Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a rhombus. Sulution: In ΔAOB and ΔAOD, BO = DO [∵ Given] ∠AOB = ∠AOD [∵ Each 90°] AO = AO [∵ Common] Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule] AB = AD [∵ CPCT] Similarly, AB = BC and BC = CD Now, all the fourRead more
Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°.
To prove: ABCD is a rhombus.
Sulution: In ΔAOB and ΔAOD,
BO = DO [∵ Given]
∠AOB = ∠AOD [∵ Each 90°]
AO = AO [∵ Common]
Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule]
AB = AD [∵ CPCT]
Similarly, AB = BC and BC = CD
Now, all the four sides of Quadrilateral ABCD are equal.
Hence, ABCD is a rhombus.
Given: ABCD is a square. To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°. Solution: ΔBAD and ΔABC, AD = BC [∵ Opposite sides of a square] ∠BAD = ∠ABC [∵ Each 90°] AB = AB [∵ Common] Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule] BD = AC [∵ CPCT] In ΔAOB and ΔCOD, ∠OAB = ∠OCD [∵ Alternate angles] AB =Read more
Given: ABCD is a square.
To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°.
Solution: ΔBAD and ΔABC,
AD = BC [∵ Opposite sides of a square]
∠BAD = ∠ABC [∵ Each 90°]
AB = AB [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule]
BD = AC [∵ CPCT]
In ΔAOB and ΔCOD,
∠OAB = ∠OCD [∵ Alternate angles]
AB = CD [∵ Opposite sides of a square]
∠OBA = ∠ODC [∵ Alternate angles]
Hence, ΔBAD ≅ ΔABC [∵ ASA Congruency rule]
AO = OC, BO = OD, [∵ CPCT]
In ΔAOB and ΔAOD.
OB = OD [∵ Proved above]
AB = AD [∵ Sides of a square]
OA = OA [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SSS Congruency rule]
∠AOB = ∠AOD [∵ CPCT]
But. ∠AOB + ∠AOD = 180° [∵ Linear pair]
⇒ 2∠AOB = 180° [∵ ∠AOD = ∠AOB]
⇒ ∠AOB = (180/2) = 90°
Hence, the diagonals of a square are equal and bisect each at right angles.
Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a square. Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus . Hence, AB = BC = CD = DA In ΔBAD and ΔABC, AD = BC [∵ Proved above] BD = AC [∵ Given] AB =Read more
Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°.
To prove: ABCD is a square.
Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus .
Hence, AB = BC = CD = DA
In ΔBAD and ΔABC,
AD = BC [∵ Proved above]
BD = AC [∵ Given]
AB = AB [∵ Common]
Hence, ΔBAD ΔABC [∵ SSS Congruency rule]
∠BAD = ∠ABC [∵ CPCT]
But, ∠BAD + ∠ABC = 180° [∵ Co-interior angles]
⇒ 2∠ABC = 180° [∵ ∠BAD = ∠ABC]
⇒ ∠ABC = (180°/2) = 90°
Hence, if the diagonals of a quadrilateral are equal and bisect each other at right angles, than it is a square.
Given: ABCD is a parallelogram with AC = BD. To Prove: ABCD is a rectangle. Solution: In ΔABC and Δ BAD, BC = AD [∵ Opposite sides of a parallelogram are equal] AC = BD [∵ Given] AB = AB [∵ Common] Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule] ∠ABC = ∠BAD [∵ CPCT] But, ∠ABC + ∠BAD = 180° [∵ Co-interiorRead more
Given: ABCD is a parallelogram with AC = BD.
To Prove: ABCD is a rectangle.
Solution: In ΔABC and Δ BAD,
BC = AD [∵ Opposite sides of a parallelogram are equal]
AC = BD [∵ Given]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule]
∠ABC = ∠BAD [∵ CPCT]
But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]
⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]
⇒ ∠BAD = 180°/2= 90°
A parallelogram with one of its is 90° is a rectangle. Hence, ABCD is a rectangle.
Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD. Hence, ar (ABD) = ar(ABC) [∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.] Subtracting ar (ABO) from both of sides ar(ABD) - ar(ABO) = ar(ABC) - ar(ABO) ⇒ ar(AOD) = ar(BRead more
Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD.
Hence, ar (ABD) = ar(ABC)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar (ABO) from both of sides
ar(ABD) – ar(ABO) = ar(ABC) – ar(ABO)
⇒ ar(AOD) = ar(BOC)
Let ABCD be the Itwaari's plot. Join BD and Throught C draw a line CF parallel to BD which meet AB Produced at F. Now join D and F. Triangles CBD and FBD are on the same base BD and between same parallels BD ∥ CF. Hence, ar(CBD = ar(FBD) [∵ Triangles on the same base (or equal bases) and between theRead more
Let ABCD be the Itwaari’s plot.
