Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th
by

1 Answer

Answer :

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3}. (i) Since (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive. (ii) (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric. (iii) (1, 2) ∈ R, (2, 3) ∈ R and (1, 3) ∈R ⇒ R is transitive. ∴ Option (a) is the right answer.
by

Related questions

Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Description : Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Last Answer : Given: A = {All books in a library of a school} R = {(x, y): x and y have the same number of pages} Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A Symmetric: Since books x and y have the same number of pages ... and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. Hence, R is an equivalence relation.

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Description : Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Description : The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Last Answer : (c) Symmetric onlyAs (1, 1), (2, 2) ∉ R so R is not reflexive. A relation a R b is said to symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R. Here (1, 2) ∈ R and (2, 1) ∈ R so it is symmetric. Also, as (1, 2) ∈ R, but (2, 3), (1, 3) ∉ R, so R is not transitive.

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Description : Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Last Answer : (c) Reflexive and transitive onlyHere A = {3, 6, 9, 12} and R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} ∵ (3, 3), (6, 6), (9, 9), (12, 12) all belong to R ⇒ {(a, a): ∈R V a ∈A} ... 12) ∈ R and (3,12) ∈ R ⇒ (3, 12) ∈ R (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R Hence R is transitive.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Description : Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Last Answer : (d) 7A = {1, 2, 3}. R = {(1, 2), (2, 3)} To make R an equivalence relation, it should be: (i) Reflexive: So three more ordered pairs (1, 1), (2, 2), (3, 3) should be added to R to make it ... 2, 1), (3, 2), (1, 3), (3, 1)} is an equivalence relation. So minimum 7 ordered pairs are to be added.

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Last Answer : (c) Symmetric only Let a ∈N. Then (a, a) ∉R as the GCD of a' and a' is a' not 2. R is not reflexive Let a, b ∈N. Then, (a, b) ∉R ⇒ GCD of a' and b' is 2 ⇒ GCD of b' and a' is 2 ⇒ (b, a) ∈R ∴ R ... , let a = 4, b = 10, c = 12 GCD of (4, 10) = 2 GCD of (10, 12) = 2 But GCD of (4, 12) = 4.

On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Last Answer : (a) Reflexive and symmetric only(a, a) ∈ R ⇒ 1 + a . a = 1 + a2 > 0 V real numbers a ⇒ R is reflexive (a, b) ∈ R ⇒ 1 + ab > 0 ⇒ 1 + ba > 0 ⇒ (b, a) ∈ R ⇒ R is symmetricWe observe that \(\big(1,rac{1}{2}\big) ... }{2},-1\big)\) ∈ Rbut (1, - 1) ∉ R as 1 + 1 (-1) = 0 \( ot>\) 0 ⇒ R is not transitive.

If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Description : Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Last Answer : (b) Reflexive and transitive but not symmetricLet A = {1, 2, 3, 4} • ∵ (1, 1), (2, 2), (3, 3) and (4, 4) ∈R ⇒ R is reflexive • ∵ (1, 2) ∈ R but (2, 1) ∉ R ; (1, 3) ∈ R and (3, 1) ∉ R ; (3, 2) ∈R and (2, 3) ∉ R ⇒ R is not symmetric • (1, 3) ∈ R and (3, 2) ∈ R and (1, 2) ∈ R ⇒ R is transitive.

If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Last Answer : (d) \(2^{n^2}\)Set A has n elements ⇒ n(A) = n ⇒ A × A has n × n = n2 elements ∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Description : If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Last Answer : (i) R is reflexive. For all (a, b) ∈ N N we have (a, b) R (a, b) because ab = ba ⇒ R is reflexive. (ii) R is symmetric. Suppose (a, b) R (c, d) Then (a, b) R (c, d) ⇒ ... ) R (e, f) ⇒ R is transitive. Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation on N N.

Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Description : Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Last Answer : (d) Symmetric and transitive| a - a | = | 0 | = 0 so (a, a) ∉R ⇒ R is not reflexive(a, b) ∈ R ⇒ | a - b | > 0 ⇒ | b - a | > 0 ⇒ (b, a) ∈R (∵ | a - b | = | b - a |) ⇒ R is symmetric (a, b) ∈ R ... a, b, c. ∴ | a - b | > 0 and | b - c | > 0 ⇒ | a - c | > 0 ⇒ (a, c) ∈ R ⇒ R is transitive.

For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

Description : For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

Last Answer : (d) an equivalence relationGiven, a R b ⇒ sin2a + cos2b = 1 Reflexive: a R a ⇒ sin2 a + cos2 a = 1 ∀ a ∈ R (True) Symmetric: a R b ⇒ sin2 a + cos2 b = 1 ⇒ 1 - cos2 a + 1 - sin2 b = 1 ⇒ sin2 b + ... + cos2 b + sin2 b + cos2 c = 2 ⇒ sin2 a + cos2 c = 1 ⇒ a R c (True)∴ R is an equivalence relation.

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Description : Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Last Answer : (a) 1For the relation R to become transitive: (1, 2) ∈R and (2, 3) ∈R should imply (1, 3) ∈R ∴ Minimum one ordered pair (1, 3) should be added to R.

Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Description : Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Description : Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Last Answer : Reflexive : A ≅ A True Symmetric : if A ≅ B then B ≅ A True Transitive : if A ≅ B and B ≅ C, then A ≅ C True Therefore, the relation ‘≅’ is an equivalence relation.

Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Description : Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Last Answer : (c) {(1, 1), (2, 2), (3, 1), (1, 3), (3, 3)} Option (c) satisfies all the conditions of an equivalence relation. (1, 1), (2, 2), (3, 3) ∈ R ⇒ (a, a) ∈ R V a ∈ A ⇒ R is reflexive (3, 1) ∈ R and (1, 3) ∈ R ... R and (1, 1) ∈ R ⇒ (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R V a, b, c ∈ A ⇒ R is transitive.

The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Description : The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Last Answer : (d) An equivalence relationLet R = {(x, y) : line x is parallel to line y, x y ∈ set of coplanar straight lines}. Every line is parallel to itself. So, if x ∈S, then (x, x) ∈R ⇒ R is ... | z ⇒ (x, z) ∈R ⇒ R is transitive ∴ R being reflexive, symmetric and transitive, it is an equivalence relation.

The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Description : The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Last Answer : (c) Only transitiveLet N be the set of natural numbers. Then R = {(a, b) : a < b, a, b ∈N}A natural number is not less than itself ⇒ (a, a)∉R where a ∈N ⇒ R is not reflexive V a, b ∈N, (a, b) ∈R ⇒ a < b \( ot\ ... ∈N, (a, b) ∈R and (b, c) ∈R ⇒ a < b and b < c ⇒ a < c (a, c) ∈R ⇒ R is transitive.

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Description : The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Last Answer : Solution :-

If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

Define the term of Relation with example. -Maths 9th

Description : Define the term of Relation with example. -Maths 9th

Last Answer : Relation: If A and B are any two non-empty sets, then any subset of A B is defined as a relation from A to B. For example, Suppose A = {1, 2, 3} and B = {1, 2, 3, 4}. Then {(2, 3), ... 3)} is a relation in A B. Many more relations (subsets) can be selected at random from our product set A B.

Define the term of Domain, codomain and range of a relation: -Maths 9th

Description : Define the term of Domain, codomain and range of a relation: -Maths 9th

Last Answer : Let R be a relation from set A to set B. Then, the set of first element of the ordered pairs in R is called the domain and the set of second elements of the ordered pairs in R is called the range. The second set B is called ... 16, 25, 36}, Range of R = {4, 5, 6} and Codomain of R = {1, 4, 5, 6}.

Define the term of Inverse of a relation. -Maths 9th

Description : Define the term of Inverse of a relation. -Maths 9th

Last Answer : For any binary relation R, a second relation can be constructed by merely interchanging first and second components in every ordered pair. The relation thus obtained is called the inverse of the first one and designated as R-1. Thus, R-1 = {(y, x ... {(1, 2), (2, 3), (3, 4), (5, 4)}. So (R-1)-1 = R.

Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Description : Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Last Answer : (i) is parallel to' is reflexive, because any line is parallel to itself. Symmetric, because if line l is parallel to line m, then line m is parallel to line l. Transitive, because if l || m ... y cannot be a factor of x. (v) No, because the relation is reflexive and transitive but not symmetric.

If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

Description : If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

Last Answer : answer:

The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th

Description : The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th

Last Answer : Let median through C be CF. AF=FB=c2 CF=122(a2+b2)−c2−−−−−−−−−−−−√=3c2 CG=c where G is the centroid and GF=c2 34( ... +M2b+9c24 c2=(23M2a)+(23M2b) BC2=AG2+BG2 So medians through A and B are perpendicular.

If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Description : If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Last Answer : answer:

Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Description : Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Last Answer : answer:

In partial order plan. A. Relationships between the actions of the behavior are set prior to the actions B. Relationships between the actions of the behavior are not set until absolutely necessary Choose the correct option. a) A is true b) B is true c) Either A or B can be true depending upon situation d) Neither A nor B is true

Description : In partial order plan. A. Relationships between the actions of the behavior are set prior to the actions B. Relationships between the actions of the behavior are not set until absolutely necessary Choose the ... ) Either A or B can be true depending upon situation d) Neither A nor B is true

Last Answer : a) A is true

A) Knowledge base (KB) is consists of set of statements. B) Inference is deriving a new sentence from the KB. Choose the correct option. a) A is true, B is true b) A is false, B is false c) A is true, B is false d) A is false, B is true

Description : A) Knowledge base (KB) is consists of set of statements. B) Inference is deriving a new sentence from the KB. Choose the correct option. a) A is true, B is true b) A is false, B is false c) A is true, B is false d) A is false, B is true

Last Answer : a) A is true, B is true

Mark the correct option a) All statements which the court permits or requires to be made before it by witnesses, in relation to matters of fact under inquiry such statements are called oral evidence b) All documents including electronic records produced for the inspection of the court called as documentary evidence c) Both the A and B are correct d) Neither A nor B are correct.

Description : Mark the correct option a) All statements which the court permits or requires to be made before it by  witnesses, in relation to matters of fact under inquiry such statements are called  oral evidence b) ... documentary evidence c) Both the A and B are correct d) Neither A nor B are correct.

Last Answer : c) Both the A and B are correct

If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Description : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Last Answer : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2) ... symmetric B. reflexive C. transitive D. an equivalence relation.

A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD→E, BE→F, B→C, AF→G} Which of the following is a candidate key? (A) A (B) AB (C) ABC (D) ABD

Description : A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD→E, BE→F, B→C, AF→G} Which of the following is a candidate key? (A) A (B) AB (C) ABC (D) ABD

Last Answer : Answer: D

A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Description : A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Last Answer : ∴ Required number of ways of seating = 9!

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm