Distance and Section Formulae

Distance and Section Formulae – Derivation, Examples | How to Solve Problems on finding Distance and Section?

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

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.

Word Problems on Ratio

Word Problems on Ratio | Free Printable Ratio Word Problems with Answers

Are you searching for the word problems on ratio, this page helps you to get answers to your questions. Here, students or teachers can get many types of ratios word problems which helps to understand the concept thoroughly. The word ‘Ratio’ is a term used to compare two or more numbers or quantities.

By using ratio it is very simple to solve the problems and gives the result in the simplest form. We use ‘:‘ to denote the ratio while comparing the quantities and read as ” is to “. Let us practice some examples from this ratio word problems worksheet with answers pdf and get an idea of how to solve the problems on ratio.

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Ratio Word Problems with Solutions

Example 1: 
Find the quantity if it is divided in the ratio 6: 4, the smaller part is 48.
Solution: 
Let the quantity be ‘m’.
Then the ratio of two parts is written as \(\frac{6m}{6+4}\) and \(\frac{4m}{6+4}\).
Here, the smaller part is 48, we get
\(\frac{4m}{6+4}\) = 48
⇒ \(\frac{4m}{10}\) = 48
⇒ 4m = 48×10
⇒ 4m = 480
⇒ m = \(\frac{480}{4}\)
⇒ m = 120
Therefore, the quantity is 120.

Example 2:
A herd of 50 cows and buffaloes has 14 cows and some buffaloes. What is the ratio of buffaloes to cows?
Solution:
Given a total of 50 cows and buffaloes.
There are 14 cows given.
To get the count of buffaloes = 50 – 14 = 36.
The ratio of buffaloes to cows is 36: 14 = 18: 7.
Hence, the simplest form of ratio is 18: 7.

Example 3:
Mr. Ram’s class has 40 students, of which 18 are girls. Find the ratio of girls to boys?
Solution:
Total students in a class = 40.
No. of girls in a class = 18 (given)
No. of boys in a class = Total students – No. of girls = 40 – 18 = 22.
There are 22 boys in a class.
The ratio of girls to boys is 18: 22.
Here, there is a common factor 2. So, we can write the ratio in the simplest form by dividing it with the factor 2.
The simplest form of ratio is 9: 11.

Example 4:
If the ratio of p: q = 4: 3, then find (2p- 3q) : (5p+q)?
Solution:
Given p: q = 5: 3, then p = 5m and q = 3m (m ≠ 0 is a common multiplier).
Now, (2p- 3q) : (5p+q) = \(\frac{2p-3q}{5p+q}\)
= \(\frac{(2×5m)-(3×3m)}{(5×4m)+3m}\)
= \(\frac{10m-9m}{20m+3m}\)
= \(\frac{1m}{23m}\)
= \(\frac{1}{23}\)
= 1: 23.
Therefore, the ratio (2p- 3q) : (5p+q) = 1: 23.

Example 5:
Two numbers are in the ratio 3: 4. If 3 is added to the first number and 8 is added to the second number, they are in the ratio of 3: 5. Find the numbers?
Solution:
Given the ratio of two numbers is 3: 4.
Let the numbers be 3x and 4x.
According to the problem given,
\(\frac{3x+3}{4x+8}\) = \(\frac{3}{5}\)
⇒ 5(3x+3) = 3(4x+8)
⇒ 15x+15 = 12x+24
⇒ 15x-12x = 24-15
⇒ 3x = 9
⇒ x = \(\frac{9}{3}\)
⇒ x = 3.
Hence, the original numbers are 3x = 3×3 = 9 and 4x = 4×3 = 12.
Therefore, the numbers are 9 and 12.

Example 6:
Find the ratio of a: c from the quantities a: b = 4: 5, b: c = 2: 6?
Solution:
Given the ratios a: b = 4: 5, b: c = 2: 6.
a: b = 4: 5 ⇒ \(\frac{a}{b}\) = \(\frac{4}{5}\) ——(i)
b: c = 2: 6 ⇒ \(\frac{b}{c}\) = \(\frac{2}{6}\) ——(ii)
Now, multiply the equations (i) and (ii), we get
\(\frac{a}{b}\) × \(\frac{b}{c}\) = \(\frac{4}{5}\) × \(\frac{2}{6}\)
⇒ \(\frac{a}{c}\) = \(\frac{8}{30}\)
The ratio a: c = 8: 30 = 4: 15.
Thus, the ratio a: c = 4: 15.

Example 7:
If we have the ratio of tomatoes to apples is 2: 4. If there are 18 tomatoes. How many apples are there?
Solution:
The ratio of tomatoes to apples is 2: 4 means that for every 2 tomatoes, you have 4 apples.
Here, you have 18 tomatoes, or we can say 9 times as much.
So, you need to multiply the apples by 9.
⇒ The apples we have is 4 = 4 × 9 = 36.
Thus, there are 36 apples.

Example 8:
If the equation (2x+5y): (6x-4y) = 8: 5. Find the ratio of x: y?
Solution:
Given, (2x+5y): (6x-4y) = 8: 5
Now,
\(\frac{2x+5y}{6x-4y}\) = \(\frac{8}{5}\)
⇒ 5(2x+5y) = 8(6x-4y)
⇒ 10x+25y = 48x-32y
⇒ 10x – 48x = -32y – 25y
⇒ -38x = -57y
⇒ 38x = 57y
⇒ \(\frac{x}{y}\) = \(\frac{57}{38}\)
⇒ x: y = 57: 38.
Hence, that ratio of x: y is 57: 38.

Example 9:
Manoj leaves $ 2461600 behind. Manoj’s wish was the money is to be divided between his son and daughter in the ratio of 3: 2. Find the money received by his son and daughter?
Solution:
Manoj has money of $ 2461600 and is to be shared with his son and daughter in the ratio of 3: 2.
We know if a quantity x is divided in the ratio of a: b, then the two parts are look alike. i.e., \(\frac{ax}{a+b}\) and \(\frac{bx}{a+b}\).
Now, the money received by his son = \(\frac{3}{3+2}\) × $ 2461600
= \(\frac{3}{5}\) × $ 2461600
= 3 × $ 492320
= $ 1476960
Next, the money received by his daughter = \(\frac{2}{3+2}\) × $ 2461600
= \(\frac{2}{5}\) × $ 2461600
= 2 × $ 492320
= $ 984640
Thus, the money received by Manoj’s son is $ 1476960 and by his daughter is $ 984640.