Join BD and Throught C draw a line CF parallel to BD which meet AB Produced at F.
Now join D and F.
Triangles CBD and FBD are on the same base BD and between same parallels BD ∥ CF.
Hence, ar(CBD = ar(FBD)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar(BDM) from both the sides
ar(CBD) – ar(BDM) = ar(FBD) – ar(BDM)
⇒ ar(CMD) = ar(BFM)
Hence, in place of ΔCMD, if ΔBFM be given to Itwaari, his plot become triangular (ΔADF).
Construction: Join CX. Triangles ADX and ACX are on the same base AX and between same parallels AB ∥ DC. Hence, ar(ADX) = ar(ACX) ...(1) [∵ Triangles on the same base (or equal bases) and between same parallels are equal in area.] Similarly, triangles ACY and ACX are on the same base AC and betweenRead more
Construction: Join CX.
Triangles ADX and ACX are on the same base AX and between same parallels AB ∥ DC.
Hence, ar(ADX) = ar(ACX) …(1)
[∵ Triangles on the same base (or equal bases) and between same parallels are equal in area.]
Similarly, triangles ACY and ACX are on the same base AC and between same parallels AC ∥ XY.
Hence, ar(ACY) = ar(ACX) …(2)
From the equation (1) and (2), ar(ADX) = ar(ACY).
Let the first angle = 3x Therefore, the cecond angle = 5x, Third angle = 9x and Fourth angle = 13x Sum of all angles of a Quadrilateral is 360°. Therefore, 3x + 5x + 9x + 13x = 360° ⇒ 30x = 360° ⇒ x = 360°/30 = 12° Hence, the first angle = 3 × 12° = 36°, The second angle = 5 × 12° = 60°, Third angleRead more
Let the first angle = 3x
Therefore, the cecond angle = 5x,
Third angle = 9x and
Fourth angle = 13x
Sum of all angles of a Quadrilateral is 360°. Therefore, 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360° ⇒ x = 360°/30 = 12°
Hence, the first angle = 3 × 12° = 36°,
The second angle = 5 × 12° = 60°,
Third angle = 9 × 12° = 108°
The forth angle = 13 × 12 = 156°
In ΔABC, AD is median. [∵ Given] Hence, ar(ABD) = ar(ACD) ⇒ ar(ABD) = 1/2 ar(ABC) ...(1) [∵ A median of a triangle divides it into two triangles of equal areas.] Similarly, in ΔABD, BE is medium. [∵ E is the mid-point of AD] Hence, ar(BED) = ar (ABE) ⇒ ar(BED) = 1/2 ar(ABD) ⇒ ar (BED) = 1/2[(1/2)ar(Read more
In ΔABC, AD is median. [∵ Given]
Hence, ar(ABD) = ar(ACD)
⇒ ar(ABD) = 1/2 ar(ABC) …(1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔABD, BE is medium. [∵ E is the mid-point of AD]
Hence, ar(BED) = ar (ABE)
⇒ ar(BED) = 1/2 ar(ABD)
⇒ ar (BED) = 1/2[(1/2)ar(ABC)] [∵ ar (ABD) = (1/2)ar(ABC)]
⇒ ar (BED) = 1/4 ar(ABC)
Diagonals of paralllelogram bisect each other. therefore, PO = OR and SO = OQ In ΔPQS, PO is median. [∵ SO = OQ] Hence, ar (PSO) = ar(PQO) ...(1) [∵ A median of a triangle divides it into two triangles of equal areas.] Similarly, in ΔPQR, QO is median [∵ PO = OR] Hence, ar(PQO) = ar(QRO) ...(2) AndRead more
Diagonals of paralllelogram bisect each other.
therefore, PO = OR and SO = OQ
In ΔPQS, PO is median. [∵ SO = OQ]
Hence, ar (PSO) = ar(PQO) …(1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔPQR, QO is median [∵ PO = OR]
Hence, ar(PQO) = ar(QRO) …(2)
And in ΔQRS, RO is median. [∵ SO = OQ]
Hence, ar (QRO) = ar((RSO) …(3)
From the equations (1), (2) and (3), we get
ar (PSO) = ar(PQO) = ar(QRO) = ar(RSO)
Hence, in parallelogram PQRS, diagonals PR and QS Divide it into four triangles in equal area.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a rhombus. Sulution: In ΔAOB and ΔAOD, BO = DO [∵ Given] ∠AOB = ∠AOD [∵ Each 90°] AO = AO [∵ Common] Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule] AB = AD [∵ CPCT] Similarly, AB = BC and BC = CD Now, all the fourRead more
Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°.
