Distance Properties of Geometrical Figures

1. Why is the distance formula important?

The distance formula is a formula used to find the distance between two distinct points on a plane.

2. Why are geometric constructions important?

Geometric constructions help us to draw lines, angles, and shapes with simple tools.

3. How useful are geometric figures in solving problems related to design?

Architects use geometry to study and divide space as well as draft detailed building plans.

Distance Formula

I think you guys are familiar with the chapter coordinate geometry. We can see the distance formula in the coordinate geometry to find the distance between the coordinate points. In this article, we will learn what is distance formula is and the distance between two points. For example to find the distance between the lengths of the triangle in the coordinate plane we will use the distance formula. Scroll down this page to know more interesting points about the distance formula.

See More: Distance and Section Formulae

Distance Formula – Definition

Distance formula in coordinate geometry can be used to find the distance between two points, distance between two parallel lines, and distance between the points in a plane. The distance between the two parallel lines is the shortest distance from one point to the other point.

How do you find the Distance between the Points?

The distance between two points of the coordinate plane can be found using the distance formula. (x, y) are the ordered pairs that represent coordinates of the points. x-coordinate is the distance of the point from the x-axis and the y-coordinate is the distance of the point from the y-axis. We can find the distance between the points using the relevant formulas. We will find the distance between the two points in 2D and 3D planes using the Pythagoras theorem.

Derivation of Distance Formula

By using the Pythagoras theorem we can derive the distance formula.
AB² = AC² + BC²
d² = (x2 – x1)² + (y2 – y1)²
Squaring on both sides
d = √(x2 – x1)² + (y2 – y1)²
Hence proved

Distance Formula for two points

The distance formula for two points is used to find the distance between two points A(x1, y1) and B(x2, y2).
d = √(x2 – x1)² + (y2 – y1)²
where,
d = distance between the points
(x1, y1) and (x2, y2) are the ordered pairs of the two coordinate planes.

Distance Formula for Three Points

The distance formula for two points is used to find the distance between two points A(x1, y1, z1) and B(x2, y2, z3).
d = √(x2 – x1)² + (y2 – y1)² + (z2 – z1)²
where,
d = distance between the points
(x1, y1, z1) and (x2, y2, z2) are the ordered pairs of the two coordinate planes.

Distance from Point to Line

Consider Ax + By + C = 0 be an equation of a line and P be any point in the cartesian-coordinate plane having coordinates P(x1, y1).
PQ = |Ax1 + By1 + C|/√A² + B²

Distance between Two Parallel Lines

The distance between two parallel lines is equal to the Perpendicular distance between them.
let y = mx + c1 and y = mx + c2 be the two parallel lines
The formula to find the shortest distance between two non-intersecting lines is as follows
d = |c1 – c2|/√A² + B²

Distance Formula Examples

Example 1.
Find the distance between the points (3,4) and (0, 5)
Solution:
The formula for the distance D between two points (a,b) and (c,d) is given by
D = √(c – a)² + (d – b)²
Apply the formula given above to find distance D between the points (3,4) and(0,5) as follows.
D = √(0 – 3)² + (5 – 4)²
√(-3)² + (1)²
√4 + 1
√5 = 3.23

Example 2.
Find the distance point (2,4) and the midpoint of the line segment joining (1,3) and (4,5).
Solution:
We first find the coordinates of the midpoint M of the segment joining (2,3) and (4,5).
We know that
Midpoint formula = [a1 + a2/2 , b1 + b2/2]
M = [1 + 4/2, 3 + 5/2]
M = [5/2, 4]
We now use the distance formula to find the distance between the points (2,4) and (5/2,4)
D = √(a2 – a1)² + (b2 – b1)²
D = √(5/2 – 2)² + (4 – 4)²
D = √ (5/2 – 2)²
D = √(0.5)²
D = √0.25
D = 0.5.
Therefore the distance is 0.5

Example 3.
Find a relationship between x and y so that the triangle whose vertices are given by (x,y), (2,2) and (4,3) is a right triangle with the hypotenuse defined by the points (2,2) and (4,3).
Solution:
Let us use the distance formula to find the length of the hypotenuse.
h = √(a2 – a1)² + (b2 – b1)²
h = √(4 -2)² + (3 -2)²
h = √(2)² + (1)²
h = √4 + 1
h = √5 = 2.23
We now use the distance formula to find the sizes of the two other sides a and b of the triangle.
a = √(x – 2)² + (y – 2)²
b = √(x – 4)² + (y – 3)²
Pythagorean theorem gives
(2.23)² = (x – 2)² + (y – 2)² + (x – 4)² + (y – 3)²
Expand the squares, simply and complete the squares to rewrite the above relationship between x and y follows
(2.23)² = x² – 4x – 4 + y² – 4y + 4 + x² – 8x + 16 + y² – 6y + 9
(2.23)² = 2x² + 2y² – 12x – 10y + 25
2x² + 2y² – 12x – 10y = 4.97 – 25
2x² + 2y² – 12x – 10y = – 20.03

Example 4.
The town network is mapped on a coordinate grid with the origin being at city hall. Evans's house is located at point (2,6) and Billy's house is located at (1,4). How far is it from Evans' house to Billy's house?
Solution:
Using the distance formula
D = √(x2 – x1)² + (y2 – y1)²
D =√ (1 – 6)² + (4 – 6)²
D = √(-5)² + (-2)²
D = √25 + 4
D = √29
D = 5.38

Example 5.
Find a relationship between x and y so that the distance between the points (2,4) is equal to 6.
Solution:
We apply the distance formula
6 = √(2 – x)² + (4 – y)²
Square both sides
36 = (2 – x)² + (4 – y)²
The above relationship between x and y is the equation of the circle.
36 = 4 – 2x + x² + 16 – 8y + y²
36 = x² + y² – 2x – 8y + 20
x² + y² – 2x – 8y = 36 – 20
x² + y² – 2x – 8y = 16

FAQs on Distance Formula

1. What is the Distance Formula in Coordinate Geometry?

The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).
d = √(x2 – x1)² + (y2 – y1)²

2. What is the distance between A and B?

The distance from A to B is the same as the distance from B to A. In order to derive the formula for the distance between two points in the plane, we consider two points A(a,b) and B(c,d).