Worksheet on Comparison on Ratios

Worksheet on Comparison on Ratios | Comparing Ratios Worksheet with Answers PDF

In this worksheet on comparison on ratios, students and teachers can find various ways of comparing ratios problems between two or more ratios. The term ratio defines the quantitative relationship of two numbers or amounts when there are three or more quantities that come together, then a comparison of ratios is required.

To compare two ratios, we need to adapt them into equivalent such as fractions. Every student should aware of the procedure of comparing ratios, which makes it easier to compare the ratios. Refer to this free worksheet for ratio comparison and solve the basic to complex problems with ease. The Printable Worksheet on Comparing Ratios is fun to practice and gives you deeper insight into the concept.

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How to Compare Ratios?

  • Firstly, check the second terms of the given ratios.
  • If they are not the same take the L.C.M of both the second terms and divide the L.C.M with the consequent of both the ratios.
  • Now, multiply the quotient with both the terms of ratio.
  • After that, we decide which ratio of the first term or antecedent is greater than that of the new ratios obtained.

Practice the questions of comparison on ratios in detail from the below comparing ratios worksheet with answers.

Comparing Ratios Worksheet PDF

Example 1.
Compare the ratio 4: 7 and 2: 14.

Solution: 

The given ratio to compare is 4: 7  and 2: 14.
Now, take the L.C.M of both the second terms 7 and 14 is 14.
Divide the L.C.M with the second term. i.e., 14\(\div\)7=2 and 14\(\div\)14=1.
Next, multiply the quotient with the antecedent and the consequent of the ratios.
Therefore, \(\frac{4}{7}\) = \(\frac{4×2}{7×2}\) = \(\frac{8}{14}\).
\(\frac{2}{14}\) = \(\frac{2×1}{14×1}\) = \(\frac{2}{14}\).
As we can see, 8 > 2, \(\frac{4}{7}\) > \(\frac{2}{14}\) i.e., 4: 7 > 2: 14.
Hence, 4: 7 is greater than the 2: 14 by the comparison rules of ratio.


Example 2.
Which ratio is greater 2\(\frac{1}{3}\): 1\(\frac{2}{5}\) and 5: 2.

Solution: 

Given ratios is 2\(\frac{1}{3}\): 1\(\frac{2}{5}\) and 5: 2.
Convert the mixed fractions into proper fractions. i.e., 2\(\frac{1}{3}\): 1\(\frac{2}{5}\)=\(\frac{7}{3}\): \(\frac{7}{5}\).
\(\frac{7}{3}\): \(\frac{7}{5}\) and \(\frac{50}{10}\): \(\frac{20}{10}\)
= \(\frac{7}{3}\)×15: \(\frac{7}{5}\)×15 and \(\frac{50}{10}\)×10: \(\frac{20}{10}\)×10
= 35: 21 and 50: 20
= \(\frac{35}{21}\) and \(\frac{50}{20}\)
= \(\frac{5×7}{3×7}\) and \(\frac{5×10}{2×10}\)
= \(\frac{5}{3}\) and \(\frac{5}{2}\)
= 5: 3 and 5: 2
Now, compare the ratios 5:3 and 5: 2.
The L.C.M of the second terms 3 and 2 is 6.
Now, divide with the second terms of both the ratios. ie., 6\(\div\)3=2 and 6\(\div\)2=3.
Therefore, \(\frac{5}{3}\) = \(\frac{5×2}{3×2}\) = \(\frac{10}{6}\).
\(\frac{5}{2}\) = \(\frac{5×3}{2×3}\) = \(\frac{15}{6}\).
So, 15>10, \(\frac{5}{2}\) > \(\frac{5}{3}\), 5: 3 < 5: 2.
Therefore, the ratios 5: 2 > 2\(\frac{1}{3}\): 1\(\frac{2}{5}\).


Example 3.
Arrange the following ratios in descending order.
(i) 1: 2, 4: 2, and 3:5
(ii) 4:1, 3:6, 7:5, and 2:4

Solution: 

(i) Given ratios to arrange in descending order are 1: 2, 4: 2, and 3:5.
The ratios can also be written as \(\frac{1}{2}\), \(\frac{4}{2}\), and \(\frac{3}{5}\).
L.C.M of all the denominators 2, 2,  and 5 is 10.
Now, divide the L.C.M
10\(\div\)2=5, 10\(\div\)2=5, and 10\(\div\)5=2.
Multiply the quotients with the ratios.
\(\frac{1}{2}\) = \(\frac{1×5}{2×5}\) = \(\frac{5}{10}\),
\(\frac{4}{2}\) = \(\frac{4×5}{2×5}\) = \(\frac{20}{10}\),
and \(\frac{3}{5}\) = \(\frac{3×2}{5×2}\) = \(\frac{6}{10}\).
Therefore, \(\frac{20}{10}\) > \(\frac{6}{10}\) > \(\frac{5}{10}\).
⇒ \(\frac{4}{2}\) > \(\frac{3}{5}\) > \(\frac{1}{2}\).
The ratios in descending order is 4: 2, 3: 5, and 1: 2.

(ii) Given ratios to arrange in descending order is 4:1, 3:5, 7:5, and 2:4.
The L.C.M of the second terms of the ratios 1, 5, 5, and 4 = 20.
Divide the L.C.M with all the consequent of the ratios. i.e., 20\(\div\)1=20, 20\(\div\)6=, 20\(\div\)5=4, and 20\(\div\)4=5.
Now,
\(\frac{4}{1}\) = \(\frac{4×20}{1×20}\) = \(\frac{80}{20}\),
\(\frac{3}{5}\) = \(\frac{3×6}{6×6}\) = \(\frac{18}{36}\),
\(\frac{7}{5}\) = \(\frac{7×4}{5×4}\) = \(\frac{28}{20}\),
and \(\frac{2}{4}\) = \(\frac{2×5}{4×5}\) =\(\frac{10}{20}\).
Here, compare the numerators for which ratio is greater.
Since, \(\frac{80}{20}\) > \(\frac{28}{20}\) > \(\frac{18}{36}\) > \(\frac{10}{20}\)
⇒ \(\frac{4}{1}\) > \(\frac{7}{5}\) > \(\frac{3}{5}\) > \(\frac{2}{4}\).
Therefore, the ratios in descending order are 4: 1, 7: 5, 3: 5, and 2: 4.


