2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).
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Step-by-step explanation:
Given
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Prove
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)Construction
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.Proof
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BC
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BCAH = BF and AH || BF ( H and F are the mid-points of AD and BC)
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BCAH = BF and AH || BF ( H and F are the mid-points of AD and BC)Therefore, ABFH is a parallelogram.
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BCAH = BF and AH || BF ( H and F are the mid-points of AD and BC)Therefore, ABFH is a parallelogram.Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BCAH = BF and AH || BF ( H and F are the mid-points of AD and BC)Therefore, ABFH is a parallelogram.Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,∴ Area (ΔHEF) = 1/2Area (ABFH) … (1)
GivenE, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.To Provear (EFGH) = ½ ar(ABCD)ConstructionLet us join HF.ProofIn parallelogram ABCD,AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)AB = CD (Opposite sides of a parallelogram are equal)1/2 AD = 1/2 BCAH = BF and AH || BF ( H and F are the mid-points of AD and BC)Therefore, ABFH is a parallelogram.Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,∴ Area (ΔHEF) = 1/2Area (ABFH) … (1)Similarly, it can be proved that
On adding equations (1) and (2), we get
On adding equations (1) and (2), we getarea of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD
On adding equations (1) and (2), we getarea of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD⇒ area of EFGH = area of ABFH
On adding equations (1) and (2), we getarea of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD⇒ area of EFGH = area of ABFH∴ ar (EFGH) = ½ ar(ABCD)
Answer:
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