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.
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.