Example 4.
Write and arrange the following ratios in ascending order.
(i) 11: 4, 5: 1, and 9: 3
(ii) 4: 5, 5: 7, and 7: 9

Solution:

(i) Given ratios are 11: 4, 5: 1, and 9: 3.
L.C.M of the second terms 4, 1, and 3 is 12.
Now,
\(\frac{11}{4}\) = \(\frac{11×3}{4×3}\) = \(\frac{33}{12}\),
\(\frac{5}{1}\) = \(\frac{5×12}{1×12}\) = \(\frac{60}{12}\),
and \(\frac{9}{3}\) = \(\frac{9×4}{3×4}\) = \(\frac{36}{12}\).
Since, \(\frac{33}{12}\) < \(\frac{36}{12}\) < \(\frac{60}{12}\)
⇒ \(\frac{11}{4}\) < \(\frac{9}{3}\) < \(\frac{5}{1}\).
Thus, the ascending order of the ratio is 11: 4, 9: 3, and 5: 1.

(ii) Given ratios are 4: 5, 5: 7, and 7: 9
As per the comparison rules of ratio, we take the L. C. M of the second term. i.e., 5, 7, and 9 is 315.
Next, divide the L.C.M with the second term of all the ratios.
315\(\div\)5=63, 315\(\div\)7=45, and 315\(\div\)9=35.
Now, multiply the quotient with both the terms of the ratios. We get,
\(\frac{4}{5}\) = \(\frac{4×63}{5×63}\) = \(\frac{252}{315}\),
\(\frac{5}{7}\) = \(\frac{5×45}{7×45}\) = \(\frac{225}{315}\),
and \(\frac{7}{9}\) = \(\frac{7×35}{9×35}\) = \(\frac{245}{315}\).
Since, \(\frac{225}{315}\) < \(\frac{245}{315}\) < \(\frac{252}{315}\)
⇒ \(\frac{5}{7}\) < \(\frac{7}{9}\) < \(\frac{4}{5}\)
⇒ 5: 7, 7: 9, and 4: 5
Therefore, the ratios in ascending order are 5: 7, 7: 9, and 4: 5.


Worksheet on Types of Ratios

Worksheet on Types of Ratios | Types of Ratios Worksheet with Answers PDF

Kids who are looking for various types of ratio numbers to practice can take the help from this Worksheet on Types of Ratios. Here, we will discuss different types of ratios that make students calculate the ratios in any manner. Without knowing the types of ratios, it becomes difficult to solve their relative sizes so go for these types of ratios practice worksheets and make your calculations so perfect.

The ratio is the mathematical expression of two or more quantities of the same units by dividing the one ratio by the other part of the ratio and must have the same unit measurements. The following different types of ratios worksheet with answers helps the students & teachers to perceive the various expression easily. Practice and express the below ratios in the respected kind of ratio.

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Different Types of Ratios Worksheet with Answers

Example 1.
Find the compound ratio of the following ratios.
(i) 4: 5 and 3: 2
(ii) b: d and g: f
(iii) 12: 8 and 7: 15 and 5: 9

Solution: 

(i) Given ratio is 4: 5 and 3: 2
To find the compound ratio, if we multiply two or more ratios termwise, then the ratio is obtained as a compound ratio.
Now, 4: 5 and 3: 2 = 4×3 : 5×2 = 12: 10.
So, the compound ratio is 12: 10.

(ii) Given ratio is b: d and g: f.
The compound ratio for b: d and g: f = b×g : d×f = bg : df.
Thus, the ratio is bg: df.

(iii) Given ratio is 12: 8 and 7: 15 and 5: 9.
To find the compound ratio for three ratios, we use the ratio formula
if m: n, p: q, and r: s, the compound ratio is (m× p× r) : (n× q× s).
Now, the ratio of 12: 8 and 7: 15 and 5: 9 = (12× 7× 5) : (8× 15× 9) = 420: 1080 = 7: 18.
Therefore, the compound ratio is 7: 18.


Example 2.
Get the duplicate ratio of the following expressions.
(i) √3: √4
(ii) a2: b2
(iii) 5√p: √20q

Solution:

(i) Given ratio is √3: √4
The duplicate ratio is the ratio of two equal ratios. For example, m: n = m2: n2.
Now, the ratio for √3: √4 = (√3)2: (√4)2 = 3: 4.
Thus, the duplicate ratio is 3: 4.

(ii) Given ratio is a2: b2
Now, duplicate ratio a2: b2 = (a²)² : (b²)² = a4: b4.
a4: b4 is the duplicate ratio of a2: b2.

(iii) Given ratio is 5√p: √20q.
The duplicate ratio of 5√p: √20q = (5√p)²: (√20q)² = 5²×√p² : √20q×√20q = 25p: 20q = 5p: 4q.
So, the duplicate ratio is 5p: 4q.


Example 3.
Find the triplicate ratio of the following expressions.
(i) ³√9m: 4
(ii) a/3: ³√a
(iii) p/2: q/4

Solution: 

(i) Given ratio is ³√9m: 4.
The triplicate ratio is the compound ratio of three equal ratios. For example, m: n = m³: n³.
Now,  ³√9m: 4 = (³√9m)³ : 4³ = 9m: 64.
The triplicate ratio is 9m: 64.

(ii) Given ratio is a/3: ³√a.
To find the triplicate ratio we apply the formula a: b = a³: b³.
a/3: ³√a = (a/3)³: (³√a)³ = a³/3³: (³√a)³ =a³/9: a = a²: 9.
So, the ratio is a²: 9.

(iii) Given ratio is p/2: q/4.
Now, the triplicate ratio of p/2: q/4 = (p/2)³: (q/4)³ = p³/8: q³/64 = 64p³: 8q³ = 8p³: q³.
Therefore, the triplicate ratio is 8p³: q³.


Example 4.
Find the subduplicate ratio of (a+b)²: (a-b)4.

Solution: 

Given ratio (a+b)²: (a-b)4.
The subduplicate ratio of p: q is √p: √q.
Now, we find the subduplicate ratio of (a+b)²: (a-b)4.
(a+b)²: (a-b)4 = √(a+b)²: √(a-b)4 = (a+b): (a-b)².
Thus, the subduplicate ratio of (a+b)²: (a-b)4 is (a+b): (a-b)².


Example 5.
Find the subtriplicate ratio of 343x³: 8y³.

Solution: 

The given ratio is 343x³: 8y³.
The subtriplicate ratio of a: b is ³√a: ³√b.
Now, the subtriplicate ratio of 343x³: 8y³
343x³: 8y³ = ³√(343x³): ³√(8y³) = 7x: 2y (since, 7³=343, 2³=8).
Hence, the subtriplicate ratio of 343x³: 8y³ is 7x: 2y.


Example 6.
Find the reciprocal ratio of a/9: b/6.

