 The area of parallelogram is 153.6 cm^{2}. The base measures 19.2 cm. what is the measurement for the height of the parallelogram?
Solution:
Area of the parallelogram = bh = 153 cm^{2}
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:
Area of parallelogram = bh
Area ABCD = 50 × 22cm^{2}
If the altitude from B to AD is h,
And ABCD = 25 × h cm^{2}
25 x h = 50 × 22
\h = 15 ×2225 = 44cmArea of parallelogram = bh
Area ABCD = 50 × 22cm^{2}
If the altitude from B to AD is h,
And ABCD = 25 × h cm^{2}
25 x h = 50 × 22
Therefore, h = ^{(15 ×22)}/_{25} = 44cm
 In a parallelogram ABCD, AB = 4x and AD = 2x + 1. If the perimeter is 38cm and area 60cm^{2} find the length of the altitude from D on to AB.
Solution:
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
 Let ABCD be a parallelogram and consider its diagonal AC. Draw perpendiculars BK and DL to AC.
Prove that BK = DL.
Solution:
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
 
 Let ABCD be a parallelogram. Prove that 2 area (ABCD) ≤AC × BD.
Solution:
Given: ABCD is a parallelogram
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:
Given: ABCD and ΔABF stand on the same base and are between the same
Parallel l 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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