Question

A point O inside a rectangle ABCD is joined to the vertices . prove that the sum of the areaa of a pair of opposite triangles so formed is equal to the sum of the areas of the other pair of triangles.​

Answer 1

Answer:

I

Step-by-step explanation:

A point O inside a rectangle ABCD is joined to the vertices . prove that the sum of the areaa of a pair of opposite triangles so formed is equal to the sum of the areas of the other pair of triangles.​

Answer 2

$\huge\underline{\underline{\texttt{{Question:}}}}$

• A point O inside a rectangle ABCD is joined to the vertices . prove that the sum of the areaa of a pair of opposite triangles so formed is equal to the sum of the areas of the other pair of triangles.

$\huge\underline{\underline{\texttt{{Given:}}}}$

• $\small\bold{a \: rect \: ABCD \: and \: O \: is \: a \: point \: inside \: in \: it. \: OA, \: OB, \: OC, \: and \: OD \: have \: been \: joined}$

$\huge\underline{\underline{\texttt{{To\:Prove:}}}}$

• $\small\bold{ar(\triangle \: AOD) + ar(\triangle \: BOC) = ar(\triangle \: AOB) + ar(\triangle \: COD).}$

$\huge\underline{\underline{\texttt{{Construction:}}}}$

• $\small\bold{draw \: EOF \: || \: AB \: and \: LOM || \: AD. }$

$\huge\underline{\underline{\texttt{{Proof:}}}}$

$\small\bold{EOF \: || \:AB \: and \: DA \: cuts \: them.}$

$\small\bold{\therefore \: \angle \: DEO \: = \: \angle \: EAB = 90\degree \: \: }$

$\small\bold{\therefore \: OE \: \bot \: AD.}$

Similarly, OF ⊥ BC ; OL ⊥ AB and OM ⊥ DC.

$\small\bold{\therefore \: ar(\triangle \: AOD) + ar(\triangle \: BOC)}$

$\small\bold{ = ( \frac{1}{2} \times AD \times OE) + ( \frac{1}{2} \times BC \times OF) }$

$\small\bold{ = \frac{1}{2}AD \times (OE + OF) \: \: (\therefore \: BC = AD) }$

$\small\bold{ = \frac{1}{2} \times AD \times EF = \frac{1}{2} \times AD \times AB \: \: \: (\therefore \: EF = AB) }$

$\small\bold{ = \frac{1}{2} \times ar(rect. \: ABCD) }$

$\small\bold\purple{again \: ar(\triangle \: AOB) + ar(\triangle \: COD)}$

$\small\bold{ = ( \frac{1}{2} \times AB \times OL) + ( \frac{1}{2} \times DC \times \: OM) = \frac{1}{2} \times Ab \times (OL + OM) \: \: \: (\therefore \: DC = AB) }$

$\small\bold{ = \frac{1}{2} \times AB \times LM = \frac{1}{2} \times AB \times AD \: \: \: \: (\therefore \: LM = AD)}$

$\small\bold{ = \frac{1}{2} \times ar(rect. \: ABCD).}$

$\small\bold\purple{\therefore \: ar(\triangle \: AOD) + ar(\triangle \: BOC) = ar(\triangle \: AOB) + ar(\triangle \: COD).}$

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