Solution: 

The given ratio is a/9: b/6.
The reciprocal ratio of p: q (p≠0, q≠0) is the ratio 1/p: 1/q = q: p.
Now, to find the reciprocal ratio of a/9: b/6
a/9: b/6 = 1/(a/9): 1/(b/6) = 9/a: 6/b = 9b: 6a = 3b: 2a.
Simply, we can say the reciprocal ratio is the inverse ratio. For, example 4: 5 is the inverse ratio of 5: 4.
Thus, the reciprocal ratio of a/9: b/6 is 3b: 2a.


Example 7.
Find x if (2x + 1) : (3x + 6) is the duplicate ratio of 4 : 6.

Solution: 

(2x + 1) : (3x + 6) is the duplicate ratio of 4: 6.
Also, the duplicate ratio of 4: 6 is 4²: 6² = 16: 36.
Now, (2x+1)/(3x+6) = 16/36
⇒ 36(2x+1) = 16(3x+6)
⇒ 72x+36 = 48x+96
⇒ 72x- 48x = 96-36
⇒ 24x = 60
⇒ x = 60/24
⇒ x = 2.5
Therefore, the value of x is 2.5.


Successive Discounts

Successive Discounts – Definition, Formula, Tips & Tricks | How do you Calculate Successive Discounts?

A successive discount is a discount that is given on the selling price of the product that has already had the discount on the marked price. While you cross a garment store, you come across the offers like 20% or 30% in blocks. The percentage we find on it is the discount offered by the shopkeeper to their customers.

For example, Suppose that a shopkeeper bought a shirt from the retailer at $500 and he decided to sell it at $800 and again put the discount of 20% where the final selling price will be $640. The customer feels that he got it for 20% off, but the shopkeeper sells it for a profit of $140 even after offering the discount. In short, we can say that, if the discount is again applied to the selling price, then it is determined as successive discounts. Successive discount is the amount of discount which is offered on the discount.

Successive Discounts Formula

To find the total discount in successive discounts case, Suppose that the first discount is x% and the second discount is y%, then the formula can be written as:
Total discount = x + y – \(\frac {xy}{100} \)%
There is another scenario like, Consider the original price of the shirt as ‘x’ and the first discount offered as ‘y’ and again the discount offered as ‘z’ on the new price. Then the selling price of the shirt is calculated as:
x-(y + z – \(\frac {yz}{100} \)) * x

Example:
Suppose that the online shopping website sells a product and it offers a discount of 10% on that product and again it offers more than 20% on the discounted value. Know the final value of the product.

In this case, let us suppose that the initial value is 100.
Given that the shopkeeper offered a discount of 10%, therefore it is (100-10) = 90
Then the shopkeeper again offered a discount of 20%, therefore it is 90 – 2(9) = 72
Hence the final value of the product is 72.
We can also calculate the final value by using the formula
Total discount = (x + y – xy / 100)%
x = 10% and y = 20%
Total discount = (10 + 20 – (10)(20) / 100)%
= (30 – 200/100)% = 28%

Successive Discounts Examples

Example 1:
Store is offering a t-shirt at the price of Rs.950. Successive discounts offered by the store are 30% and 50%. Calculate the selling price and total discount offered by the store?

Solution:
Given that, the price of the t-shirt = Rs. 950
Successive discounts are 30% and 50%
Total discounts = (x + y – xy / 100)%
= (30 + 50 – (30)(50) /100)% = 80 – 1500/100% = 65%
Discount = 65% of Rs. 950 = 65/100 * 950 = Rs. 617.5
Therefore, the selling price of the shirt = Rs. 950 – 617.5 = Rs. 332.5
Hence, the total discount offered is Rs. 617.5

The selling price is Rs. 332.5

Example 2:
Successive discounts on the product are 5%, 10%, and 15%. The price of the product in the store is $1000. Calculate the overall selling price and discount of the product?

Solution:
Given that, the price of the product = $1000
Successive discounts are 5%, 10% and 15%
Total discount of 5% and 10% are (x + y – xy / 100)% = 5 + 10 – (5)(10) / 100% = 15 – (50)/100% = 14.5%
Overall discount due to 14.5% and 15% = 14.5 + 15 – (14.5) * (15) /100%
= 29.5 – (217.5)/100% = 27.325%
Discount = 27.325 of $1000 = $273. 25
Total selling price of the product = Store price – overall discount
= $1000 – 273.25 = $726.75
Hence, the total discount offered = $273.25

The total selling price of the product = $726.75

Example 3:
The price of the product is $2250. The successive discounts are 10% and 20%. Find the selling price?

Solution:
Given that, the price of the product is $2250
The successive discounts are 10% and 20%
Total discount = (x + y – xy / 100)%
x = 10% and y = 20%
Total discount = 10 + 20 – (10)(20) 100% = (30 -200/100)%
Total discount = 28%
Discount = 28% of 2250 = (28/100) * 2250
Discount = 630
Therefore, the discount earned is 630
Selling Price = Marked Price – Discount = 2250 – 630 = 1620
Selling Price = 1620

Therefore, Selling Price = 1620
Discount = 630

Example 4:
Successive discounts of 30% and 20% are offered by the trader. Find the total discount offered?

Solution:
Given that, the successive discounts are 30% and 20%
Suppose that MRP = 100
Discount 1 = 30% of 100
Discount = 30
Hence, the price is 100-30 = 70
Now, the discount amount is 20%
Discount = 20% of 70
Discount = 14
Hence, the final price is 70-14 = 56
Total discount offered by the trader = MRP – Selling Price
= 100 – 56 = 44

Hence, the total discount offered = Rs. 44

Example 5:
3 successive discounts of 50%, 20%, and 10% are offered by a trader. Find the total discount percent?

Solution:
Suppose that MRP = 100
Given that, First discount = 50% of 100
Discount = 50
Hence, the price = 100-50 = 50
Second discount = 20% of 50 = 10
Hence, the price = 50 – 10 = 40
Final Discount = 10% of 40 = 4
Hence, the total price = 40 – 4 = 36
Therefore, total discount = MRP – Selling Price = 100 – 36 = 64
Total discount = 64/100 * 100 = 64%

Hence the total discount offered = 64%

We have mentioned all the tricks and tips to solve successive discount problems. Practice all the questions and improve your skills in solving the problems on successive discounts.

Frequently Asked Questions on Successive Discounts

1. How are successive discounts calculated?

Suppose that successive discounts are d1, d2 and d3, then the selling price is SP = (1-d1/100) * (1-d2/100) * (1-d3/100) * Marked Price

2. Is 30 or 40 successive discounts better?

The discount of 70% is better than the successive discounts of 30% and 40%.