3. What is the distance formula?

The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).

Distance and Section Formulae

Attention Reader!! This article is the one-stop solution for all the students who are feeling distance and section formulae is a tough chapter. We are providing the information deeply about the Distance and Section Formulae Syllabus. All the details seen in this article are prepared by the math experts. So, use this opportunity to learn more about Distance Formula, Section Formula with examples, worksheets, and problems.

Topics Covered in Distance and Section Formulae

  • Distance Formula
  • Distance Properties in some Geometrical Figures
  • Conditions of Collinearity of Three Points
  • Problems on Distance Formula
  • Distance of a Point from the Origin
  • Distance Formula in Geometry
  • Section Formula
  • Midpoint Formula
  • Centroid of a Triangle
  • Worksheet on Distance Formula
  • Worksheet on Collinearity of Three Points
  • Worksheet on Finding the Centroid of a Triangle
  • Worksheet on Section Formula

Distance Formula

The distance formula is used to calculate the distance between the points on the two-dimensional and three-dimensional coordinate plane. You can find the distance between the points by substituting the distance formula.
d = √(x2 – x1)² + (y2 – y1)²
where,
d = distance between the points
(x1, y1) and (x2, y2) are the ordered pairs of the two coordinate planes.
Distance between three points formula:
The distance formula for two points is used to find the distance between two points A(x1, y1, z1) and B(x2, y2, z3).
d = √(x2 – x1)² + (y2 – y1)² + (z2 – z1)²
where,
d = distance between the points
(x1, y1, z1) and (x2, y2, z2) are the ordered pairs of the two coordinate planes.

Section Formula

The Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. This formula is used to find the centroid, incenter, and excenters of a triangle.

Do Check: Co-ordinate Geometry

Distance and Section Formulae Solved Problems

Example 1.
Given a triangle ABC in which A = (4,-4) B = (0,5) and C(5,6) A point P lies on BC such that BP : PC = 3:2 find the length of line segment AP.
Solution:
Given BP : PC = 3 : 2
Using the section formula
The coordinates of the point p(x,y) divide the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio m1: m2.
[m1x2 + m2x1/m2 + m1 , m1y2 + m2y1/m2 + m1]
m1 = 3
m2 = 2
x1 = 0
x2 = 5
y1 = 5
y2 = 10
[3×5 + 2×0/3 + 2, 3×10 + 2×5/3 + 2]
[15/2 , 40/5]
(3,8)
Using distance formula we have
AP = √(3 – 4)² + (8 + 4)²
√1 + 144
√145
= 12.04

Example 2.
A(10,0) and B(5,15) are two fixed points. Find the coordinates of the point P is AB such that 5PB = AB also find the coordinates of some other points Q in AB such that AB = 6AQ.
Solution:
Given that,
5PB = AB
AB/PB = 5/1
AB – PB/PB = 5 – 1/1
AB/PB = 4/1
Using section formula coordinates of P are
P(x,y) = P(4×10 + 1×5/4 + 1 , 4×15 + 1 × 0/4 + 1
(40 + 5/5 , 60/5)
(9,12)
Given AB = 6AQ
AQ/AB = 1/6
AQ/AB – AQ = 1/6 – 1
AQ/QB = 1/5
Using section formula coordinates of Q are
Q(x,y) = Q(1×10 + 5×5/1 + 5, 1×15 + 5×0/1 + 5
Q(x,y) = Q(10 + 25/6, 15/6)
Q(35/6, 15/6)

Example 3.
Coordinates of the points which Divide the join of (1,7) and (4,3) in the ratio of 2:3 is.
Solution:
We know that the coordinates of the point dividing (x1, y1) and (x2,y2) in the ratio m:n is given by
(m1x1 + m2xx1/m1 + m2 , m1y2 + m2y1/m1 + m2)
Given that
(x1,y1) = (1,7)
(x2,y2) = (4,3)
(m1, m2) = 2 : 3
Coordinates of point of intersection of line segment is
(2×4 + 3×1/2 + 3, 2×3 + 3×7/2 + 3)
(8 + 3/5, 6 + 21/5)
(11/5, 27/5)

Example 4.
Using the section formula, the points A(7,5) B(9,3) and C(13,1) are collinear.
Solution:
If three points are collinear, then one of the points divides the line segment joining the order points in the ratio r : 1.
If the P is between the line A and B and AP/PB = r.
Distance AB = √(x1 – x2)² + (y2 – y1)²
= √(9 – 7)² + (3 – 5)²
= √(7)² + (-2)²
= √49 + 4
= √53
= 7.28
Distance BC = √(13 – 9)² + (1 – 3)²
= √(4)² + (-2)²
= √16 + 4
= √20
To find r
r = AB/BC = 7.28/4.47
The line divides in the ratio of 7.28 : 4.47
A line divides internally in the ratio m : n the point P = (mx1 + nx1/m + n, my2 + ny2/m + n)
m = 7.28, n = 4.47
x1 = 7
x2 = 13
y1 = 5
y2 = 1
By point B = 7.28×1 + 4.47×7/7.28 + 4.47
(94.64 + 31.29/11.75, 7.28 + 22.35/11.75)
The three points AB and C are collinear
(125.93/11.75, 29.63/11.27)
(10.7, 2.6)

