THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

  1.  The area of parallelogram is 153.6 cm2. The base measures 19.2 cm. what is the measurement for the height of the parallelogram?

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Area of the parallelogram = bh = 153 cm2
Base = b = 19.2cm

b x h = 19.2 × h = 153.6
h = 153.6/19.2 = 1536/192 = 8cm


2. In parallelogram ABCD, AD = 25cm and AB = 50 CM. if the altitude from a vertex D on to AB measure 22 cm, what is the altitude from a vertex B on to AD?
Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Area of parallelogram = bh
Area ABCD = 50 × 22cm2
If the altitude from B to AD is h,
And ABCD = 25 × h cm2
25 x h = 50 × 22
\h = 15 ×2225 = 44cmArea of parallelogram = bh
Area ABCD = 50 × 22cm2
If the altitude from B to AD is h,
And ABCD = 25 × h cm2
25 x h = 50 × 22
Therefore, h = (15 ×22)/25 = 44cm


  1. In a parallelogram ABCD, AB = 4x and AD = 2x + 1. If the perimeter is 38cm and area 60cm2 find the length of the altitude from D on to AB.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Perimeter = 2 (l + b) = 38

l + b = 38/2 = 19

4x + 2x + 1 = 19

6x = 19 – 1 = 18

x = 18/6 = 3

AB = 4x = 4 × 3 = 12cm

Let the altitude DE = h

Area = bh = 12 × h = 60

⸫h = 6012 = 5cm 


  1. Let ABCD be a parallelogram and consider its diagonal AC. Draw perpendiculars BK and DL to AC.

Prove that BK = DL.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Given: ABCD is a parallelogram. BK and DL are perpendiculars drawn

To diagonals AC.

Proof:

 

Statement Reason
In quadrilateral.

ABCD DL ^ AC, BK ^ AC

given
ΔADC @ ΔABC Diagonal divides a parallelogram into two congruent triangles
⸫  Area Δ ABC = Area Δ ADC

 

i. e. 1/2 AC × BK = 1/2 AC × DL

 

BK = DL

 


  1. Let ABCD be a parallelogram. Prove that 2 area (ABCD) ≤AC ×  BD.

Solution:

Given: ABCD is a parallelogram

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

 

To prove: 2 areas (ABCD) ≤AC × BD.

Construction: Draw DF and BG perpendicular to AC

Proof:

Area ABCD = area ΔABC + area ΔADC

1/2 AC × DF + 1/2 AC × BG

1/2 AC (DF + BG)

2 area (ABCD) = 2 × 1/2 AC (DF + BG)

= AC (DF + BG)

In right ΔDFE, DF <  hypotenuse DE

Similarly in ΔBGE, BG < BE

Thus 2 area ABCD = AC (DF + BG)

≤ AC (DE + BE)

i.e. 2 areas ABCD ≤ AC × BD

2 areas ABCD will be equal to AC × BD if AC ^ BD AC will be perpendicular to BD if ABCD is a rhombus or a square.

⸫  2 area ABCD = AC × BD in a rhombus or in a square.


6. Prove that the area of triangles standing on the same base or equal bases and between same parallels are equal in area.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Given: ABCD and ΔABF stand on the same base and are between the same

Parallel and m

To prove: area ΔABD = area ΔABF

Construction: drew parallelograms ABCD and ABFE

Proof:
Statement Reason
1.. Area of parallelogram ABCD and ABFE are equal They stand on the same base and are between the same parallels.

 

2. Area ΔABD = 1/2 area ΔABCD Area of triangles is equal to half the area of a parallelogram stand base and between the same parallels
3. Similarly area

 

ΔABF = 12 area ABFE

 

4. Area ΔABD = area ΔABF from (1), (2) and (3)

 

 

 

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