3. How do you calculate two successive discount formulas?

The two discount formula is (x + y – xy / 100)%
Where x is the first discount and y is the second discount

4. What is the meaning of successive discounts?

Successive discount is the discounted price on the already given discount which is similar to compound interest.

Understanding Discount and Markups

Understanding Discount and Markup – Definition, Examples | How do you do Discount and Markup?

If you are wondering what is the difference between discounts and markup then you have come the right way? Here we will give you insight on discount and markups definition, formulae, and solved examples explaining step by step on how to approach. Follow the concept and know the real-time examples of discount and markup by checking the below sections for detailed information.

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What is meant by Discount?

Discount is the reduction in the price/rate of some product/item. The main purposes of providing discounts are:

  • Increase in sales
  • Clear the old stock
  • Encourage distributors
  • Reward potential customers

Discount is considered the easiest way to increase the product demand and plays an important role in online products. We can check amazing offers while shopping for various products. These offers mainly concentrate to attract customers which are named as discounts. It is considered as the value/price of the total amount/quantity which is generally less than the original value. In general words, we can tell that the total amount is sold at a particular discount to attract customers.

Discount Formula

The discount formula is as follows:
Discount = Marked Price – Selling Price
where Marked Price (M.P) is the actual value of the product without the discount value.
Selling Price is what the customers pay for the product.
Discount is the percentage of the marked price.

Markup

Markup is considered as the total profit or gross on a particular service or commodity. It is defined as the percentage over cost price. For suppose, if the product cost is Rs. 100 and its selling price is Rs. 150, then the markup will be 50%. It is also defined as the difference between CP(cost price) and SP(selling price) of the product. It estimates the profit and loss of the business.

What is meant by Markup Price?

Markup is the difference between the retail price of a commodity and cost. The amount that is added to the cost determines the retail prices of particular items. Markup is combined with total C.P to meet the business costs and generation of profit. Markup represents the percentage or fixed amount of selling price or cost price.

Markup Formula

As mentioned above, markup value is the difference between the SP(selling price) and CP(cost price) of the product.
Markup = Retail – Cost

Markup Percentage

The formula to calculate the markup percentage is:
Sale Price = Cost * (1 + Markup) or Markup = 100 * (Sale Price – Cost Price)/Cost

Markup and Discount Examples

Example 1.
Lara wanted to gift her mom a dress. The cost of the dress was Rs. 450 and it is marked as 20% off on the dress. How much is the discount amount?

Solution:
Given that, Cost of the dress = Rs. 450
Discount Percentage = 20%
Discount Amount = Original Price * Discount Rate
Discount Amount = 450 * 20% = 450 * (20/100) = 90
Therefore, the discount amount is Rs. 90
Hence the final cost of the dress is Cost Price – Discount Amount
= 450 – 90 = 340

The final cost is 340.

Example 2.
Soheal bought a watch which was originally Rs. 5000. He got a discount of 50% on the total amount. How much is the discount amount?

Solution:
Given that, the cost of the watch = Rs. 5000
Discount he got = 50%
Discount Amount = Original Price * Discount Rate
Discount Amount = 5000 * 50%
= 5000 * (50/100)
= 2500
Therefore, the discount amount is 2500
Hence the final cost of the watch is Cost Price – Discount Amount
= 5000 – 2500
= 2500

The final cost of the watch is 2500.

Example 3.
The original price of the book is Rs. 400. I got 20% on the total amount. How much is the discount amount?

Solution:
Given that, the cost of the book is Rs. 400
The discount I got = 20%
Discount Amount = Original Price * Discount Rate
Discount Amount = 400 * 20% = 400 * (20/100) = 80
Therefore, the discount amount is Rs. 80
Hence the final cost of the book is Cost Price – Discount Amount
= 400 – 80 = 320

The final cost of the book is Rs. 320

Example 4.
A car dealer advertises a 7% markup over cost. Find the selling price of the car that cost the dealer $13,000.

Solution:
Given that, Markup = 7% of 13,000 = (7/100) * 13000 = 910
The markup price is 910
Selling Price = Cost Price + Markup Price
=13,000 + 900
=13,900

The selling price of car = 13,900

Example 5.
A ring that costs the jeweler $360 sells for $630. Find the markup rate?

Solution:
Given that, Cost of the ring = $360
Selling Price = $630
Markup Price = Selling Price – Cost Price
= 630 – 360 = 270
The markup price = $270
Markup Percentage = Markup Price / Cost Price
= 270/360 = 75%

Therefore, the markup percentage = 75%

Example 6.
A person got a loan from a bank at a rate of 3% per year for some period. In how much period of time his loan of Rs. 65,000 will become Rs. 68,000.

Solution:
Given that, Principal amount = Rs. 65,000
Profit for bank = Rs. 68,000 – 65000 = 3900
Markup rate = 3/100per year
Profit per year = 650 * 3 = Rs. 1,950
The total amount of time period to clear a loan is 3900 – 1950 = 1950

Therefore, it takes 2 years to become Rs. 68,000 from Rs. 65,000 is 2 years.

Hence, the complete information is given on discount and markup. Check our page for more new concepts and information. Discount and Markup are used in day-to-day life and solve various problems of business.

Understanding Overheads Expenses

Understanding Overheads Expenses – Definition, Types, Examples | How do you Calculate Overhead Expenses?

Overhead expenses are business and other costs which are not related to direct materials, labor, and production. Overhead expenses are some of the indirect costs which are not related to particular business activities. Calculating the overhead expenses is not only important for budgeting but also to determine the charge or investment for a product or service. Suppose that you have a good business that is service-based. Apart from the direct investments or costs, indirect costs like insurance, rent, utilities are considered as overheads expenses.

To understand it better we will consider another example, ie., Suppose a person bought a TV at the cost price of Rs 12,000. Now, he took cable connection for the TV. He has to pay the cable bill every month which is considered as an overhead expense.

What are Overhead Expenses?

Overhead Expenses support the business but they do not generate any revenue. These expenses are mandatory and you have to pay them irrespective of your revenue. The main examples of overhead expenses are property taxes, utilities, office supplies, insurance, rent, accounting and legal expenses, advertising expenses, government licenses and fees, depreciation, and property taxes.

Types of Overhead Expenses

Among the overhead expenses, not all the expenses are the same or equal. These expenses are divided into 3 categories. Know the different expenses and their types which can create a meaningful budget for the business. The different types of overhead expenses are:

  • Fixed Overhead
  • Variable Overhead
  • Semi-Variable Overhead

1. Fixed Overhead Expense

These expenses are something which won’t change from month to month. If a fixed overhead has to change then it changes only annually during the renewal period. Examples of fixed overhead are insurance, salaries, rent. These overhead expenses are easy to budget and plan. These fixed overheads are tough to reduce or restrict the cash flow.

