THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 – Class IX

  1. Prove the statement of example 4 using the following ideas: Join DF and product it to meet AB produced in G. Show the two parallelograms ABFE and EFCD have equal area using the converse of mid-point theorem.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Data: ABCD is a parallelogram E and F is the mid-points of AD and BC

Respectively.

To prove: Area of parallelogram ABFE = area of | | gm FEDC

Proof: 
In Δ DCF and Δ BFG, 
∠DCF = ∠FBG(alternate angles, DC | | AB)

 

∠DFC = ∠BFG(V.O.A)

 

FC = FB(F is midpoint)

 

⸫  Δ DCF ≈ Δ BFC(ASA postulate)
⸫  DC = BG

 

But DC = AB

 

⸫  AB = BG

 

Now area of parallelogram ABEF = 2 area of BGF (1)

 

(∵Parallelogram and Δ standing on equal base and between same parallel)

Now Area of parallelogram EDCF = 2 area of Δ DCF

But Δ DCF ≈ Δ BGFS

⸫  Area of parallelogram EDCF = 2 area of Δ BGF (2)

From (1) and (2)

Area of parallelogram ABEF = Area of parallelogram EDCF


  1. Prove the converse of the mid-point theorem following the guidelines given below.

Consider a triangle ABC with D as the min-point of AB. Draw DE | | BC to intersect AC in E. let E1 be the mid-point of ac. Use mid-point theorem to get DE1 | | BC and DE1 = 𝐁𝐂/2 . Conclude E = E1 and hence E is the mid-point of AC.

Solution:

PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Given: in Δ ABC, D is the mid-point of AB. DE | | BC

To prove: E is the midpoint of AC

Proof:
StatementReason

 

1. In Δ ABC, D is the mid-point of AB. DE | | BC

 

Given

 

Let E, be the mid-point of AC.

Join DE, D is the mid-point of

AB and E1 is the mid-point of AC

∴ DE1 | | BC

Mid-point theorem parallels.

 

But DE | | BCGiven

 

This is possible only if E and E1 consider (Through a given point, only one line can be drawn | | to be a given line)

∴ E and E1 coincide

i.e. DE1 is the same as DE

Thus a line drawn through the mid-point of a side of a triangle and parallel to another bisects the third side.


3. In a rectangle ABCD, P, Q, R and S are the mid-point of the sides AB, BC, CD and DA respectively. Find the area of PQRS in terms of area of ABCD.

Solution:

PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Given: ABCD is a rectangle, P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To find: the areas (PQRS) in terms of area (ABCD)

Proof: S and Q are the mid-point of AD and BC

∴ SQ divides rectangles ABCD into two rectangles equal in area.

Area (ABQS) = area (SQCD) = 1/2 area (ABCD)

Δ SRQ and parallelogram (rectangles) SQCD stand on the same base SQ and are between the same parallel.

∴ Area Δ SRQ = 1/2 area (SQCD)

Similarly area Δ SPQ = 1/2 area (SABQ)

Adding, area Δ SRQ + area Δ SPQ = 1/2 area (SQCD) + 1/2 area (SABQ)

i.e., area (PQRS) = 1/2 area (ABCD)


4. Suppose x, y and z are the mid-point of the sides PQ, QR and RP. Respectively of a triangle PQR. Prove that XYPZ is a parallelogram.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Given: x, y and z are the mid-point of the sides PQ, QR and RP of Δ PQR, xy and xz are joined.

Prove: XYRZ is a parallelogram.

Proof:
StatementReason

 

In PQR, x, y, and z are the

Mid-point of PQ QR and RP

Respectively.

 

XY | | PRmid-point theorem

 

XY = 1/2 PRmid-point theorem

 

ZR = 1/2 PRGiven

 

Therefore, XY RZ is a Parallelogram.

 

one pair of opposite sides parallel and equal

  1. Suppose EFGH is a trapezium in which EF is parallel to HG. Through X, the mid-point EH, XY is drawn parallel to EF meeting FG at Y. prove that XY bisects FG.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Given: EFGH is a trapezium with EF | | HG. Through X the mid-point of EH,

Is drawn parallel to EF

To prove: XY bisects FG

Construction: Join HF to intersect XY at P

Proof
StatementReason

 

In Δ HEF XP | | EFconverse of mid-point theorem

 

∴ XP bisects HF PY | | HG

 

HY | | EF and EF | | HG

 

Δ HGF, PY passes through mid-point of HF and is parallel to HG

∴ PY bisects FG

converse of mid-point theorem

 


6. Prove that if the mid-points of the opposite sides of a quadrilateral are joined, they bisect each other.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 - Class IX

Data: ABCD is a quadrilateral P, Q, R and S are mid-points of AB, BC, CD and DA respectively.

To prove: PR and SQ bisect each other.

Construction: join AC.

Proof:
StatementReason

 

In Δ ABC, PQ | | AC, and PQ = 12 ACmid-point theorem ……. (1)

 

 

In Δ ADC, SR | | AC, and SR = 12 ACmid-point theorem ……. (2)

 

∴ PQ | | SR and PQ = SRfrom (1) and (2) opposite sides equal and parallel

 

∴PQRS is a parallelogram PR and SQ are diagonals PR bisect SQ

 


 

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