SEE Compulsory Math Question Paper With Solution
SEE 2077 (2021)
If initial population of any place is X_{0} and annual population growth is R%, then what is the population of that place after Y years?
Solution,
Here, initial population (P) = X_{0}
Population growth rate (R) = R% p.a
Population after Y years (P_{T}) = ?
By using population growth formula,
P_{T} = \(P\left(1+\frac R{100}\right)^T\)
= \(X_0\left(1+\frac R{100}\right)^Y\)
∴ Population of that place after Y years is \(X_0\left(1+\frac R{100}\right)^Y\)
Write the area of an isosceles triangle having length of equal sides 'm' cm and third side 'n' cm,
Write down the relation between the area of rectangle and triangle standing on bases MN and between the parallel lines MN and AB.
Write the relation between opposite angles of a cyclic quadrilateral.
The opposite angles of a cyclic quadrilateral are supplementary or their sum is 180°.
Cyclic Quadrilateral is a quadrilateral which has all its four vertices lying on a circle. It is also know as inscribed quadrilateral. The sum of the opposite angles of a cyclic quadrilateral is supplementary.
Let ∠A, ∠B, ∠C and ∠D are the four angles of an inscribed quadrilateral. Then,
∠A + ∠C = 180°
∠B + ∠D = 180°
Therefore, an inscribed quadrilateral also meet the angle sum property of a quadrilateral, according to which, the sum of all the angles equals 360 degrees. Hence,
∠A + ∠B + ∠C + ∠D= 360°
What is the price of price of a bag costing Rs. 1800 after levying value added tax VAT? Find it.
The present price of a car is Rs. 2000000. If its price reduces by 10% annually, after how many years it's price will be Rs. 1458000? Find it?
Solution:
Given:
Present price (P) = Rs. 20,00,000
Rate of compound depreciation (R) = 10% pa
Price after depreciation (P_{T}) = Rs. 14,58,000
To find:
Time taken to reach the new reduced price (T) = ?
Using Formula:
\(P_t=P\left(1\frac R{100}\right)^T\)
Or, Rs, 1458000 = \(Rs\;2000000\left(1\frac{10}{100}\right)^T\)
Or, \(\frac{Rs\;1458000}{Rs\;2000000}={(10.1)}^T\)
Or, \(0.729=\left(0.9\right)^T\)
Or, \(\left(0.9\right)^3=\left(0.9\right)^T\)
Or, T = 3
∴ Required time (T) = 3 years.
Find the area of triangle MNO in the given figure
Solution,
Given:
In ΔMNO,
MN (a) = 24 cm
NO (b) = 25 cm
MO (c) = 7 cm
To find,
Area of triangle ΔMNO,
Here,
SemiPerimeter (s) \(=\frac{a+b+c}2\)
\(=\frac{24cm+25cm+7cm}2\)
\(=\frac{56cm}2\)
=28cm
Now using formula,
Area of triangle (A) = \(\sqrt{s(sa)(sb)(sc)}\)
= \(\sqrt{28(2824)(2825)(287)}cm^2\)
= \(\sqrt{28x4x3x21}cm^2\)
=\(\sqrt{7056}cm^2\)
=\(84cm^2\)
∴ Area of triangle MNO is 84 cm^{2}.
If the volume of a hemisphere is 486π cubic cm, then find its radius.
Solution,
Given:
Volume of a hemisphere(V) =486π cm^{3}.
To find:
Radius of the hemisphere (R) = ?
By formula,
Volume of hemisphere(V) = \(\frac23\pi R^3\)
Or, 486π cm^{3} = \(\frac23\pi R^3\)
Or, \(R^3\) = \(\frac{3\times486\mathrm\pi}{2\mathrm\pi}\)
Or, \(R^3\) = 729 cm^{3}.
^{∴ R = 9 cm}
Hence, the radius of hemisphere is 9cm.