2. Variable Overhead Expense

These expenses are mostly affected by business activities and not by sales. Some of the examples of variable overhead expenses are office supplies, legal expenses, repairs, advertising expenses, and maintenance expenses. It is no guarantee that office supplies will not change according to sales volume. In the same way, advertising expenses may increase during peak sales. The drawback of variable expense is that it is difficult to predict while budgeting.

3. Semi-Variable Expenses

These expenses are also not the same from month to month. These semi-variable expenses are also not completely unpredictable and some examples of these expenses are many utilities, hourly wages, some commissions, and vehicle expenses.

Also, See:

Calculating Overhead Rate

Calculating overhead rate is an important factor in the business. It determines the exact amount of sales that goes into overhead expenses. To calculate it, we have to add all the overhead expenses and divide that number with your sales. The formula of overhead rate is:

Overhead Rate = Overhead Expenses / Sales

Overhead Charges Examples

Example 1.
An industry estimated the factory overhead for the period of 10 years at 1,60,000. The estimation of materials produced for 40,000 units is 200,000. Production requires 40,000 hours of man work at the estimated wage cost of 80,000. Machines will run for 25,000 hours approximately. Calculate the overhead rate on each of the following bases:
i. Direct labor cost
ii. Machine hours
iii. Prime Cost

Solution:

To find the direct labour cost, machine hours and prime cost we have to calculate the overhead rate.
(i) Direct Labour Cost
= (Estimated Factory OverHead / Estimated Direct Labour Cost) * 100
= (1,60,000 / 80,000) * 100
= 200%
(ii) Machine hours
= (Estimated Factory Overhead / Estimated Machine hours)
= 1,60,000 / 25,000
= 6.40 per machine hour
(iii) Prime Cost Basis
= Estimated Factory Overhead / Estimated prime cost
= 1,60,000 / (2,00,000 + 80,000)) * 100
= 89%

Example 2.
A shopkeeper purchased a second hand car for Rs. 1,40,000. He spent Rs. 15,000 on its repair and painting and then sold it for Rs. 17,000. Find his profit or loss?

Solution:
Cost Price = Rs. 1,40,000
Overhead Charges = Rs. 15,000
C.P.N = (1,40,000 + 15,000) = Rs. 1,55,000
S.P = Rs. 17,000
Therefore, S.P > C.P
Hence, it is profit
Profit Percent = S.P – C.P
P = 170000 – 155000
P = Rs. 15000
P% = P / C.P.N * 100
P% = 15000/155000 * 100
P% = 300/31%
Therefore, the profit percentage is 300/31%

Example 3.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300. Determine his profit percentage?

Solution:
Cost Price of radio = Rs. 225
Overhead expenses = Rs. 15
Selling Price of radio = Rs. 300
Net Cost Price = Rs. 225 + 15 = 240
Profit % = Selling Price – Cost Price / Cost Price * 100
P% = 300 – 240 / 240 * 100
P% = 60/240 * 100
P% = 25%
Therefore, the profit percentage = 25%

Frequently Asked Questions on Overhead Costs

1. What are the examples of overhead expenses?

The examples of overhead expenses are interest, labor, advertising, insurance, accounting fees, travel expenditure, telephone bills, supplies, utilities, taxes, legal fees, repairs, legal fees, etc.

2. What are the types of overhead?

The types of overheads are fixed overhead expense, semi-variable overhead expense, and variable overhead expense.

3. What is the minimum percentage for overhead?

The minimal percentage is that it should not exceed 35% of the total revenue. In growing or small businesses, the overhead percentage factor is usually considered as the critical figure which is of concern.

4. How will overhead affect profit?

Overhead represents the supporting costs of production or service delivery. If there is an increase in overhead, it reduces profits by the exactly same amount.

Worksheet on Ratios in Lowest Term

Worksheet on Ratio in Lowest Term | Simplifying Ratios Worksheet with Answers PDF

In this Worksheet on the Ratio in Lowest Term or the simplest form, all students can learn many practice questions to make the topic easy. This reducing ratios to lowest form worksheet provide various types of lowest term ratios questions with solutions. In short, we can say that ratio shows their relative sizes.

The ratio of two or more quantities of the same units of measurement is a comparison got by dividing one quantity by the other and must have the same units of measurement. The following Worksheet on Ratio in Simplest Form helps the students and teachers to perceive the ratio in the lowest term. Practice the questions on express the following ratios in the simplest form worksheet with answers pdf.

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Reduce Ratios to Lowest Term Worksheet with Answers

1. Find the ratio of 30 min and 1½ hr in the simplest form.

Solution:

The given ratio is 30: 1½.
First, we have to convert the time 1½ hr to minutes.
1½ hr = 60min + 30min = 90min
Therefore, the ratio = 30min : 90min.
Dividing by 30, we get the ratio of 1:3.
Hence, the simplest form of the given ratio is 1:3.


2. Find the ratios of the following in the simplest form.
(i) 2.4 kg to 400g
(ii) 425 and 220
(iii) 4 dozens to 2 scores
(iv) 1kg 100g and 2kg 300g

Solution:

(i) Given 2.4 kg to 400g
Convert the kilograms to grams
We know 1kg = 1000g.
Now, 2.4kg = 2.4 × 1000 = 2400gms.
The ratio of 2400g: 400g is reduced to the lowest term by diving both the numbers by 400.
When we divide the ratio by 400, we get
2400g : 400g = 6:1.
So, the simplest form is 6:1.

(ii) Given 425: 220
Divide the ratio by 5, we get
425/5: 220/5 = 85: 44.
The simplest form of ratio 425: 220 is 85: 44.

(iii) Given 4dozens and 2scores
We know, 4dozens = 4×12 = 48.
2 scores = 2×20 = 40 (since, 1 score = 20 of something).
Now, the ratio of 4dozens and 2scores is 48: 40
Divide by 8, we get
48/8 : 40/8 = 6:5.
The simplest form of ratio is 6:5.

(iv) Given 1kg 100g and 2kg 300g.
First, have to convert them into grams.
1kg 100g = 1000g × 100g = 1100g.
2kg 300g = 2000g × 300g = 2300g.
Now, the ratio is 1100g: 2300g.
The G.C.F of 1100 and 2300 is 100.
When we divide by 100, we get
1100g/100 : 2300g/100 = 11:23.
The simplest form of given ratio 1kg 100g and 2kg 300g is 11:23.