Example 5.
Find the values of a such that PQ = QR, where P, Q, and R are the points whose coordinates are (1, – 1), (1, 3), and (a, 8) respectively.
Solution:
We know that
Distance AB = √(x1 – x2)² + (y2 – y1)²
PQ = √(1−1)²+(−1−3)²
=√0+(−4)²
=√ 0+16
= √16
QR =√ (1−a)²+(3−8)²
= √(1−a)²+(−5)²
= √(1−a)²+25
Therefore, PQ = QR
√16 =√ (1−a)²+25
16 = (1 – a)² + 25
(1 – a)² = 16 – 25
(1 – a)² = -9
1 – a = ±3
a = 1 ±3
a = (4, -2)

FAQs on Distance and Section Formulae

1. Which is section formula?

Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter, and excenters of a triangle.

2. What are the distance formula section formula and midpoint formula?

The distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2+b2=c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse.

3. Is the section formula and distance formula the same?

The distance formula is used to find the distance between two defined points on a graph. The section formula gives the coordinates of a point that divides the line joining two points in a ratio, internally or externally.

Straight Line Drawn from the Vertex of a Triangle to the Base

This article provides you with the concept of a Straight line drawn from the vertex of a triangle to the base. Children are aware of the topics triangles and straight lines. Here, we will learn and prove how a triangle bisects the opposite side by a straight line with the help of the midpoint theorem. A triangle is a polygon shape formed by three line segments with three midpoints respectively.

By the end of this page, kids will perceive how a line bisects the other side. Without any delay let us discuss the concept and prove the statement. On this page, we have also given a glimpse on the topics called median and the centroid of a triangle for a better understanding of the Straight Line Drawn from the Vertex of a Triangle to the Base.

Also, Refer:

  • Similar Triangles
  • Midpoint Theorem
  • Converse of the Midpoint Theorem

Median

A median is defined as the line segment joining a vertex of the opposite side of a triangle and bisects the side. In other words, a straight line drawn from one vertex to the opposite of the vertex of a triangle is called a median and every triangle has three medians. Learn the concept of the median of raw data and gain proper knowledge on how to solve it easily.

The Centroid of a Triangle

A triangle is formed by joining three dots by a line segment connected to each other and forms a triangle. A triangle has three medians with three angles. A centroid is defined as "the point of intersection of three medians of a triangle".

For example, if we take a triangle ABC we have three midpoints D, E, and F respectively. Here, O is the point of intersection of three medians and O is a centroid of a triangle. The below figure gives how a centroid of a triangle looks like.

Centroid of a triangle

Properties of a Centroid

The properties of a centroid are as follows

  • It is a point of congruency.
  • Centroid is the intersection of three medians and bisects on the opposite side.
  • A centroid is always inside of a triangle.
  • It divides each median in the ratio of 2:1.

Statement

A straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the midpoints of the other two sides of the triangle.

Triangle with midpoints

To Prove:

Consider a triangle PQR.

Given M, N, and O are the midpoints of PQ, PR, and QR respectively.

We have to prove that MN bisects PO i.e., PX = XO.

Construction:

From the above △PQR, join the midpoints M, N, and O of a triangle and name the intersection point of a triangle as X.

Traingle

The above figure is the constructed triangle PQR, to prove the theorem statement.

Proof of the Theorem:

Given PQR is a triangle.

Let M be the midpoint of PQ, N be the midpoint of PR, and O be the midpoint of QR.

O is the midpoint of QR then PO is the straight line that bisects MN at point O.

Since M and N are the midpoints of the sides PQ and PR of a △PQR, then

MN ∥ QR by the midpoint theorem.

since MN ∥ QR, then MX ∥ QO.

In △PQO, we know that M is the midpoint of PQ and X be the midpoint of PO.

By converse of midpoint theorem

MX ∥ QO

⇒ PX=XO.

Thus, MN bisects PO.

Hence, the statement Straight Line Drawn from the Vertex of a Triangle to the Base is proved.

FAQ's on Prove that a Line Drawn from the Vertex to its Base is a Straight Line

1. What do you call the straight line joining the vertex with the center of the base?
In mathematics, a triangle is a line segment that joins the vertex to the midpoint of the opposite side, it bisects the side. Every triangle has three medians from each vertex and all three midpoints intersect each other and it is known as the centroid of a triangle.

2. What is called the line joining vertex and the midpoint of the base of the triangle?
The line segment joining a vertex of a triangle to the midpoint of a triangle is known as the median. Every triangle has three medians and three altitudes.

3. What is the base of a triangle?
The base of a triangle is the bottom line of a triangle, and it is one of the three sides of a triangle. But in a triangle, one side is a base side and the remaining two sides are the height and the hypotenuse side of a triangle.

4. What is the vertex of a triangle?
The vertex or vertices are the corners of a triangle. There will be three vertices for every triangle.

Simple Interest Worksheet

Calculating Simple Interest is an essential skill for anyone who maintains a bank account or wants to apply for a loan. Free Printable Simple Interest Worksheet PDF will improve your homeschool math skills as well as helps you to become better at calculations. Solutions provided for all the Questions in the Simple Interest Word Problems Worksheet will make it easy for you to learn the concept. Learn the Formulae, Tips & Tricks associated with the Simple Interest Concept easily by solving the problems on Simple Interest.

