Question

explain parallelogram law of addition . for 5 mark from 11th std​

Answer 1

Parallelogram law of addition:-

Statement: The sum of squares of length of four sides of parallelogram is equal to sum of squares of lengths of the two diagonals.

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For a parallelogram its opposites are equal.

Let ABCD is a parallelogram then,

→ AD = BC & AB = DC

According to the parallelogram law of addition,

→ 2(AB)² + 2 (BC)² = (AC)² + (BD)²

Proof :-

Let AD=BC = x, AB = DC = y, and ∠ BAD = α.

We know:

Law of cosines,

$\mapsto \sf c=\sqrt{a^2 +b^2-2ab.cos\gamma }$

Where,

=>c =length of side c

=>a=length of side a

=>b=length of side b

=> γ=angle opposite c

Now,

Apply the law of cosines in the Δ BAD

⇒ x² + y² - 2xy cos(α) = BD²......(1)

And,

As  the adjacent angles are supplementary,

∠ADC = 180 – α  

Now,

Apply the law of cosines in the Δ ADC

⇒x² + y² - 2xy cos(180 – α) = AC² ....(2)

We know:

→cos(180 – x) = – cos x

Substituting this in eqn (2) we get:

⇒  x² + y² + 2xy cos(α) = AC²

Now,

⇒BD² + AC²  = x² + y² - 2xycos(α) + x² + y² + 2xy cos(α)

⇒BD² + AC² =2x² + 2y².....(3)

We know AD=BC = x, AB = DC = y,

⇒BD² + AC² = 2(AB)² + 2(BC)²

Hence proved !

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Know more:-

Parallelogram:A parallelogram is a simple quadrilateral with two pairs of parallel sides.

• Area: base × height
• Perimeter: 2 x (sum of lengths of adjacent sides)

Parallelogram law of vector addition:-

If two vectors are  adjacent sides of a parallelogram, the resultant of the two vectors is given by the vector which is diagonal passing through the point of contact of two vectors.

Let  A and B are the two vectors which are adjacent sides of a parallelogram. let the parallelogram is OPTQ. Then the resultant is,

$\mapsto \sf \bar{R}=\bar{A}+\bar{B}$

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