In the given figure, the area of rectangular surfaces of the prism is 180 sq. cm, ∠PQR = 90°, PR = 5 cm, and P'Q' = 3cm. Find the length of PP'
Solution,
Given:
Area of the rectangular faces of a prism (L.S.A) = 180 cm^{2}., ∠PQR = 90°, PR = 5cm
To find:
The length of PP'
Here, PQ = P'Q' = 3cm
Then, in the right angles triangle PQR,
PQ^{2} + QR^{2} = PR^{2}
Or, (3cm)^{2} + QR^{2} = (5cm)^{2}
Or, 9cm^{2} + QR^{2} = 25cm^{2}
Or, QR^{2} = 16cm^{2}
Or, QR = 4cm
Again,
Perimeter of triangle PQR = PQ + QR + PR
= 3 cm + 4 cm + 5 cm
= 12 cm
Now, by formula
L.S.A of a prism = perimeter of the base x height
Or, 180 cm^{2} = 12 x PP'
Or, \(\frac{180cm^2}{12cm}=PP'\)
Or, 15 cm = PP'
∴ The length of PP' = 15 cm
Rationalize the denominator of : \(\frac{\sqrt3+\sqrt2}{\sqrt3\sqrt2}\)
Solution:
Here,
\(\frac{\sqrt3+\sqrt2}{\sqrt3\sqrt2}\)
Multiplying both numerator and denominator by \(\sqrt3+\sqrt2\) , we get
\(\frac{\sqrt3+\sqrt2}{\sqrt3\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\)
= \(\frac{{(\sqrt3+\sqrt2)}^2}{{(\sqrt3)}^2{(\sqrt2)}^2}\)
= \(\frac{{(\sqrt3)}^2+2\times\sqrt3\times\sqrt2+{(\sqrt2)}^2}{32}\)
= \(\frac{3+2\times\sqrt6+2}{32}\)
= \(\frac{5+2\times\sqrt6}{1}\)
= \(5+2\sqrt6\)
Is the required answer
If the sum of two consecutive even numbers is 32, find the numbers.
Solution:
Let, the smaller even number be x. Then the consecutive greater even number will be x + 2.
Now, according to the question,
x + (x + 2) = 34
Or, 2x + 2 = 34
Or, 2x = 34 2
Or, 2x = 32
Or, x = 32/2
Or, x = 16,
Then, x + 2 = 16 + 2 = 18
∴ The two required consecutive even numbers whose sum is 34 are 16 and 18.
In the given figure, WXYZ is a parallelogram. If ∠XAY (angle XAY) = 90°, AY = 8 cm and the area of parallelogram WXYZ is 64 sq. cm, find the length of AX.
Solution:
Given,
angle XAY (∠XAY ) = 90°,
AY = 8cm
and Area of parallelogram WXYZ = 64cm^{2}
To find,
The length of AX = ?
Here,
Area of triangle ΔAXY = 1/2 area of parallelogram WXYZ [ ∵ Both are standing on same base between the same parallel lines.]
Or, Area of ΔAXY = 1/2 x 64 cm^{2}
Or, Area of ΔAXY = 32 cm^{2}
Now using formula,
Area of ΔAXY = 1/2 x base x height
Or, 32 = 1/2 x AX x AY
Or, 32 = 1/2 x AX x 8
Or, AX = 32/4 = 8 cm
∴ The length of AX = 8 cm.
In the given figure, M is the center of the circle. If ∠GEF = 2x° and ∠EGF = 3x°, find the value of ∠FGH.
Solution:
Given,
M is the center of the circle.
∠GEF = 2x° and ∠EGF = 3x°
To find:
The value of ∠FHG
Here,
∴ The value of ∠FGH = 36°
In the given figure, X is the center of the circle, AB and AC are two tangents to the circle. If angle ∠BAC = 105°, find the value of ∠BAC
Solution:
Given:
X is the center of the circle,
AB and AC are two tangents to the circle.
∠BXC = 105°
To find:
The value of ∠BAC
Here,
∴ The value of ∠BAC is 75°.
In triangle ΔDEF, DE = 12√3 cm, ∠DEF = 60° and the area of ΔDEF is 36 sq cm, find the measurement of EF.
Solution:
Given, In ΔDEF, DE = 12√3 cm, 
To find:
The measurement of EF
By formula,
Area of ΔDEF = \(\frac12DE\times EF\times\sin\left(60\right)\)
Or, 36 cm^{2} = \(\frac1212\sqrt3\times EF\times\frac{\sqrt3}2\)
Or, 36 cm^{2} = 9cm x EF
Or, EF = 4cm
∴ The measurement of EF is 4cm is the required answer
In a continuous series the average marks of some students is 40 and the sum of their marks is 1200. Then find the number of students.
Solution:
Given: In a continuous series,
Average marks (\(\overline x\)) = 40
and sum of their marks (∑fm) = 1200.
To find: No of students (N)
By formula,
\(\overline x=\frac{\sum fm}N\)
Or, \(40=\frac{1200}N\)
Or, \(40\times N=1200\)
Or, N = 30
∴ The number of students (N) = 30 is the required answer
If a dice is rolled and coin is tossed at the same time, find the probability of occurring odd number on on dice and head on the coin.
