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Equality of area of geometric shapes

Theorem. Two parallelograms are equal in area if. Equality of area of geometric shapes. 1) They have a Common base . 2) Between two parallel lines one of the one of them carry this base. E. C. D. F. B. A. Area of parallelogram (ABCD) = Area of parallelogram ( ABEF ) . Corollary.

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Equality of area of geometric shapes

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  1. Theorem Two parallelograms are equal in area if Equality of area of geometric shapes 1) They have a Common base 2) Between two parallel lines one of the one of them carry this base E C D F B A Area of parallelogram (ABCD) =Area of parallelogram (ABEF)

  2. Corollary parallelograms and rectangle are equal in area if 1) They have a Common base 2) Between two parallel lines one of the one of them carry this base E C D F B A Area of parallelogram (ABCD) =Area of rectangle (ABEF)

  3. Corollary parallelograms with equal bases and between two parallel lines Are equal in area H G C D E B F A Area of parallelogram (ABCD) =Area of parallelogram (EFGH)

  4. Solved Examples Solved Examples Solved Examples Solved Examples Solved Examples Solved Examples Solved Examples

  5. A B X Y 1) p. (74) no (1) Using the opposite figure: Find the area of the parallelogram ABED. 24 cm D E 12 cm Proof: // , is a common base Between the parallelogram ABED and the rectangle XYED Area (ABED) = Area (XYED) A(XYED)= 24 x 12 = 288cm2 The area of the parallelogram (ABED) = 288cm2

  6. Theorem Two triangles are equal in area if they have a common base and their vertices lies on a straight line parallel to this common base Area of Δ (ABC) =Area of Δ(ABD) D C Two triangles are equal in area if they have equal bases and their vertices lies on a straight line parallel to the straight line carrying these two bases B A Z C Area of Δ (ABC) =Area of Δ (XYZ) X B A Y

  7. The median of triangle divides The triangle into two triangles equal in area A C D B Area of Δ (ABD) =Area of Δ (ACD)

  8. A 2) In the opposite figure: D, E are the mid points of and respectively ∩ = {M}, Prove that: The area of Δ(DMB) = The area of Δ(EMC) Proof: Since E, D are the midpoints of and respectively Therefore, // Since is a common base between the triangle (DBC) and the triangle (EBC) Therefore area of triangle (DBC) = area of triangle (EBC) By subtracting the area of triangle (EMD)from both sides we get, Area of triangle (DMB) = Area of triangle (EMC) D E M C B

  9. 3) p. (125) no (2) By using the opposite figure prove that: 1) Area of Δ(AMB) = area of Δ(DMC) 2) Area of the shape (ABXM) = area of the shape (DCXM) Proof: Since // , is a common base between Δ(ABC) and Δ(DBC) Therefore, area of Δ(ABC) = area of Δ(DBC) By subtracting the area of the Δ(MBC) from both sides we get: Area of Δ(AMB) = area of Δ(DMC)…………….. (1) Since is a median in triangle (MBC) Therefore, area of Δ(MBX) = area of Δ(MCX)…………… (2) By adding (1) and (2) we get Area of the shape (ABXM) = area of the shape (DCXM) A D M X B C

  10. A 4) P. (126) no (8) By using the opposite figure prove that: Area of triangle (ABY) = area of triangle (ACY) Proof: Since is a median in triangle (ABC) Therefore, area of Δ(AXB) = area of Δ(AXC)…….. (1) Since is a median in triangle (YBC) Therefore, area of Δ(YXB) = area of Δ (YXC)…………….. (2) By subtracting (2) from (1) we get Area of triangle (ABY) = area of triangle (ACY) Y C X B

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