What is the first step in the construction of a line parallel to AB through point P?

What is the first step in the construction of a line parallel to AB through point P?
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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

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)

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

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)

which correctly shows the result of the third step when constructing a line through P parallel to L using method 1?

Description : which correctly shows the result of the third step when constructing a line through P parallel to L using method 1?

Last Answer : A, B, C, p, d, m

In step 4 of the construction of a perpendicular line through a point why must the compass point be placed on the points where the arc intersects with the original line How would the construction be d?

Description : In step 4 of the construction of a perpendicular line through a point why must the compass point be placed on the points where the arc intersects with the original line How would the construction be d?

Last Answer : What is the answer ?

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

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)

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 :-

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 the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

Last Answer : answer:

AB and CD are two smooth parallel walls. A child rolls a ball along ground from A towards point P find PD so that ball reaches point B after striking

Description : AB and CD are two smooth parallel walls. A child rolls a ball along ground from A towards point P find PD so that ball reaches point B after striking

Last Answer : AB and CD are two smooth parallel walls. A child rolls a ball along ground from A towards point P find PD so that ball ... B. 0.5 m C. 1.2 m D. 0.6 m

To construct a parallel to a line through a point not on the line using folding you can perform the construction twice?

Description : To construct a parallel to a line through a point not on the line using folding you can perform the construction twice?

Last Answer : perpendicular line segment (apex)

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).

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).

To set out a parallel from a given inaccessible point to a given line AB, the following observations are made: Distance AB and angle PAM = a and angle PBA = b are measured where M is a point on the line BA produced. The perpendicular to the desired parallel line from A and B are: (A) AB/(cot b - cot a) (B) AB/(cos b - cos a) (C) AB/(cot a - cot b) (D) AB/(cot a - cos b)

Description : To set out a parallel from a given inaccessible point to a given line AB, the following observations are made: Distance AB and angle PAM = a and angle PBA = b are measured where M is a point on the line BA produced. The ... cos b - cos a) (C) AB/(cot a - cot b) (D) AB/(cot a - cos b)

Last Answer : (A) AB/(cot b - cot a)

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.

is this statement true of false if point P is not on line L, the rhombus method can be used to construct a line M that is parallel to line L through P.?

Description : is this statement true of false if point P is not on line L, the rhombus method can be used to construct a line M that is parallel to line L through P.?

Last Answer : 1

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

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.

The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Description : The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Last Answer : answer:

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Last Answer : answer:

In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Description : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Last Answer : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

In the given figure, line DE is parallel to line AB. CD = 3 while DA = 6. Which of the following must be true? -Maths 9th

Description : In the given figure, line DE is parallel to line AB. CD = 3 while DA = 6. Which of the following must be true? -Maths 9th

Last Answer : answer:

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Last Answer : answer:

ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Description : ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Last Answer : answer:

Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

Description : AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

Last Answer : AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

Which of the following should be the FIRST step in removing a generator from parallel operation? A. Trip the generator off the switchboard. B. Turn off all electrical equipment. C. Remove the load from the 'off going' generator. D. Increase the cycles of the generator staying on the line.

Description : Which of the following should be the FIRST step in removing a generator from parallel operation? A. Trip the generator off the switchboard. B. Turn off all electrical equipment. C. Remove the load from the 'off going' generator. D. Increase the cycles of the generator staying on the line.

Last Answer : Answer: C

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

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.

Let ABCD be a parallelogram. P is any point on the side AB. If DP and CP are joined in such a way that they bisect the angles -Maths 9th

Description : Let ABCD be a parallelogram. P is any point on the side AB. If DP and CP are joined in such a way that they bisect the angles -Maths 9th

Last Answer : answer:

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If `sqrt(B

Description : Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If `sqrt(B

Last Answer : Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ... 4)` C. `(2)/(9)` D. `(4)/(9)`

How would the construction be different if you changed the compass setting in the next step of the perpendicular bisector construction?

Description : How would the construction be different if you changed the compass setting in the next step of the perpendicular bisector construction?

Last Answer : What is the answer ?

In the construction of the incenter why is step 5 necessary for drawing the inscribed circle?

Description : In the construction of the incenter why is step 5 necessary for drawing the inscribed circle?

Last Answer : What is the answer ?

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Last Answer : Solution :- x= -4

Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Description : Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Last Answer : Solution :- y = -4

The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.

Which line is parallel to y = 2 and goes through the point (3, 4)?

Description : Which line is parallel to y = 2 and goes through the point (3, 4)?

Last Answer : The line will have the same slope but with a different yintercept

is this statement true or falseIf line x is parallel to line y, then point Z is the only point through which y passes.?

Description : is this statement true or falseIf line x is parallel to line y, then point Z is the only point through which y passes.?

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

How do you write an equation that is parallel to a given line and passes through the given point?

Description : How do you write an equation that is parallel to a given line and passes through the given point?

Last Answer : Parallel straight line equations have the same slope but withdifferent y intercepts

Dew point of an air-water vapor mixture (A) Decreases with decrease in pressure (B) At constant humidity & total pressure is fixed (C) Corresponding to any point on the humidity chart is obtained by projecting a line through this point parallel to the temperature axis and to the saturation curve (D) All 'a', 'b' & 'c'

Description : Dew point of an air-water vapor mixture (A) Decreases with decrease in pressure (B) At constant humidity & total pressure is fixed (C) Corresponding to any point on the humidity chart is obtained by projecting a ... to the temperature axis and to the saturation curve (D) All 'a', 'b' & 'c'

Last Answer : (D) All 'a', 'b' & 'c'

A line AB with end point A (2,3) & B (7,8) is to be rotated about origin by 300 inclockwise direction. Determine the coordinates of end points S of rotated line. a.(3.232, 2.598) b.(5.232, 3.598) c.(3.232, 1.298) d.(3.232, 1.598)

Description : A line AB with end point A (2,3) & B (7,8) is to be rotated about origin by 300 inclockwise direction. Determine the coordinates of end points S of rotated line. a.(3.232, 2.598) b.(5.232, 3.598) c.(3.232, 1.298) d.(3.232, 1.598)

Last Answer : d.(3.232, 1.598)

In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Description : In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Last Answer : Solution of this question