Solution,
Let, the event of occurring odd number on dice be A and the event of occurring head on a coin be H.
Then, when a dice is rolled,
S = {1, 2, 3, 4, 5, 6} and A = {1, 3, 5}
∴ n(S) = 6 and n(A) = 3
By formula P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac36\)
= \(\frac12\)
And,
When a coin is tossed,
P(H) = \(\frac12\)
To find: P(A ∩ H)
here, A and H are independent events
∴ P(A ∩ H) = P(A) x P(H)
= \(\frac12\times\frac12\)
= \(\frac14\)
∴ The probability of occurring odd number on dice and head on a coin is 0.25 or \(\frac14\)
Two cards drawn randomly from a well shuffled deck of 52 cards in succession without replacement. Show the probabilities of possible outcomes of getting Ace and not getting Ace in tree diagram.
In a survey of people of a community, it was found that 70% liked curd, 60% liked milk, 20% don't like both curd and milk and 550 liked both curd and milk then,
Solution:
Let, the set of people who like curd be C and the set of people who like milk be M. Then,
n(U) = 100%
n(C) = 70%
n(M) = 60%
n(\(\overline{C\cup M}\)) = 20%
and n(C ∩ M) = 550
To find,
Again let, n(C ∩ M) = x%
Representing above information in venndiagram we get,
From above venndiagram,
no(C) + no(M) + n(C ∩ M) + n(\(\overline{C\cup M}\)) = n(U)
Or, (70% x%) + (60%  x%) + x% + 20% = 100%
Or, 150%  x% = 100%
Or, x% = 50%
∴ n(C ∩ M) = 50%
Here,
50% of n(U) = n(C ∩ M)
Or, 0.5 x n(U) = 550
∴ n(U) = 550/0.5 = 1100
Again, from the venn diagram above, we get
n_{o}(C) = (70%  x%) of n(U)
= (70%  50%) x 1100
= 20% x 1100
= 0.2 x 1100
= 220
∴ Total number of people participated in survey = 1100 and the number of people who like curd only are 220.
A businessman exchanged Nepali currency Rs. 6,60,000 into US dollar at the rate of $1 = NRs. 110. After four days Nepali currency is revaluated by 10% and he exchanged the dollars into Nepali currency again. What is his gain or loss? Find it.
Solution:
Given,
US $1 = NRs. 110
Exchanged currency = NRs. 6,60,000
Revaluated rate of nepali currency = 10%
To find,
Profit or loss when US dollar is reexchanged into Nepali currency
According to the initial rate of exchange,
NRs 110 = US $1
Or, NRs 1 = US $1/110
Or, NRs 6,60,000 = $6,60,000/110
= $6000
Here, when Nepali currency is revaluated by 10%, then the new rate of exchange becomes
$1 = NRs 110  10% of NRs 110
= NRs 110  NRs 11
= NRs 99
Now, according to the new rate of exchange
US $1 = NRs. 99
∴ $6000 = NRs. 5,94,000
Here, Nepali currency received by reexchange is less than the initial currency. So, there is a loss
∴ loss = Initial currency  currency received after reexchange
= NRs 6,60,000  NRs 5,94,000
= NRs. 66,000
∴ The loss for businessman is NRs. 66,000
The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the solid object is 6 cm, length of the cylindrical part is 116 cm and the total length of the solid object is 120 cm. Find the total surface area of the solid.
Solution:
Given,
In a combined solid made of a cylinder and a cone, diameter of the base (d) = 6 cm,
length of cylindrical part = 116 cm,
Total length of the solid = 120 cm
To find:
Total surface area of the combined solid.
