Prove that the diagonal divides a parallelogram into two congruent triangles. -Maths 9th

Prove that the diagonal divides a parallelogram into two congruent triangles. -Maths 9th
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Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Description : Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

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Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

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Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

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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.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

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Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Last Answer : . Solution: (i) In ΔADC and ΔCBA, AD = CB (Opposite sides of a parallelogram) DC = BA (Opposite sides of a parallelogram) AC = CA (Common Side) , ΔADC ≅ ΔCBA [SSS congruency] Thus, ∠ACD = ∠CAB by ... are equal) Also, AB = BC = CD = DA (Opposite sides of a parallelogram) Thus, ABCD is a rhombus.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th

Last Answer : Join QS. Let diagonals PR and QS intersect each other at T. We know, that diagonals of a parallelogram bisect each other . ∴ T is the mid - point of QS. Since a median of a triangle divides it into two triangles of equal ... ar(△PTS) + ar( △STO) = ar(△PQT) = ar( △ QTO ) ⇒ ar(△PSO) = ar(△PQO)

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th

Last Answer : Join QS. Let diagonals PR and QS intersect each other at T. We know, that diagonals of a parallelogram bisect each other . ∴ T is the mid - point of QS. Since a median of a triangle divides it into two triangles of equal ... ar(△PTS) + ar( △STO) = ar(△PQT) = ar( △ QTO ) ⇒ ar(△PSO) = ar(△PQO)

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).

In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Description : PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Last Answer : Solution :- As diagonal of the parallelogram divides it into two triangles of equal area. Since, area (△SRQ ) = 1/2 area(PQRS) area (△SRQ ) = 1/2 x 180 ... = 90 cm2 (Given) This is not possible unless area (△SRQ ) = area (△ASR ) So, the given statement is false.

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

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Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

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Which arrangement cannot be used to prove two triangles congruent?

Description : Which arrangement cannot be used to prove two triangles congruent?

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what- If you are given or can prove that two triangles are congruent, then you may use CPCTC to prove that the angles or sides are?

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Is this statement true or falseTo prove triangles similar using the Side-Side-Side Similarity Theorem, you must first prove that corresponding angles are congruent?

Description : Is this statement true or falseTo prove triangles similar using the Side-Side-Side Similarity Theorem, you must first prove that corresponding angles are congruent?

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Is this statement true or falseThere is enough information to prove the triangles congruent using HL?

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1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

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Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

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In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

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If two parallelogram PQAD and PQBC arw on the opposite sides of PQ prove that ABCD is a paralellogram -Maths 9th

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

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

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Last Answer : Suppose AL and PM are the altitudes corresponding to equal bases AB and PQ of ||gm ABCD and PQRS respectively . Since the ||gm are between the same parallels PB and SC. ∴ AL = PM Now, ar(||gm ABCD) = AB AL ar(|| ... PM But, AB = PQ [given] AL = PM [proved] ∴ ar(||gm ABCD) = ar(||gm PQRS)