Do Refer:

  • Word Problems on Simple Interest
  • Compound Interest

Calculating Simple Interest Worksheet

I. Find the amount and S.I of the following
a) If p=7000, R=5%, T=3 years
b) If p= 9000, R=3%, T=5 years

Solution:

Given that,
p=7000, R=5%, T=3 years
We know that S.I=ptr/100
S.I=7000 × 3 × 5/100
S.I=105000/100
S.I=1050
Amount=P+S.I
Amount=7000 + 1050=8050
Therefore, S.I is Rs 1050 and the Amount is Rs 8050.
b)If p= 9000, R=3%, T=5 years
Given that,
p= 9000, R=3%, T=5 years
We know that S.I=ptr/100
S.I=9000 × 5 × 3/100
S.I= 1350
Amount=Principle+S.I
Amount=9000 + 1350
=10350
Therefore, S.I is Rs 1350, Amount is Rs 10350.

II. a)If Principle=6000, S.I=250. Find the amount?
b) If S.I=300, Amount=4800. Find the Principle?
c) If Principle=1700,Amount=2000. Find the S.I?

Solution:

a) Given that,
Principle=Rs 6000, S.I=Rs 250
We know that Amount=Principle + S.I
Amount= 6000 +  250
Amount=Rs 6250
Therefore, Amount=Rs 6250.
b) Given that,
S.I=Rs 300, Amount=Rs 4800
We know that Amount=Principle + S.I
Principle=Amount-S.I
principle= 4800 –  300
=Rs 4500
Therefore, Principle is Rs 4500.
c) Given that,
Principle=Rs1700,
Amount=Rs 2000
We know that Amount=principle + Interest
Interest=Amount- Principle
Interest=2000 – 1700
Interest=Rs 300
Therefore, Simple Interest is Rs 300.

III. Find Time when, Principal is Rs 5000, Interest is Rs 200, Rate is 2% p.a?

Solution:

Given p=5000,
I=Rs 160
R=3%
We know that S.I=ptr/100
T = SI×100 / P × R
T=200 × 100/5000 × 3
T=20000/5000 × 3
T=12
Therefore, Time=12 years.

Iv. Find Rate when Principal is Rs 7000, Interest is Rs 700 and Time is 4 years?

Solution:

Given that,
Principal=7000
Interest=700
Time=4 years
We know that S.I=ptr/100
r=SI×100 / P × t
r=700 ×100 /7000 ×4
r=70000/28000
r=2.5
Therefore, rate=2.5%.

V. Find the principal which gives an Amount of Rs 3500 at the rate of 8% for 5 years?

Solution:

Given that,
Amount=3500
rate=8%
T=5 years
We know that S.I=ptr/100
p=SI×100 / t × r
p= 3500 ×100 / 5 × 8
p=350000/40
p=Rs 8750
Therefore, the Principal is Rs 8750.

VI. Sarath deposited Rs 10000 for 2 years and 6 months at the rate of 5% p.a. Find the amount at maturity?

Solution:

Given that,
Sarath deposited p=Rs10000
T=2 years 6 months
r=5%
First, the compound interest for 2 years will be calculated and then the amount after 2 years will be considered as principal and on this principle, the simple interest for 6 months will be calculated.
Compound interest for 3 years
A = P (1 + R/100)ⁿ
A=10000(1 + 5/100)2
=10000(1+1/20)2
=10000(21/20)2
=Rs 1102.5
Now for the next 6 months, the principle will be Rs 1102.5.
Simple Interest = (P*R*T)/100
S.I=1102.5 × 5 × 6/100
S.I=Rs 330.75
Amount=Principle + interest
Amount=1102.5 + 330.75
Amount=Rs 1433.25
Hence, the maturity amount will be Rs 1433.25.

VII. Arun deposited Rs 15000 at a rate of 12% (p.a.) for 175 days. Find the amount he got back after 175 days.

Solution:

Given that,
P=Rs 15000,
r=12%,
T=175 days=5 months 15 days=5 1/2=11/2
Simple Interest = (P×R×T)/100
=15000 × 12 × 11/100 × 2 × 12
=825
Amount=Principle + Interest
=15000 + 825
=15825
Therefore, Amount is Rs 15825.

VIII. Kishore deposited Rs. 500 per month for 12 months in a banks' recurring deposit account. If the bank pays interest at the rate of 9% per annum, find the amount (in Rs.) he gets on maturity?

Solution:

Given that,
p=500,
T=12 months
R=9%
Equivalent principal for 12 months = 100× n(n+1)/2
=500 × 12(13)/2
=39000
Interest=PRT/100
=39000 × 9 ×1/100 × 12
=292.5
Maturity amount=P × T+I
=500 × 12 + 292.5
=6000+292.5
=6292.5
Therefore, the maturity amount is Rs 6292.5.

​VIII. Mahesh borrowed Rs 2000 at the rate of 5% (p.a.) for 5 years and 5 months. Find the amount he paid back?

Solution:

principle = Rs2000
time = 5.5 years
rate of interest = 5%
Simple Interest = P×T×R ÷ 100
= 2000×5.5×5 ÷ 100
= 550
Amount = Principle + Simple Interest
= 2000+550
= 2550
Therefore, Amount is Rs 2550.

IX. Find Principal when, Time is 2 years, Interest is Rs 400 and Rate is 3% p.a?

Solution:

Given that,
Time=2 years,
I=Rs 400
Rate=3%
Pricipal=(100 × Interest)/(Rate × Time)
=(100 × 400)/(3 × 2)
=40000/6
=6666.66
Therefore, Principle is Rs 6666.66.