Here,
Radius of the base (r) = \(\frac d2\)
= \(\frac{6cm}2\)
= 3 cm
Height of the cone (h) = 120 cm  116 cm = 4 cm
And,
slant height of cone (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{{(3cm)}^2+{(4cm)}^2}\)
= \(\sqrt{9cm^2+16cm^2}\)
= \(\sqrt{25cm^2}\)
= 5 cm
Now, by formula
Area of circular base = πr^{2}
= \(\frac{22}7{(3cm)}^2\)
= \(\frac{22}7\times9cm^2\)
= 28.29 cm^{2}
CSA of cylinder = 2πrh
= \(2\times\frac{22}7\times3cm\times116cm\)
= 2187.43 cm^{2}
And,
CSA of cone= πrl
= \(\frac{22}7\times3cm\times5cm\)
= 47.14 cm^{2}
∴ TSA of the combined solid = 28.29cm^{2} + 2187.43cm^{2} + 47.14cm^{2} = 2262.86 cm^{2} is the required answer
Find the HCF of: p^{2} + 4pq + 4q^{2} , p^{4} + 8pq^{3} and 3p^{4}  10p^{2}q^{2} + p^{3}q
Solution:
Here, first expression
= \(p^2+4pq+4q^2\)
= \(p^2+2.p.2q+{(2q)}^2\)
= \({(p+2q)}^2\)
= (p+2q)(p+2q)
Second expression
= \(p^4+8pq^3\)
= \(p(p^3+8q^3)\)
= \(p(p^3+{(2q)}^3)\)
= \(p(p+2q)(p^22pq+4q^2)\)
Third expression
= \(3p^4+10p^2q^2+p^3q\)
= \(p^2(3p^210q^2+pq)\)
= \(p^2(3p^2+pq10q^2)\)
= \(p^2(3p^2+(65)pq10q^2)\)
= \(p^2(3p^2+6pq5pq10q^2)\)
= \(p^2(3p(p+2q)5p(p+2q)\)
= \(p^2(p+2q)(3p5q)\)
∴ HCF is (2+2q)
Simplify: \(\frac1{1+p+p^2}\frac1{1+p+p^2}\frac{2p}{1p^2+p^4}\)
Solution:
Here,
\(\frac1{1+p+p^2}\frac1{1+p+p^2}\frac{2p}{1p^2+p^4}\)
= \(\frac{1\left(1+p+p^2\right)1\left(1p+p^2\right)}{\left(1p+p^2\right)\left(1+p+p^2\right)}\frac{2p}{1p^2+p^4}\)
= \(\frac{1+p+p^21+pp^2}{\left\{\left(1+p^2\right)p\right\}\left\{\left(1+p^2\right)+p\right\}}\frac{2p}{1p^2+p^4}\)
= \(\frac{2p}{\left(1+p^2\right)^2p^2}\frac{2p}{1p^2+p^4}\)
= \(\frac{2p}{1+2p^2+p^4p^2}\frac{2p}{1p^2+p^4}\)
= \(\frac{2p}{1+p^2+p^4}\frac{2p}{1p^2+p^4}\)
= \(\frac{2p\left(1p^2+p^4\right)2p\left(1+p^2+p^4\right)}{\left(1+p^2+p^4\right)\left(1p^2+p^4\right)}\)
= \(\frac{2p2p^3+2p^52p2p^32p^5}{\left\{\left(1+p^4\right)+p^2\right\}\left\{\left(1+p^4\right)p^2\right\}}\)
= \(\frac{4p^3}{\left(1+p^4\right)^2\left(p^2\right)^2}\)
= \(\frac{4p^3}{1+2p^4+p^8p^4}\)
= \(\frac{4p^3}{1+p^4+p^8}\)
In the given figure, EF//GH, EI//FG and FJ//GI, then prove that:
Solution:
Given: EF//GH, EI//FG and FJ//GI
To prove:
Proof:
SN  Statements  Reasons 
1. 
In ΔEFJ and ΔHGI


2.  ∴ ΔEFJ ≅ ΔHGI  By A.A.S statement. 
3.  Area of ΔEFJ = area of ΔHGI  Areas of congruent triangles, from statement 2 
4.  Area of Trapezium EFGI  ΔHGI = Area of Trapezium EFGI  area of ΔEFJ  Subtracting equal triangles from Trapezium EFGI. 
5.  ∴ Area of parallelogram EFGH = Area of parallelogram FGIJ  From statement (4), by whole part axiom. 
Hence proved.
Construct a quadrilateral MNOP in which NO = MN = 4.2 cm, OP = PM = 5.2 cm, and ∠NOP = 75°. Also construct the ΔPQO which is equal in area to the quadrilateral MNOP.
Verify experimentally that the circumference angle BED is half of the central angle BCD standing on the same arc BD of a circle. (Two circles having radii at least 3 cm are necessary).
Solution:
Statement:
The circumference angle BED is half of the central angle BCD standing on the same arc BD of a circle.
Construction of figures:
Two circles of different radii with center C are drawn. In each circle, circumference angle BED and central angle BCD are drawn which are standing on the same arc BD.
