A transversal intersects two parallel lines. -Maths 9th

A transversal intersects two parallel lines. -Maths 9th
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Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and  ∠CQP, respectively.
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A transversal intersects two parallel lines. -Maths 9th

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Description : If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Last Answer : Solution :-

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Description : If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Last Answer : answer:

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 and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

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.

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.

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

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

Last Answer : Solution :-

is this statement true or falseA transversal is a line that intersects another line.?

Description : is this statement true or falseA transversal is a line that intersects another line.?

Last Answer : Answers is the place to go to get the answers you need and to ask the questions you want

is this statement true or falseIf a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other.?

Description : is this statement true or falseIf a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other.?

Last Answer : 1

Is this statement true or falseIf two lines are intersected by a transversal so that consecutive interior angles are congruent, then the lines are parallel.?

Description : Is this statement true or falseIf two lines are intersected by a transversal so that consecutive interior angles are congruent, then the lines are parallel.?

Last Answer : Answers is the place to go to get the answers you need and to ask the questions you want

Is this statement true or falseIf two lines are intersected by a transversal so that the alternate interior angles are congruent, then the lines are parallel.?

Description : Is this statement true or falseIf two lines are intersected by a transversal so that the alternate interior angles are congruent, then the lines are parallel.?

Last Answer : 1

If two parallel lines are intersected by a transversal, then the alternate exterior angles are complementary.?

Description : If two parallel lines are intersected by a transversal, then the alternate exterior angles are complementary.?

Last Answer : Answers is the place to go to get the answers you need and to ask the questions you want

If two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.?

Description : If two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.?

Last Answer : 1

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

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

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

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.

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

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

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

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.

Find the co-ordinate where the equation 2x + 3y = 6 intersects x-axis. -Maths 9th

Description : Find the co-ordinate where the equation 2x + 3y = 6 intersects x-axis. -Maths 9th

Last Answer : Solution :-

ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

Description : ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

Last Answer : Solution :-

In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Last Answer : Solution :-

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

Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

Description : Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

Last Answer : We have ∠BED = ∠BAD (Angles in the same segment) ⇒ ∠BED = 1/2∠A ...(i) Also, ∠BEF = ∠BCF (Angles in the same segment) ⇒ ∠BEF = 1/2∠C ...(ii) From (i) and (ii) ∠BED + ∠BEF = 1/2∠A + 1/2∠C ∠DEF ... ∠A + ∠C) ⇒ ∠DEF = 1/2(180° - ∠B) (Since, ∠A + ∠B + ∠C = 180°) ⇒ ∠DEF = 90° - 1/2∠B

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

In Fig.6.5,which of the two lines are parallel? -Maths 9th

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Description : Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Last Answer : Solution :-

In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Last Answer : Solution :-

For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Last Answer : Solution :-

Whether the pair of given lines are parallel or not give reason. -Maths 9th

Description : Whether the pair of given lines are parallel or not give reason. -Maths 9th

Last Answer : Where are the lines please tell first

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.

If m parallel lines in a plane are intersected by a family of n parallel lines, find the number of parallelograms formed? -Maths 9th

Description : If m parallel lines in a plane are intersected by a family of n parallel lines, find the number of parallelograms formed? -Maths 9th

Last Answer : answer:

Show that the equation of the parallel line midway between the parallel lines -Maths 9th

Description : Show that the equation of the parallel line midway between the parallel lines -Maths 9th

Last Answer : ∵ Length of perpendicular from point (x1, y1) to line a\(x\) + by + c = 0 = \(rac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\)∴ Length of perpendicular from (0, 0) to \(rac{x}{a}\) + \(rac{y}{b}\) = 1 ⇒ \(rac{\big|rac{1}{a} imes0+ ... {1}{b^2}+rac{1}{b^2}}\) ⇒ \(rac{1}{p^2}\) = \(rac{1}{a^2}\) + \(rac{1}{b^2}\)

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Description : A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Last Answer : (c) 3x - 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (-5, 0) having slope = m is y - 0 = m(x - (-5)) or mx - y + 5m = 0 ...(i) Its ... (rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x - 4y + 15 = 0.

The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

Is this statement true or falseIf two lines are intersected by a transversal so that the alternate exterior angles are congruent, then the lines are perpendicular.?

Description : Is this statement true or falseIf two lines are intersected by a transversal so that the alternate exterior angles are congruent, then the lines are perpendicular.?

Last Answer : Answers is the place to go to get the answers you need and to ask the questions you want