See lessTo prove: ABCD is a rhombus.
Sulution: In ΔAOB and ΔAOD,
BO = DO [∵ Given]
∠AOB = ∠AOD [∵ Each 90°]
AO = AO [∵ Common]
Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule]
AB = AD [∵ CPCT]
Similarly, AB = BC and BC = CD
Now, all the four sides of Quadrilateral ABCD are equal.
Hence, ABCD is a rhombus.
Show that the diagonals of a square are equal and bisect each other at right angles.
Given: ABCD is a square. To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°. Solution: ΔBAD and ΔABC, AD = BC [∵ Opposite sides of a square] ∠BAD = ∠ABC [∵ Each 90°] AB = AB [∵ Common] Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule] BD = AC [∵ CPCT] In ΔAOB and ΔCOD, ∠OAB = ∠OCD [∵ Alternate angles] AB =Read more
Given: ABCD is a square.
See lessTo prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°.
Solution: ΔBAD and ΔABC,
AD = BC [∵ Opposite sides of a square]
∠BAD = ∠ABC [∵ Each 90°]
AB = AB [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule]
BD = AC [∵ CPCT]
In ΔAOB and ΔCOD,
∠OAB = ∠OCD [∵ Alternate angles]
AB = CD [∵ Opposite sides of a square]
∠OBA = ∠ODC [∵ Alternate angles]
Hence, ΔBAD ≅ ΔABC [∵ ASA Congruency rule]
AO = OC, BO = OD, [∵ CPCT]
In ΔAOB and ΔAOD.
OB = OD [∵ Proved above]
AB = AD [∵ Sides of a square]
OA = OA [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SSS Congruency rule]
∠AOB = ∠AOD [∵ CPCT]
But. ∠AOB + ∠AOD = 180° [∵ Linear pair]
⇒ 2∠AOB = 180° [∵ ∠AOD = ∠AOB]
⇒ ∠AOB = (180/2) = 90°
Hence, the diagonals of a square are equal and bisect each at right angles.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a square. Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus . Hence, AB = BC = CD = DA In ΔBAD and ΔABC, AD = BC [∵ Proved above] BD = AC [∵ Given] AB =Read more
Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°.
See lessTo prove: ABCD is a square.
Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus .
Hence, AB = BC = CD = DA
In ΔBAD and ΔABC,
AD = BC [∵ Proved above]
BD = AC [∵ Given]
AB = AB [∵ Common]
Hence, ΔBAD ΔABC [∵ SSS Congruency rule]
∠BAD = ∠ABC [∵ CPCT]
But, ∠BAD + ∠ABC = 180° [∵ Co-interior angles]
⇒ 2∠ABC = 180° [∵ ∠BAD = ∠ABC]
⇒ ∠ABC = (180°/2) = 90°
Hence, if the diagonals of a quadrilateral are equal and bisect each other at right angles, than it is a square.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Given: ABCD is a parallelogram with AC = BD. To Prove: ABCD is a rectangle. Solution: In ΔABC and Δ BAD, BC = AD [∵ Opposite sides of a parallelogram are equal] AC = BD [∵ Given] AB = AB [∵ Common] Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule] ∠ABC = ∠BAD [∵ CPCT] But, ∠ABC + ∠BAD = 180° [∵ Co-interiorRead more
Given: ABCD is a parallelogram with AC = BD.
See lessTo Prove: ABCD is a rectangle.
Solution: In ΔABC and Δ BAD,
BC = AD [∵ Opposite sides of a parallelogram are equal]
AC = BD [∵ Given]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule]
∠ABC = ∠BAD [∵ CPCT]
But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]
⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]
⇒ ∠BAD = 180°/2= 90°
A parallelogram with one of its is 90° is a rectangle. Hence, ABCD is a rectangle.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD. Hence, ar (ABD) = ar(ABC) [∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.] Subtracting ar (ABO) from both of sides ar(ABD) - ar(ABO) = ar(ABC) - ar(ABO) ⇒ ar(AOD) = ar(BRead more
Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD.
See lessHence, ar (ABD) = ar(ABC)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar (ABO) from both of sides
ar(ABD) – ar(ABO) = ar(ABC) – ar(ABO)
⇒ ar(AOD) = ar(BOC)
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Let ABCD be the Itwaari's plot. Join BD and Throught C draw a line CF parallel to BD which meet AB Produced at F. Now join D and F. Triangles CBD and FBD are on the same base BD and between same parallels BD ∥ CF. Hence, ar(CBD = ar(FBD) [∵ Triangles on the same base (or equal bases) and between theRead more
Let ABCD be the Itwaari’s plot.