To verify:
By using a protractor, ∠BED and ∠BCD are measured and the results are tabulated below.
Figure  ∠BED  ∠BCD  ∠BCD  Results 
1.  40°  80°  40°  ∠BED = 1/2 ∠BCD 
2.  63°  126°  63°  ∠BED = 1/2 ∠BCD 
Conclusion:
Hence, the circumference angle of a circle is half of the central angle standing on the same arc.
Hence verified.
The heights of house and temple are 13 m and 25 m respectively. If a man observes the roof of a temple he finds the angle of depression 45°, find the distance between the house and the temple.
Solution:
Here, CF = CD  FD
= 25 m  13 m
= 12 m
Now, in the right angled triangle ΔCFA,
tan(45°) = CF/AF
Or, 1 = 12m/BD
Or, BD = 12 m
∴ The distance between the house and the temple is BD = 12 m
Find the median from the data given below:
Class Interval  4050  5060  6070  7080  8090 
Frequency  8  6  5  4 
Solution:
Here,
Preparing a cumulative frequency table to calculate median from the given data, we get
Class (X)  Frequency (f)  Cumulative frequency (cf) 
4050  7  7 
5060  8  7+8=15 
6070  6  15+8=21 
7080  5  21+5=26 
8090  4  26+4=30 
N=30 
Using formula,
Median (M_{d}) = The value of (N/2)^{th} item
= (30/2)^{th} item
= 15^{th} item
here,
The corresponding class of cf 15 is 5060
∴ Median class = 5060
Again, by formula
Actual median (M_{d}) = \(L+\frac{\left(\frac{N}{2}cf\right)}{f}\cdot i\)
Where,
L = 50,
N/2 = 15
cf = 7
f = 8
i = 10
So, Actual median (M_{d})
= \(50+\frac{\left(157\right)}{8}\cdot10\)
= 60
∴ Median of the given data is 60.
A person deposited Rs. 55,000 in bank 'p' for 2 years at 10% compound interest compounded annually. But after one year the bank changed the policy and decided to pay semiannually, compound interest at the same rate. What is the percentage difference between the compound interest of the first and second year? Give reason with calculation.
Solution:
Here,
For the first year:
Principal (P1) = Rs. 55,000
Rate of interest (R1) = 10% p.a.
Time (T1) = 1 year
Annual compound interest (CI1) =?
By formula,
CI1 = \(p1\left\{\left(1+\frac{R1}{100}\right)^{T1}1\right\}\)
= \(55000\left\{\left(1+\frac{10}{100}\right)^11\right\}\)
= \(55000\left\{\left(1.1\right)1\right\}\)
= Rs. 5500
∴ Compound interest of the first year (CI1) = Rs. 5500
And, compound Amount after 1 year (CA1) = P1 + CI1 = Rs. 60,500
Again,
For the second year:
Principal (P2) = CA1 = Rs. 60,500
Rate of interest (R2) = 10% p.a.
Time (T2) = 1 year
Semiannual interest (CI2) = ?
By formula,
Semiannual compound interest (CI2)
= \(P2\left\{\left(1+\frac{R2}{200}\right)^{2\cdot T2}1\right\}\)
= \(60500\left\{\left(1+\frac{10}{200}\right)^21\right\}\)
= \(60500\left\{1.10251\right\}\)
= \(\left\{60500\cdot0.1025\right\}\)
= Rs. 6201.25
Here, CI2  CI1 = Rs. 6201.25  Rs. 5500
= Rs. 701.25
∴ The interest of the second year is more than the interest of the first year by Rs. 701.25
Now, the difference in interest in percentage = \( \frac{diffininterest}{CI1}\cdot100\%\)
= \(\frac{Rs701.25}{Rs5500}\cdot100\%\)
= 12.75%
∴ The interest of the second year is 12.7% more than the interest of the second year.
The general formula for compound interest.
The formula for compound interest is:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal amount (the initial investment/loan amount)
r = the annual interest rate (expressed as a decimal)
n = the number of times the interest is compounded per year
t = the number of years the money is invested/borrowed for
This formula calculates the amount of interest earned on an investment or the amount of interest owed on a loan, taking into account the compounding effect of earning interest on both the principal amount and any previously earned interest. The more frequently the interest is compounded, the greater the impact on the overall amount of interest earned.
Two pillars of height 8 feet each with four faces shown of the gate of a stadium have oneone pyramid of height 4 feet each having same base on their tops. The base of each pillar is 6 ft x 6 ft. If the pillars with pyramid are painted at the rate of Rs. 75 per square feet, what will be the total cost?
