Question

200 केजी आफ सूगर वास परचेस्ड एट द रेट ऑफ रुपीस फिफ्टीन पर केजी एंड सोल्ड अट ए प्रॉफिट आफ 5% कंप्यूट द सेलिंग प्राइस ऑफ शुगर पर केजी​

Answer 1

Answer:

Appropriate Question :-

In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF

and BC || EF. Vertices A, B and C are joined to

vertices D, E and F respectively. Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) Δ ABC ≅ Δ DEF

\large\underline{\sf{Solution-}}

Solution−

Given that,

\begin{gathered}\sf \: AB \: \parallel \:DE \: and \: AB = DE \\ \\ \end{gathered}

AB∥DEandAB=DE

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

\begin{gathered}\sf\implies \bf \: ABED \: is \: a \: parallelogram. \\ \\ \end{gathered}

⟹ABEDisaparallelogram.

\begin{gathered}\sf\implies\sf \: AD\: \parallel \:BE \: \: and \: \: AD = BE - - - (1) \\ \\ \end{gathered}

⟹AD∥BEandAD=BE−−−(1)

Further given that,

\begin{gathered}\sf \: BC\: \parallel \:EF \: and \: BC = EF \\ \\ \end{gathered}

BC∥EFandBC=EF

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

\begin{gathered}\sf\implies \bf \: BEFC \: is \: a \: parallelogram. \\ \\ \end{gathered}

⟹BEFCisaparallelogram.

\begin{gathered}\sf\implies \sf \: BE\: \parallel \:CF \: \: and \: \: BE = CF - - - (2) \\ \\ \end{gathered}

⟹BE∥CFandBE=CF−−−(2)

From equation (1) and (2), we concluded that

\begin{gathered}\sf\implies \bf \: AD\: \parallel \:CF \: \: and \: \: AD = CF \\ \\ \end{gathered}

⟹AD∥CFandAD=CF

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

\begin{gathered}\sf\implies \bf \: ACFD \: is \: a \: parallelogram. \\ \\ \end{gathered}

⟹ACFDisaparallelogram.

\begin{gathered}\sf\implies \sf \: AC\: \parallel \:DF \: \: and \: \: AC = DF \\ \\ \end{gathered}

⟹AC∥DFandAC=DF

Now,

\begin{gathered}\sf \: In \: \triangle \: ABC \: and \: \triangle \: DEF \\ \\ \end{gathered}

In△ABCand△DEF

\begin{gathered}\qquad\sf \: AB = DE \: \: \: \: \{given \} \\ \\ \end{gathered}

AB=DE{given}

\begin{gathered}\qquad\sf \: BC = EF \: \: \: \: \{given \} \\ \\ \end{gathered}

BC=EF{given}

\begin{gathered}\qquad\sf \: AC = DF \: \: \: \: \{proved \: above \} \\ \\ \end{gathered}

AC=DF{provedabove}

\begin{gathered}\sf\implies \bf \: \triangle \: ABC \: \cong \: \triangle \: DEF \: \: \: \{SSS \: Congruency \: rule \} \\ \\ \end{gathered}

⟹△ABC≅△DEF{SSSCongruencyrule}