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

O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th
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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 area.  ∴ In △PQS , PT is its median .  ⇒  ar(△PTS) = ar(△PQT) ---i) In △SQO, OT is its median .  ⇒  ar(△STO) = (△QTO) ---ii)  Adding (i) and (ii) , we have   ar(△PTS) + ar( △STO) = ar(△PQT) = ar( △ QTO )   ⇒ ar(△PSO) = ar(△PQO)
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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).

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)

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).

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.

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.

If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th

Description : If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th

Last Answer : if pqrs is a trapezium so pq and RS are parallel is you draw a diagonal you will divide the trapezium into two parts such that two equal triangle so

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Description : In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Last Answer : We know that opposite angles of a paralleloram are equal . ∴ ∠P = ∠R ⇒ 3x - 4 = 56 - 3x ⇒ 6x = 60 ⇒ x = 10 Thus, ∠P = ∠R = 3 10 - 4 = 26° Also, ∠P + ∠Q = 180° [ ... Hence, the four angles of the parallelogram PQRS are 26°, 154°,26° and 154°. We should care our earth to have good eco balance.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Description : In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Last Answer : We know that opposite angles of a paralleloram are equal . ∴ ∠P = ∠R ⇒ 3x - 4 = 56 - 3x ⇒ 6x = 60 ⇒ x = 10 Thus, ∠P = ∠R = 3 10 - 4 = 26° Also, ∠P + ∠Q = 180° [ ... Hence, the four angles of the parallelogram PQRS are 26°, 154°,26° and 154°. We should care our earth to have good eco balance.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

In parallelogram PQRS, PQ = 10cm. The altitudes corresponding to the sides PQ and SP are respectively 6cm and 8cm. Find SP. -Maths 9th

Description : In parallelogram PQRS, PQ = 10cm. The altitudes corresponding to the sides PQ and SP are respectively 6cm and 8cm. Find SP. -Maths 9th

Last Answer : Solution :-

The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ

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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

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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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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

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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

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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

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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...

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.

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

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

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

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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

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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

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

Last Answer : (i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A. ∴ ∠DAC=∠BAC ---- ( 1 ) Now, AB∥DC and AC as traversal, ∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 ) AD∥BC and AAC as traversal, ∴ ∠DAC= ... ---- ( 2 ) From ( 1 ) and ( 2 ), ⇒ AB=BC=CD=DA Hence, ABCD is a rhombus.

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

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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

The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Description : The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Description : ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Last Answer : Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median ... AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 8 = 16 cm2

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Description : The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Description : ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Last Answer : Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median ... AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 8 = 16 cm2

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

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Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Last Answer : Its given ABCD is a IIgram and AC and BD are its diagonals intersecting at point O. . Given : angle BOC = 900 angle BDC = 500 To find : angle OAB Answer : i) ...

ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Description : ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

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PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

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