X. In what time will Rs 500 amount to Rs 570 at the rate of 5% p.a. simple interest?

Solution:

Given that,
Principle = 500
Amount = 570
A = SI + P so,
SI = A – P
= 570-500
= 70
Now, T = SI * 100/P*R
= 70*100/500*5
= 7000/2500
= 2.8 years
Therefore, Time taken will be 2.8 years.

Area and Perimeter of a Semi Circle

Area and Perimeter of Semi-circle Problems help the students to explore their knowledge of semi-circle Problems. Solve all the Problems to learn the formula of the Area of the semi-circle and the Perimeter of a semi-circle. To know the definition, Problems with Solutions, Formulas of Semi-circle you can visit our website. We have given the complete Semi-circle concept along with examples. Check out the Area and Perimeter of Semi-circle Problems and know the various strategies to solve problems in an easy and understandable way.

Semi-Circle – Definition

A semi-circle is defined as a half-circle that is formed by cutting a whole circle into two halves along with a diameter line. The Semi-circle has only one line of symmetry which is the reflection symmetry. A line segment is known as the diameter of a circle cuts the circle into two equal semicircles. It is also referred to as half-disk. A semi-circle will be half of the circle that is 360 degrees, the arc of the semicircle always measures 180 degrees. The below figure shows the Semi-circle.

Area and Perimeter of a Semi-Circle – Definitions and Formulas

Area of a Semi-Circle: The area of a semicircle is half of the area of a circle. We know that the area of a circle is πr2. So, the area of a semicircle is,
Area of a Semi-circle = 1/2πr2
where r will be the radius.
The value of π is constant. So, the value is 3.14 or 22/7.

The perimeter of a Semi-Circle: The perimeter of a semicircle is defined as the total length of its boundary. It is also known as the Circumference of a Semicircle. It is calculated as adding the length of the diameter and half the circumference of a circle. The perimeter of a circle unit is expressed in linear units like inches, feet, meters or centimeters and etc.
The Perimeter of a semicircle is, πr+2r.
Where r is the radius and π is a constant value that is 3.14.

The perimeter of the semicircle is P = Half of the Circumference of the original circle + Length of the Diameter.
The Circumference of a circle is 2πr.
The half of the circumference of the circle is 1/2 x the circumference of a circle.
=1/2 x 2πr = πr =π(d/2).
The length of the diameter is d = 2r.
In terms of radius, the circumference of the semicircle is P = C = πr + d = πr + 2r = (π + 2)r
In terms of Diameter, the circumference of a semicircle is P = C = π(d/2) + d.

Also Read :

  • Area and Circumference of a Circle
  • Worksheet on Area and Perimeter of Triangle
  • Worksheet on Area and Perimeter of Squares

Solved Problems on Area and Perimeter of a Semicircle

Problem 1: Find the area of a semicircle. If the perimeter of a semicircle is 122 units. Consider the π value is  22/7.

Solution:
Given the values in the question,
The perimeter of a semicircle is 122 units.
We know that the area of a semicircle is 1/2(πr2).
Using the perimeter of a semicircle value, find the radius (r) value.
The perimeter of a semicircle is πr + 2r = 122units.
πr + 2r = 122 units.
(22/7+2)r = 122 units
(36/7)r = 122 units
r = (7/36) x 122 = 23.7 units.
Now, we will find the area of a semicircle.
A = 1/2(πr2).
A = (22/7 × (23.72))/2 = (3.14)x(561.6)/2 = 881.7 square units.
Therefore, the area of the semicircle is 881.7 sq. units.

Problem 2: If the radius of a semicircle is 4units, then using the semicircle formula find its perimeter?

Solution:
As given in the question, the radius of the semicircle is 4units.
Now, we have to find the perimeter of a semicircle using the formula.
The perimeter or circumference of a semicircle is, πr + d = πr + 2r
Substitute the radius (r) value in the above formula, we get
P = πr + 2r =(π+ 2)r = (3.14 + 2)(4) = 20.56 units
Thus, the perimeter of a semicircle is 20.56 units.

Problem 3: Using the below figure, find the Perimeter and Area of a semicircle?

Solution:
Given in the question, the figure consists of radius value, r = 2cm.
Using the radius value, find the area and perimeter of a semicircle.
We know that the Area of a semicircle is (πr2)/2.
So, the value is, A = {(22/7) (2)2} /2 cm2.
A = (3.14) (4) /2 cm2= 6.28cm2
Now, the perimeter of a semicircle is (π + 2)r.
Substitute the values in the above formula. We get,
P = (π + 2)r = (3.14 + 2)(2) = (5.14)(2) = 10.28 cm.
Thus, the perimeter and area of a semicircle are 6.28cm2 and 10.28 cm.

Problem 4: Find the area of semicircle using the below figure. The below figure consists of a semicircle and equilateral triangle.

Solution:
As given in the question, the figure consists of a semicircle and triangle.
The radius of a semicircle is 4cm.
We know that the area of a semicircle is 1/2(πr2).
Substitute the given values in the above formula, we get
Now, we will the area of a semicircle.
A = 1/2(πr2).
A = (22/7 × (42))/2 = (3.14)x(16)/2 = 25.12 cm2.
Therefore, the area of the semicircle is cm2.