Solution:
Given:
In pillars formed by prism and pyramid,
Length of the square base (l) = 6 feet
Height of prism = 8 feet
Height of pyramid = 4 feet
No of pillars = 2
Rate of painting = Rs. 75/sq.ft
To find: Total cost of painting on two pillars.
Here,
Slant height of the square based pyramid (h1)
= \(\sqrt{h^2+\left(\frac{l}{2}\right)^2}\)
= \(\sqrt{4^2+\left(\frac{6}{2}\right)^2}\)
= 5 feet
And, Perimeter of the square base = 4l = 4 x 6 feet = 24 feet
By formula, L.S.A of a square based prism
= Perimeter of base x height
= 24 feet x 8 feet
= 192 sq. ft
Again, by formula
LSA of a square based pyramid = 2lh_{1}
= 2 x 6 ft x 5 ft
= 60 sq ft
Here, Surface area of each pillar
= LSA of prism + LSA of pyramid
= 192 sq ft + 60 sq ft
= 252 sq ft
∴ Surface area of two pillars = 2 x 252 sq ft = 504 sq ft
Now, total cost of painting = Cost per sq ft x total area = Rs 75 x 504 = Rs. 37,800
∴ The total cost of painting two pillars = Rs. 37,800
The area of a rectangular land is 3000 sq meters and perimeter is 220 meters. Out of length or breadth which one is to be decreased by what percent to make it square? Find it.
Solution:
Given:
Area of a rectangular land = 3000 sq m
and perimeter of the land = 220 m
To find:
Percentage of length or breadth to be decreased to make the square land.
Let, length (l) = x m and breadth (b) = y m
Then, by formula, area of rectangle = l x b i.e., xy = 3000  ( equation 1)
and , perimeter of a rectangle = 2(l + b) = 2 ( x + y) = 220 i.e., x + y = 110  (equation 2)
Substituting the value of y = 110  x from equation 2 in equation 1 we get,
x (110  x) = 3000
Or, 110x  x^{2} = 3000
Or, x^{2}  110x + 3000 = 0
Or, x^{2}  (60x + 50x) + 3000 = 0
Or, x^{2}  60x 50x + 3000 = 0
Or, x(x  60)  50(x60) = 0
Or, (x  50)(x  60) = 0
Either, x 60 = 0  (equation 3)
Or, x  50 =0  (equation 4)
From, eqn (3), x  60 = 0 => x = 60
Now, putting the value of 'x' in equation 2 we get
y = 110  60 = 50
∴ If x = 60, then y = 50
Again, from equation 4
x  50 =0 => x = 50. Putting x in eqn (2) we get, y = 110  x = 110  50 = 60
∴ If x = 50, then y = 60
But assuming that length is longer than breadth, we get length(l) = 60 m and breadth (b) = 50 m
∴ Difference between length and breadth = 60 m  50 m = 10 m
So, to make the land square the length has to be decreased by 10 m.
Here, Percentage decreased in length = \(\frac{difference}{length}\cdot100\%\)
= \(\frac{10m}{60m}\cdot100\%\)
= \(\frac{50}{3}\%\)
= 16.66%
∴ To change the given rectangular land into a square, length has to be decreased by 16.66%
Points K,L,M and N are concyclic such that arc KL = arc NM. If the chords KM and LN are intersected at a point P, then prove that.
Solution:
Given:
Points K, L, M and N are concyclic such that arc \(\overset\frown{KL}\) and arc \(\overset\frown{NM}\) are intersected at a point P.
To Prove:
Proof:
Statements  Reasons  
1. 
In triangle ΔKPL and triangle ΔNPM,


2.  ∴ triangle ΔKPL ≅ ΔNPM  By A.S.A statement 
3.  ∴ Area of ΔKPL = Area of ΔNPM  Area of congruent triangles are equal. 
4.  arc \(\overset\frown{KL}\) = arc \(\overset\frown{NM}\)  Given 
5.  \(\overset\frown{KL}\) + \(\overset\frown{LM}\) = \(\overset\frown{LM}\) + \(\overset\frown{MN}\)  Adding arc MN on both sides of statement (4) 
6.  (\overset\frown{KLM}\) = (\overset\frown{LMN}\)  From statement (5), by whole part axiom. 
7.  ∴ Chord KM = Chord LN  From statement (6), chords subtended by equal arcs in the circle. 
Hence proved
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