See lessJoin BD and Throught C draw a line CF parallel to BD which meet AB Produced at F.
Now join D and F.
Triangles CBD and FBD are on the same base BD and between same parallels BD ∥ CF.
Hence, ar(CBD = ar(FBD)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar(BDM) from both the sides
ar(CBD) – ar(BDM) = ar(FBD) – ar(BDM)
⇒ ar(CMD) = ar(BFM)
Hence, in place of ΔCMD, if ΔBFM be given to Itwaari, his plot become triangular (ΔADF).
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
Construction: Join CX. Triangles ADX and ACX are on the same base AX and between same parallels AB ∥ DC. Hence, ar(ADX) = ar(ACX) ...(1) [∵ Triangles on the same base (or equal bases) and between same parallels are equal in area.] Similarly, triangles ACY and ACX are on the same base AC and betweenRead more
Construction: Join CX.
See lessTriangles ADX and ACX are on the same base AX and between same parallels AB ∥ DC.
Hence, ar(ADX) = ar(ACX) …(1)
[∵ Triangles on the same base (or equal bases) and between same parallels are equal in area.]
Similarly, triangles ACY and ACX are on the same base AC and between same parallels AC ∥ XY.
Hence, ar(ACY) = ar(ACX) …(2)
From the equation (1) and (2), ar(ADX) = ar(ACY).
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Let the first angle = 3x Therefore, the cecond angle = 5x, Third angle = 9x and Fourth angle = 13x Sum of all angles of a Quadrilateral is 360°. Therefore, 3x + 5x + 9x + 13x = 360° ⇒ 30x = 360° ⇒ x = 360°/30 = 12° Hence, the first angle = 3 × 12° = 36°, The second angle = 5 × 12° = 60°, Third angleRead more
Let the first angle = 3x
See lessTherefore, the cecond angle = 5x,
Third angle = 9x and
Fourth angle = 13x
Sum of all angles of a Quadrilateral is 360°. Therefore, 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360° ⇒ x = 360°/30 = 12°
Hence, the first angle = 3 × 12° = 36°,
The second angle = 5 × 12° = 60°,
Third angle = 9 × 12° = 108°
The forth angle = 13 × 12 = 156°
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar (ABC).
In ΔABC, AD is median. [∵ Given] Hence, ar(ABD) = ar(ACD) ⇒ ar(ABD) = 1/2 ar(ABC) ...(1) [∵ A median of a triangle divides it into two triangles of equal areas.] Similarly, in ΔABD, BE is medium. [∵ E is the mid-point of AD] Hence, ar(BED) = ar (ABE) ⇒ ar(BED) = 1/2 ar(ABD) ⇒ ar (BED) = 1/2[(1/2)ar(Read more
In ΔABC, AD is median. [∵ Given]
See lessHence, ar(ABD) = ar(ACD)
⇒ ar(ABD) = 1/2 ar(ABC) …(1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔABD, BE is medium. [∵ E is the mid-point of AD]
Hence, ar(BED) = ar (ABE)
⇒ ar(BED) = 1/2 ar(ABD)
⇒ ar (BED) = 1/2[(1/2)ar(ABC)] [∵ ar (ABD) = (1/2)ar(ABC)]
⇒ ar (BED) = 1/4 ar(ABC)
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Diagonals of paralllelogram bisect each other. therefore, PO = OR and SO = OQ In ΔPQS, PO is median. [∵ SO = OQ] Hence, ar (PSO) = ar(PQO) ...(1) [∵ A median of a triangle divides it into two triangles of equal areas.] Similarly, in ΔPQR, QO is median [∵ PO = OR] Hence, ar(PQO) = ar(QRO) ...(2) AndRead more
Diagonals of paralllelogram bisect each other.
See lesstherefore, PO = OR and SO = OQ
In ΔPQS, PO is median. [∵ SO = OQ]
Hence, ar (PSO) = ar(PQO) …(1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔPQR, QO is median [∵ PO = OR]
Hence, ar(PQO) = ar(QRO) …(2)
And in ΔQRS, RO is median. [∵ SO = OQ]
Hence, ar (QRO) = ar((RSO) …(3)
From the equations (1), (2) and (3), we get
ar (PSO) = ar(PQO) = ar(QRO) = ar(RSO)
Hence, in parallelogram PQRS, diagonals PR and QS Divide it into four triangles in equal area.