3. Find the ratio of $ 2.15: $ 6.25 in the simplest form?

Solution:

The given ratio is $ 2.15: $ 6.25.
Where, $ 2.15= 215 cents and $ 6.25= 625 cents
Therefore, the required ratio is 215 cents: 625 cents.
Divide the ratio by 5, we get
215cents/5 : 625cents/5 = 43:125.
Thus, the simplest ratio is 43:125.


4. Simplify the following ratios
(i) 4/5: 7/5
(ii) 2 1/3: 4
(iii) 0.04: 0.26
(iv) 4 1/5: 2 3/10: 6/10
(v) 1.8: 2.2

Solution:

(i) Given ratio 4/5: 7/5.
By multiplying both the parts of the ratio by 5, we get
4/5: 7/5 = (4/5)×5: (7/5)×5 = 4: 7.
So, the simplest form of the ratio is 4: 7.

(ii) Given ratio 2 1/3: 4.
First, convert 2 1/3 as an improper fraction, we get 7/3.
Now, multiply both the parts by 3,
7/3: 4 = (7/3)×3: 4×3 = 7: 12.
Thus, the simplest ratio is 7:12.

(iii) Given ratio 0.04: 0.26
As we know, after the decimal part we have two units. So, multiply both the parts by 100.
Then, 4: 26
Now, divide the ratio by 2
4: 26 = 4/2: 26/2  2: 13.
Hence, the simplest ratio is 2: 13.

(iv) Given ratio 4 1/5: 2 3/10: 6/10
Convert, the mixed numbers into improper fractions,
4 1/5 = 21/5 and 2 3/10 = 33/10.
Now, the proper ratio is 21/5: 33/10: 6/10.
The L.C.D is 10.
We now have,
21/5: 33/10: 6/10 = 42/10: 33/10: 4/10 = 42: 33: 6.
Try to reduce the ratio in the simplest form.
The G.C.F for 42: 33: 6 is 3.
We get, 42/3: 33/3: 6/3 = 14: 11: 2.
Therefore, the simplest form of the given ratio is 14: 11: 2.

(v) Given ratio 1.8: 2.2.
After the decimal part we have one unit, multiply both the parts by 10.
1.8 = 18 and 2.2 = 22.
Divide by 2, we get
18: 22 = 18/2: 22/2 = 9: 11.
The simplest ratio is 9:11.


5. Reduce the following ratios to the lowest terms
(i) 3hours: 1hour 40min
(ii) 2years 2months: 4years 4months
(iii) 5weeks: 25days

Solution:

(i) Given 3hours: 1hour 40min
Convert hours into minutes.
3hrs = 3×60 = 180min.
1hr 40min = 60min + 40min = 120min.
The required ratio is 180min: 120min = 180min/60: 120min/60 = 3:2.
So, the ratio in the lowest term is 3:2.

(ii) Given 2years 2months: 4years 4months.
Convert years into months,
2years 2months = 2×12 + 2 = 24+2 = 26months.
4years 4months = 4×12 + 4 = 48+4 = 52months.
The ratio = 26months: 52months = 26/26: 52/26 = 1: 2.
The simplest ratio is 1:2.

(iii) Given the ratio 5weeks: 25days
Convert weeks into days, as we know one week = 7days.
5weeks = 5 × 7 = 35days.
Now, the ratio is 35days: 25days = 35/5: 25/5 = 7: 5.
So, the lowest term of the given ratio is 7:5.


Collinear Points Proved by Midpoint Theorem

Collinear Points Proved by Midpoint Theorem – Statement & Proof | How do you prove that Points are Collinear?

In this article, students can acquire more about the collinear points and the proof of the Collinear Points by Midpoint Theorem. Collinear points are the group of three or more points that lie on the same straight line. These points may exist on different planes but not on different lines. The property of the points being collinear is known as collinearity.

Let us learn the concept of collinear points and see how we construct the theorem to prove the given statement. We hope the information provided on this page will help you all to understand how the points are collinear and proved by the mid-point theorem.

Collinear Points – Definition

The word collinear is a mix of two Latin words ‘col’ and ‘linear’. The word ‘col’ means together and the word ‘linear’ means line. Thus, collinear points mean points together in the same or a single line. In geometry, the set of three or more points that lies on the same straight line is said to be collinear.

You can see many real-world examples of the collinearity name as a group of people or students standing in a straight line, a bunch of oranges kept in a single row, etc. Observe the following figure to get clear knowledge of the collinear points and identify the collinear points and non-collinear points.

Collinear Points

In the above figure, the set of collinear points are {A, E, B}, {B, F, C}, {C, G, D}, {A, H, D}, and {B, I, D}. The set of non-collinear points are {E, G}, {H, F}, and {G, F}, etc.

Collinear Points Proved by Midpoint Theorem Statement and Proof

In ∆ABC, the medians CM and BN are produced to the points P and Q respectively such that CM=MP and BN=NQ. Prove that the points P, A, and Q are collinear and A is the midpoint of PQ.

Collinear Points Theorem Statement

To Prove:

Consider an ∆ABC,

Given, the points M and N are the midpoints of AB and AC respectively. CM and BN are produced to P and Q respectively such that CM=MP and BN=NQ.

Now, we prove that

(i) P, A, and Q are collinear.

(ii) A is the midpoint of PQ.

Construction:

From the above triangle ABC, draw a dotted line to join the points PA, AQ, and MN.

Collinear Points Theorem Construction

Proof:

In ∆APC, M and N are the midpoints of PC and AC respectively (from given)

Therefore, MN ∥ AP and MN = ½AP (by the midpoint theorem) —-(1)

In ∆ABQ, M and N are the midpoints of AB and BQ respectively (from given)

Therefore, MN ∥ AQ and MN = ½AQ (by the midpoint theorem) —-(2)

Thus, AP ∥ MN and AQ ∥ MN. ( from (1) and (2))

Now, AP and AQ lie in the same straight line because both passes through the same point A and are parallel to the same straight line MN.

∴ The points P, A, and Q are collinear. [(i) proved]

Also, ½AP = ½AQ (from statements (1) and (2)) —-(3)

⇒ AP = AQ (from statement (3))

Therefore, A is the midpoint of PQ. [(ii) proved]

Hence, the given statement is proved and the points are collinear by a midpoint theorem.

Also, refer:

FAQs on How to Prove Points are Collinear using MidPoint Theorem

1. How do you prove the given points are collinear?
Three or more points are on the same straight line are collinear, if the slope of any two pairs of points is the same. For example, we have three points A, B, and C, the pairs formed are AB, BC, and CA respectively. If the slope of AB = Slope of BC = Slope of AC, then the points A, B, and C are collinear points.