FAQ'S on Area and Perimeter of a Semicircle

1. What are the steps for finding the perimeter of a semicircle?

The steps to determine the perimeter of a semicircle are given below:

  • Step1: First, find the product of π and the radius (r) of the semicircle.
  • Step 2: Next, find the diameter of the semicircle.
  • Step 3: Then Add the values obtained in the above two steps.
  • Step 4: Thus the value obtained is the perimeter of the semicircle.

2. What is the difference between circumference and Perimeter of Semicircle?

The perimeter of a semicircle and the circumference of a semicircle mean the same. Both are referred to the total length of the boundary of a semicircle. Therefore, the circumference of a semicircle is another name for the perimeter of a semicircle.

3. Is a semicircle is half the circle?

Yes, a semicircle is half the circle. It means a circle can be divided into two semicircles.

4. What is the shape of a semicircle?

The shape of a semicircle is obtained by cutting a circle into two equal parts along its diameter and the full arc of a semicircle is always measured 180 degrees. An example of a semicircle shape is a protractor.

5. What is the angle of the semicircle?

The angle of the semicircle is 90 degrees, the angle is made by the triangle in a semicircle is a right angle.

Square Matrix

A matrix is defined as the arrangement of elements in the form of an array. A square matrix is a type of matrix where the number of rows is equal to the number of columns. It is an effective way of arranging elements of the matrix. Learn the definition, concept, examples and operations on a square matrix in the following sections.

Definition of A Square Matrix

A square matrix is a matrix that has an equal number of rows and columns. The n x n matrix is called a square matrix. It is possible to add, subtract and multiply any two square matrices. The multiplication of two square matrices is also a square matrix. So, the number of elements of a square matrix is always a perfect square number.

The important matrices related to a square matrix are as follows:

  • Identity Matrix: It is a square matrix that has 1 as the diagonal elements and the remaining elements as zeros.
  • Scalar Matrix: It is a square matrix where the diagonal elements are constant numbers and others are equal to zero.
  • Trace of a Matrix: The total of the diagonal elements of a matrix is the trace of a matrix.

Also, Check

  • Order of a Matrix
  • Classification of Matrices
  • Row Matrix

Square Matrix Examples

Here we are giving some of the examples of a square matrix with a detailed explanation.

\( A =\left[
\begin{matrix}
11 & 22\cr
33 & 44\cr
\end{matrix}
\right]
\)

The above mentioned matrix is a square matrix of order 2 x 2. The number of rows = 2 = number of columns. So, it is called a square matrix of order 2

\( B =\left[
\begin{matrix}
1 & 2 & 3\cr
4 & 6 & 8\cr
3 & 5 & 7\cr
\end{matrix}
\right]
\)

The above matrix has an order 3 x 3. Since the number of rows and columns are equal, it is a square matrix of order 3. We can find the determinant of the square matrix.

\( C =\left[
\begin{matrix}
14 & 10 & 3 & 11\cr
15 & 1 & 5 & 15\cr
7 & 13 & 2 & 8\cr
6 & 4 & 12 & 16\cr
\end{matrix}
\right]
\)

It is a square matrix of orde 4 x 4.

Operations of Square Matrices

Mathematical operations such as addition, subtraction, multiplication can be performed across two square matrices. Here we are giving the process and examples for a better understanding.

Addition & Subtraction of Two Square Matrices:

Two square matrices can be added/subtracted in a simple way. Let us consider two square matrices.

\(\left[
\begin{matrix}
a1 & a2\cr
a3 & a4\cr
\end{matrix}
\right] \) ±\(\left[
\begin{matrix}
b1 & b2\cr
b3 & b4\cr
\end{matrix}
\right] \) = \(\left[
\begin{matrix}
a1 ± b1 & a2 ± b2\cr
a3 ± b3 & a4 ± b4\cr
\end{matrix}
\right] \)

Examples:

\(\left[
\begin{matrix}
11 & 22\cr
33 & 44\cr
\end{matrix}
\right]
\) + \(\left[
\begin{matrix}
6 & 8\cr
15 & 10\cr
\end{matrix}
\right]
\) = \(\left[
\begin{matrix}
17 & 30\cr
48 & 54\cr
\end{matrix}
\right]
\)

\(\left[
\begin{matrix}
7 & 16\cr
12 & 18\cr
\end{matrix}
\right]
\) – \(\left[
\begin{matrix}
5 & 9\cr
3 & 6\cr
\end{matrix}
\right]
\) = \(\left[
\begin{matrix}
2 & 7\cr
9 & 12\cr
\end{matrix}
\right]
\)

Multiplication of Square Matrix:

Multiplication of a constant with a square is simple.

5A = 5 x \(\left[
\begin{matrix}
2 & 5\cr
10 & 13\cr
\end{matrix}
\right]
\) = \(\left[
\begin{matrix}
10 & 35\cr
45 & 60\cr
\end{matrix}
\right]
\)

The multiplication of two square matrices involves a sequence of steps. We have to multiply the first row of the first matrix with the first column of the second matrix. The following is the detailed process.

\(A = \left[
\begin{matrix}
1 & 3\cr
2 & 4\cr
\end{matrix}
\right]
\) and \(B =\left[
\begin{matrix}
12 & 15\cr
4 & 6\cr
\end{matrix}
\right]
\)

A x B = \(\left[
\begin{matrix}
(1×12 + 2×4) & (1×15 + 3×6)\cr
(2×12 + 4×4) & (2×15 + 4×6)\cr
\end{matrix}
\right]
\) = \(\left[
\begin{matrix}
20 & 33\cr
40 & 54\cr
\end{matrix}
\right]
\)

Transpose of a Square Matrix

The transpose of a matrix is the matrix obtained by transposing the elements of rows into columns, and columns into rows. The order of the given matrix and its transpose is different for some matrices. If the order of a matrix is m x n, then its transpose order is n x m. But in the case of a square matrix, the order is the same.

