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

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
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1 Answer

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) ⇒ ad = bc ⇒ cb = da ⇒ (c, d) R (a, b) ⇒ R is symmetric. Commutivity of multiplication in N (iii) R is transitive. Suppose (a, b) R (c, d) and (c, d) R (e, f). Then ad = bc and cf = de ⇒ (ad) (cf) = (bc) (de) ⇒ af = be ⇒ (a, b) R (e, f) ⇒ R is transitive. Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation on N × N.
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Related questions

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.

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.

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.

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.

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.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.

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.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Description : ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Last Answer : Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through ... △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.

ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Description : ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Last Answer : Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through ... △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Last Answer : Solution :-

ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Last Answer : Solution :-

In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

Description : In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

Last Answer : It is given that ∠BAD=∠EAC ∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides] ∴∠BAC=∠DAE In △BAC and △DAE AB=AD (Given) ∠BAC=∠DAE (Proved above) AC=AE (Given) ∴△BAC≅△DAE (By SAS congruence rule) ∴BC=DE (By CPCT)

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 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}.

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

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.

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.

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.

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 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)}

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.

ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

Last Answer : This answer was deleted by our moderators...

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Last Answer : Solution :-

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

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in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Last Answer : This answer was deleted by our moderators...

If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

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In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Description : In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Last Answer : AD = 24, BC = 12. In ΔCBE and ΔADE, ∠CBA = ∠CDA, ∠BCE = ∠DAE (Angles in the same segment are equal) ∠BEC = ∠DEA (vertical opposite angles are equal) ⇒ ΔBCE and ΔDEA are similar Δs with sides in the ratio 1 : 2. ∴ Ratio of areas = Ratio of square of sides = 12 : 22 = 1 : 4