Question

In the adjoining figure, area of parallelogram ABCD is :

• A. AB \(\times\) BC
• B. BC \(\times\) BN
• C. DC \(\times\) DL
• D. AD \(\times\) DL

Hint:

Area of a parallelogram = Base \(\times\) Height.
The "Base" can be any side.
The "Height" must be the {perpendicular} distance to that chosen base from the opposite side. In the figure:
If you choose base = AB (or DC), the corresponding height is DL. So, Area = AB \(\times\) DL or DC \(\times\) DL.
If you choose base = AD (or BC), the corresponding height would be a perpendicular dropped from B or D onto AD (or from A or C onto BC). Option (3) DC \(\times\) DL correctly uses a base (DC) and its corresponding height (DL).

Answer 1

The Correct Option is C

Concept: The area of a parallelogram is given by the formula: Area = Base \(\times\) Corresponding Height. The "base" can be any side of the parallelogram, and the "corresponding height" is the perpendicular distance from that base to the opposite side. Step 1: Identify potential bases and corresponding heights from the figure From the provided figure for parallelogram ABCD:
If we consider side AB (or its equal opposite side DC) as the base.
The perpendicular distance from vertex D to the base AB (or its extension) is shown as DL. DL is the altitude (height) corresponding to base AB or DC.
If we consider side BC (or its equal opposite side AD) as the base.
The perpendicular distance from vertex D to the base BC (or its extension) would be a different height (let's call it DM, not explicitly shown as such but BN is related to height on AD if drawn from B to AD extended). The line segment BN in the diagram appears to be the perpendicular from C to the extension of AB, which would be a height if AB is the base. However, DL is clearly marked as an altitude to AB (or from D to AB). Step 2: Apply the area formula with identified base and height Using Base = DC and Corresponding Height = DL: Area of parallelogram ABCD = DC \(\times\) DL. Since AB is parallel and equal to DC (opposite sides of a parallelogram), we could also write: Area of parallelogram ABCD = AB \(\times\) DL. Using Base = AD and a corresponding height (let's call it \(h_{AD}\)) perpendicular to AD from B or C: Area of parallelogram ABCD = AD \(\times\) \(h_{AD}\). Similarly, using Base = BC and a corresponding height (let's call it \(h_{BC}\)) perpendicular to BC from A or D: Area of parallelogram ABCD = BC \(\times\) \(h_{BC}\). Step 3: Evaluate the given options based on the diagram
(1) AB \(\times\) BC: This is the product of two adjacent sides. This is not the formula for the area of a parallelogram unless it's a rectangle and these are length and width (in which case BC would be the height corresponding to base AB). For a general parallelogram, Area = \(AB \times BC \times \sin(\angle \text{ABC})\).
(2) BC \(\times\) BN: In the diagram, BN appears to be the perpendicular from C to the line AB extended. If this is the case, then BN is the height corresponding to base AB (or DC). So, Area = AB \(\times\) BN or DC \(\times\) BN. The option uses BC as the base. If BN were the height corresponding to base BC, this would be correct, but BN is shown as perpendicular to AB extended.
(3) DC \(\times\) DL: Here, DC is a base, and DL is shown as the perpendicular height from D to the line containing AB. Since AB is parallel to DC, DL is also the height corresponding to base DC (it's the perpendicular distance between the parallel lines AB and DC). This formula is correct: Base \(\times\) Height.
(4) AD \(\times\) DL: Here, AD is a side. DL is the height corresponding to base AB or DC, not directly to AD, unless \(\angle \text{DAB}\) is \(90^\circ\) and AD becomes the height for base DL (which is not how it's defined) or if DL also happens to be perpendicular to AD (making AD and AB perpendicular, i.e. a rectangle, where DL would be AB or DC). This is generally incorrect. Step 4: Conclusion Based on the standard formula (Base \(\times\) Corresponding Height) and the diagram: If DC is taken as the base, then DL is the corresponding perpendicular height. Thus, Area = DC \(\times\) DL. This matches option (3).