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

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
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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 :-

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

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In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

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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

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ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

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Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

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In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Description : In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Last Answer : First Find the points D and E by midpoint formula. (x₂+x₁/2 , y₂+y₁/2) For DE=1/2AB In ΔsCED and CAB ∠ECD=∠ACB and the ratio of the side containing the angle is same i.e, CD=1/2BC ⇒CD/BC=1/2 EC=1/2AC ⇒EC/AC=1/2 ∴,ΔCED~ΔCAB hence the ratio of their corresponding sides will be equal, DE=1/2AB

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : NEED ANSWER

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

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)

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

D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6

Description : D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6

Last Answer : (b) 3

In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

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Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th

Description : Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th

Last Answer : answer:

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.

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

Last Answer : answer:

In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th

Description : In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th

Last Answer : answer:

In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

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 :-

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

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

Figure abc is rectangle. segment AB is 6cm long , segment BC is 8 cm long , and segment AC is 10 cm long. what is the area of triangle abc?

Description : Figure abc is rectangle. segment AB is 6cm long , segment BC is 8 cm long , and segment AC is 10 cm long. what is the area of triangle abc?

Last Answer : 28

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

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.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units and AC=2 units What is the relationship among AB,BC and C

Description : In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units and AC=2 units What is the relationship among AB,BC and C

Last Answer : In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units ... units What is the relationship among AB,BC and CA?

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.

In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Last Answer : answer:

In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th

Description : In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th

Last Answer : answer:

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

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.

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

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 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).