Example:

\( A =\left[
\begin{matrix}
8 & 10 & 12\cr
14 & 15 & 16\cr
3 & 5 & 7\cr
\end{matrix}
\right]
\)

AT = \(\left[
\begin{matrix}
8 & 14 & 3\cr
10 & 15 & 5\cr
12 & 16 & 7\cr
\end{matrix}
\right]
\)

In a square matrix, if the given matrix and its transpose are equal, then it is called a symmetric matrix. If the transpose matrix is equal to the negative of the given matrix, then it is a skew-symmetric matrix.

Square Matrix Determinant

The determinant of a matrix is a numerical value or it is a summary value that represents the entire set of elements of the matrix. The determinant of a square matrix having the order 2 x 2 can be easily calculated using the below-given formula.

\( A =\left[
\begin{matrix}
a & b\cr
c & d\cr
\end{matrix}
\right]
\)

|A| = |ad – bc|

If |A| = 0, then it is called a singular matrix otherwise it is a non-singular matrix.

The Inverse of a Square Matrix

The inverse of a matrix is used to divide one matrix with another matrix. You have to calculate the determinant of a square matrix and its adjoint to find its inverse. The inverse of a matrix is obtained by dividing the adjoint matrix with the det of the square matrix.

\( A =\left[
\begin{matrix}
a & b\cr
c & d\cr
\end{matrix}
\right]
\)

A-1 = \(\frac { 1 }{ |ad – bc| } \) . \(\left[
\begin{matrix}
d & -b\cr
-c & a\cr
\end{matrix}
\right]
\) = \(\frac { 1 }{ |A| } \) . adj(A)

A square matrix is called an orthogonal matrix if A T = A-1

Important Properties of Square Matrix

The important properties of a square matrix are given here:

  • The number of rows is equal to the number of columns.
  • The sum of the elements of a square matrix is the trace of a matrix.
  • The order of the transpose and original matrices are the same.
  • We can perform different operations like addition, subtraction, multiplication and inverse on a square matrix.
  • The determinant also can be calculated easily.

Example Questions & Answers

Question 1:
Find the transpose of the square matrix \( A =\left[
\begin{matrix}
4 & 9\cr
11 & 35\cr
\end{matrix}
\right]
\)

Answer:
Given matrix is \( A =\left[
\begin{matrix}
4 & 9\cr
11 & 35\cr
\end{matrix}
\right]
\)
It's transpose is \(\left[
\begin{matrix}
4 & 11\cr
9 & 35\cr
\end{matrix}
\right]
\)

Question 2:
Find the multiplication of two square matrices \( A =\left[
\begin{matrix}
1 & 8\cr
7 & 5\cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
5 & 4\cr
6 & 7\cr
\end{matrix}
\right]
\)

Answer:
The given matrices are \( A =\left[
\begin{matrix}
1 & 8\cr
7 & 5\cr
\end{matrix}
\right]
\) and \( B =\left[
\begin{matrix}
5 & 4\cr
6 & 7\cr
\end{matrix}
\right]
\)
A x B = \(\left[
\begin{matrix}
1×5 + 8×6 & 1×4 + 8×7\cr
7×5 + 5×6 & 7×4 + 5×7\cr
\end{matrix}
\right]
\) = \(\left[
\begin{matrix}
53 & 60\cr
65 & 63\cr
\end{matrix}
\right]
\)

Question 3:
Find the inverse of the given square matrix \(C = \left[
\begin{matrix}
5 & 6\cr
2 & 3\cr
\end{matrix}
\right]
\)

Answer:
Given matrix is \(C = \left[
\begin{matrix}
5 & 6\cr
2 & 3\cr
\end{matrix}
\right]
\)
|C| = |ad – bc| = |15 – 12| = 3
C-1 = \(\frac { 1 }{ 3 } \) . \( \left[
\begin{matrix}
3 & -6\cr
-2 & 5\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
1 & -2\cr
-2/3 & 5/3\cr
\end{matrix}
\right]
\)

FAQ's on Square Matrix

1. What is a square matrix give an example?

A square matrix is a m x m matrix, where first m is the number of rows and second m is the number of columns. The example is \( A =\left[
\begin{matrix}
3 & 4 & 9\cr
12 & 11 & 35\cr
12 & 11 & 35\cr
\end{matrix}
\right]
\)

2. How to find a square matrix?

A square matrix can be found by checking the number of rows and columns in it. If they are equal, then it is said to be a square matrix.

3. What is squaring a matrix?

Squaring a matrix means multiplying a matrix by its own. Before multiplying the matrices, you have to check whether the number of columns of the first matrix is equal to the number of rows of the second matrix or not.

4. What are the dimensions of a square matrix?

The dimension of a square matrix is the number of rows by the number of columns. Here, both numbers should be equal.

Worksheet on Word Problems on Linear Equation

Worksheet on Word Problems on Linear Equation will be of much help in practicing problems on the linear equations. If you would like to know more about the linear equation concept then the Linear Equations Word Problems Worksheet with Answers can be of utmost help. Linear Equation Word Problems Worksheet PDF can be a great resource to practice a variety of word problems. Step by Step Solutions provided for all the problems makes it easy for you to understand the concept in a fun and engaging manner.

Do Refer:

  • Practice Test on Word Problems on Linear Equations
  • Practice Test on Linear Equations

Linear Equations Word Problems Worksheet with Answers PDF

I. Two third of a number is 52. Find 50% of the number?

Solution:

Given that,
Two third of the number is 52.
Let the number be x.
2/3(x)=52
x=52 × 3/2
=78
50% of the number=1/2(78)
=39
Therefore, 50% of the number is 39.

II. The difference between the two numbers is 12. The ratio of the two numbers is 5 : 3. What are the two numbers?

Solution:

Given that,
The difference between the two numbers is =12
The ratio of the two numbers is =5 : 3
Let the two numbers be x,y.
x-y=12 –>(1)
x/y=5/3 —>(2)
x=5/3y
substitute x in (1)
5/3y-y=12
2y/3=12
2y=36
y=18
Substitute y in Eqn(1)
x-y=12
x-18=12
x=18+12=30
Therefore, Two numbers are 30, 18.

III. Sum of the two numbers is 70. If one exceeds the other by 10, Find the numbers?

Solution:

Given that,
Sum of the two numbers is= 70
Let x be one of the two numbers.
Then, the other number is (x + 10).
So, we have
x + (x + 10) = 70
x + x + 10 = 70
2x + 10 = 70
Subtract 10 from each side.
2x + 10 – 10 = 70 – 10
2x = 60
Divider each side by 2.
2x/2 = 60/2
x = 30
x + 10 = 30 + 10
x + 10 = 40
Hence, the two numbers are 30 and 40.

IV. The perimeter of a rectangular swimming pool is 120 m. Its length is 4 m more than thrice its width. What are the length and the width of the pool?

Solution:

Given that,
The perimeter of a rectangular swimming pool is =120 m
Let l be the length and w be the width of the swimming pool.
Also given  Length is 4 m more than thrice its width.
So, the length is l = 4w + 4
The perimeter of the swimming pool is 120 m.
2l + 2w = 120
Plug l = 4w + 4
2(3w + 4) + 2w = 120
6w+8+2w=120
8W+8=120
Subtract 8 from each side
8w+8-8=120-8
8w=112
w=112/8=14 m
Therefore, length=4w+4
=4(14)+4
=56+4
=60 m
Therefore, the length and width of the pool are 14 m and 60 m.

V. Vijaya is 5 years older than her younger sister. After 10 years, the sum of their ages will be 45 years. Find their present ages?

Solution:

let the age of younger sister be X and Vijaya=x+5
after10 years younger sister age=x+10and Vijaya=x+10+5
x+10+x+15=45
2x+25=45
x=45-25/2
=20/2
=10
The age of Vijaya=x+5 years=10+5=15 years.
The age of younger sister=10years.
Therefore, the age of Vijaya is 15 years, and her younger sister is 10 years.

VI. In a class of 64 students, the number of boys is 3/5 of the number of girls. Find the number of boys and girls in the class?

Solution:

Let the number of girls be x.
So, 3/5x + x = 64
3x+5x/5=64
8x = 320
x=320/8=40
The number of girls is 40
So the number of boys is 3/5(40)=24
Therefore, the number of boys and girls in the class are 24 and 40.

VII. Among two supplementary angles, the measure of the larger angle is 42 more than the measure of the smaller. Find their measures?

Solution:

Given that,
The measure of the larger angle is 42 more than the measure of the smaller.
Let us assume supplementary be x0 and (180−x)0.
Let the larger angle be x0.
Then,
(180−x)+42=x
222-x=x
222=2x
x=222/2=111
Larger angle=1110.
Smaller angle=(108-42)=660.
Therefore, the Larger angle is 1110 and the smaller angle is 660.

VIII. In an isosceles triangle, the base angles are equal and the vertex angle is 60°. Find the measure of the base angles?

Solution:

Let the base angle be x.
Another base angle =x.
Angle sum property of triangle
2x+60=1800
2x=180-60
2x=120
x=120/2=600
First base angle=600=second base angle.
Therefore, the measure of base angles is 600.

Ix. The sum of two consecutive even numbers is 30. Find the numbers?

Solution:

Given that,
The sum of two consecutive even numbers is =30
Let's assume the two consecutive even numbers are x and x+2.
According to the question,
The Sum of x and x+2 =30.
We have,
x+(x+2)=30
2x+2=30
2x=30-2
2x=28
x=28/2=14
x+2=14+2=16
Therefore, the two consecutive numbers are 14 and 16.

X. The denominator of a fraction is greater than the numerator by 6. If the numerator is increased by 3 and the denominator is decreased by 1, the number obtained is 4/5, find the fraction?

Solution:

Let the numerator of the rational number be x.
So as per the given condition, the denominator will be x + 6.
The rational number will be x/x+6
According to the given condition,
x+3/x+6-1=4/5
x+3/x+5=4/5
5(X+3)=4(x+5)
5x+15=4x+20
5x+15-4x-20=0
x-5=0
x=5
The rational number will be x/x+6
=5/5+6
=5/11
Therefore, the rational number is 5/11.

XI. Laxman's father is 66 years old. he is 6 years older than six times adman's age. what is adman's age?

Solution:

Let Adam's father's present age be f.
Let Adam's present age be a.
f=66
f=6+6a
66=6+6a
60=6a
a=60/6=10
Hence, Adam's age is 10years.

XII. Convert the following statements into equations.
(a) 4 added to a number is 10.
(b) 2 subtracted from a number is equal to 14.
(c) 5 times a number decreased by 2 is 13.
(d) 4 times the sum of the numbers x and 5 is 40.

Solution:

(a) Let the number be x.
4+x=10
(b) Let the number be x.
x-2=14
(c) Let the number be x.
5x-2=13
(d) 4(x+5)=40