If logy x = (a. logz y) = (b. logxz) = ab, then which of the following pairs of values for (a, b) is not possible ? -Maths 9th

If logy x = (a. logz y) = (b. logxz) = ab, then which of the following pairs of values for (a, b) is not possible ? -Maths 9th
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Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal. -Maths 9th

Description : Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal. -Maths 9th

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In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

If x = logabc, y = logbca, z = logc ab, then -Maths 9th

Description : If x = logabc, y = logbca, z = logc ab, then -Maths 9th

Last Answer : (a) x + y + z + 2 = xyz.x = logabc ⇒ ax = bc ⇒ ax + 1 = abc ⇒ a = (abc)1/(x +1) Similarly, b = (abc)1/(y + 1), c = (abc)1/(z + 1)∴ abc = \((abc)^{rac{1}{x+1}+rac{1}{y+1}+rac{1}{z+1}}\)⇒ \({rac{1}{x+1}+ ... + xz + x + z + 1 + xy + y + x + 1 = xyz + xy + yz + zx + x + y + z + 1⇒ x + y + z + 2 = xyz.

If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

Description : If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

Last Answer : (c) 1Let each ratio = k and base = e ⇒ loge x = k(a2 + ab + b2) ⇒ (a - b) loge x = k (a - b) (a2 + ab + b2) ⇒ loge xa - b = k(a3 - b3) ⇒ xa - b = \(e^{k(a^3-b^3)}\) Similarly, yb-c = \(e^{k(b^3-c^3)}\), zc-a = \ ... (e^{k(b^3-c^3)}\) . \(e^{k(c^3-a^3)}\)= \(e^{k[a^3-b^3+b^3-c^3+c^3-a^3]}\) = e0 = 1.

Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c. -Maths 9th

Description : Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c. -Maths 9th

Last Answer : Sol-2(a2+b2+c2-ab-bc-ca)/2 multiplying & dividing by 2 ...

Write three pairs of form (x,y) that satisfy the equation x + y = 5 . Use these ordered pairs to draw the graph of the equation x + y = 5 . -Maths 9th

Description : Write three pairs of form (x,y) that satisfy the equation x + y = 5 . Use these ordered pairs to draw the graph of the equation x + y = 5 . -Maths 9th

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Write three pairs of form (x,y) that satisfy the equation x = y . Use these ordered pairs to draw the graph of the equation x = y . -Maths 9th

Description : Write three pairs of form (x,y) that satisfy the equation x = y . Use these ordered pairs to draw the graph of the equation x = y . -Maths 9th

Last Answer : Given equation is x−3y=4 for y=−1 x−(3×−1)=4 x=1 therefore ordered pair of points is (1,−1)

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

Description : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

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X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

Description : X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

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In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Description : In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Last Answer : Since diagonals of a rhombus bisect each other at right angle . ∴ In △AOB , we have ∠OAB + ∠x + 90° = 180° ∠x = 180° - 90° - 35° [∵ ∠ OAB = 35°] = 55° Also, ∠DAO = ∠BAO = 35° ∴ ∠y + ∠DAO + ∠BAO + ∠x ... 180° ⇒ ∠y = 180° - 125° = 55° Hence the values of x and y are x = 55°, y = 55°.

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Description : In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Last Answer : Since diagonals of a rhombus bisect each other at right angle . ∴ In △AOB , we have ∠OAB + ∠x + 90° = 180° ∠x = 180° - 90° - 35° [∵ ∠ OAB = 35°] = 55° Also, ∠DAO = ∠BAO = 35° ∴ ∠y + ∠DAO + ∠BAO + ∠x ... 180° ⇒ ∠y = 180° - 125° = 55° Hence the values of x and y are x = 55°, y = 55°.

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y as its solution? -Maths 9th

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y as its solution? -Maths 9th

Last Answer : Solution :-

If x = 3 and y = -1, find the values of each of the following using in identity: -Maths 9th

Description : If x = 3 and y = -1, find the values of each of the following using in identity: -Maths 9th

Last Answer : Formula, (a+b+c)2=a2+b2+c2+2ab+2bc+2ca Given, (2+x−2y)2=22+(x)2+(−2y)2+2(2)(x)+2(x)(−2y)+2(2)(−2y) =4+x2+4y2+4x−4xy−8y

If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

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

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

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 two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

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

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

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 two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved

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

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

Description : in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

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

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, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Description : Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Last Answer : answer:

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Description : From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Last Answer : (i) In Δ O C E ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A