2. How do you prove three points are collinear vectors?
Take the three points with position vectors a, b, and c. To prove the vectors are collinear, if and only if the vectors (a-b) and (a-c) are parallel. In other words, to prove the collinearity of the vectors, we need to show (a-b) = k(a-c), where k is a constant.

3. What is the theorem of the midpoint?
The midpoint theorem states that ” The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Area and Perimeter of Combined Figures

Area and Perimeter of Combined Figures – Definition, Facts, Examples | How do you find the Area and Perimeter of Combined Shapes?

In Mathematics, Area and Perimeter are two important properties of two-dimensional shapes which is used in our day-to-day life. The Area and Perimeter of different combined figures are explained here. Before that, you must learn about different basic shapes such as triangle, square, rectangle, circle, and sphere, etc. The Perimeter is defined as the distance of the boundary of the shape whereas the Area will explain the region occupied by it.

In this article, we will discuss the definition of Combined Figures, how to find Area and Perimeter of Combined Figures, Solved Example problems, and so on. There are various types of shapes but commonly used are square, rectangle, triangle, circle.

Also, Read :

Combined Figures – Definition

Combined Figures is defined as shapes that are composed of a combination of simpler shapes. Each shape has its own area and perimeter formula. It is applicable to any shape and size whether it is regular or irregular.

To find the perimeter, add all the outside sides of our shape, and find the area we divide our shapes into simple shapes, calculate the area of those shapes separately, and then add areas up to get total.

How to find Area and Perimeter of Combined Figures?

The following are the steps for finding the area and perimeter of combined figures:
How to find Area of Combined Shapes?
Step 1: First, divide the compound shape into a basic regular shape.
Step 2: Next, find each basic shape area separately.
Step 3: Then Add all the areas of basic shapes together.
Step 4: Now, write the final answer in square units.

How to find the Perimeter of Combined Figures?
The perimeter of the combined figure is the length of the outline of a shape. To find the perimeter of any shape like rectangle, square, and so on you have to add all the lengths of four sides. Consider x is in this case the length of the rectangle and y is the width of the rectangle.

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Area and Perimeter of Combined Shapes Examples

Problem 1: Find the area and perimeter of the following combined figures. The below- combined figure consists of a square and a triangle.

Solution: 
As given in the question, the combined figure is given.
Now, we will find the area and perimeter of a combined figure.
The combined figure consists of a square and equilateral triangle.
First, find the area of a square and the area of an equilateral triangle.
We know, the formulas of the area of a square and the area of an equilateral triangle.
The area of a square is a² and the area of an equilateral triangle is (√3/4) × a2 square units.
Substitute the value in the above formulas, we get
Area of combined figures = Area of a square + Area of an equilateral triangle.
A= a² + (√3/4) × a2
A = 8 x 8 x (1.732/4) x 8 x 8      (√3 = 1.732)
A  = 64 x 1.732 x 16 = 1774 sq.units.
So, the area of combined figures is 1774 sq. units.
Now, find the perimeter. So, the perimeter of combined figures is AB+BC+CD+DE+EA
P= 8+8+8+8+8 = 40 cm.
Hence the area and perimeter of the combined figure are 1774 sq. units and 40 cm.

Problem 2: A combined Figure is as given below. Find the area and perimeter of that Figure?

Solution: 
The given figure is a combination of a rectangle and two circles. ABCD is a rectangle and AEB, DFC is two semicircles both are having equal area.
The length of the rectangle is l = 4cm.
The breadth of the rectangle is, b = 2cm.
The diameter of a semicircle is 2 cm.
The radius of a semicircle is r = 22cm.
Now, we will find the area and perimeter of the combined figure.
So, the perimeter of the given combined figure is, AD+BC+AEB+DFC
P= 4+4+2×21 (circumference of a circle)
P= 8+ 2 x 21 x 2πr  = 8+2 x 722 x 1
P= 8+2 x 3.14 = 14.28 cm.
So, the perimeter of the given combined figure is 14.28 cm.
Next, find the area of combined figures.
Area of a combined figure = Area of a rectangle (ABCD)+ 2 x Area of a semicircle.
After substituting the values, we will get the value that is
A = l x b + 2 x 2πr2
A= 4×2+2x7x22x1x1 = 11.14 cm2
Therefore, the area and perimeter of the given combined figure are 11.14 cm2 and 14.28 cm.

Problem 3:  Find the area of a below-given combined figure.

Solution: 
As given in the question, the figures consist of a square and semicircle.
Now, we will find the area of a combined figure.
So, the Area of a combined figure = Area of a square + Area of a semicircle
A = sxs+1/2 π(rxr) = 14 X 14 + (1/2)(22/7)(7 x 7) = 77+196
A= 273 cm2
Thus the area of a given combined figure is 273 cm2.

Problem 4: Find the perimeter of a given figure.

Solution: 
In the given question consists of a figure.
Now, we will find the perimeter of that figure.
Perimeter means adding all the lengths of shapes.
Perimeter of a combined figure is,
P = 4 cm+3 cm+8 cm+5cm.
P = 20cm
Hence, the Perimeter of a given combined figure is 20cm.

Problem 5: The figure is shown in the below figure. Using figure values find the area and perimeter of a combined figure?

Solution: 
The given figure consists of four semi-circles and one square.
Now, we will find the perimeter and area of the figure.
The diameter of each semicircle is 7cm.
The length of each square side is 7cm.
Using the diameter value, we will find the radius.
Radius = d/2 = 7/2 = 3.5cm.
First, find the perimeter, we have to add the four semicircles and a square. The resultant sum is Perimeter.
So, the perimeter of a given figure = 4 x Perimeter of semicircles + perimeter of a square.
P =  4 (Πr )+ 4a = (22/7)(3.5 cm) + 4(7).
P = 4(22 x 0.5 )+ 28 = 44 cm + 28 cm = 72 cm.
Thus the perimeter of the given figure is 72cm.
Next, find the area of a given figure.
Area of a combined figure = Area of 4 semicircles + Area of a square.
The Area of 1 semicircle is  Πr²/2
=  (1/2) x (22/7) x (7/2)²=  (1/2) x (22/7) x (7/2) x (7/2)
=  (1/2) x 11 x (7/2)=  77/4= 19.25 cm²
Then, the Area of 4 semi circles = 4 (19.25)=  77cm².
The area of a square is a²=  (7)²=  7 x 7 =  49 cm².
Area of a combined figure = 19.25 cm² + 49 cm² = 126 cm².
Hence, the area and perimeter of a given combined figure are 72 cm and 126 cm².