Math, asked by ninjamri76, 9 months ago

find area of trapezium abcd in which ab parallel to cd ,ab=77cm,bc=25cm,cd=60cm and da=26cm with diagram

Answers

Answered by sauhardyachakrpe6ont
4

Answer:

1712.5 cm²

Step-by-step explanation:

Area of trapezium

= (Sum of parallel sides) × height/2

= (77 + 60) × 25/2 = 137 × 25/2

= 3425/2 = 1712.5 cm²

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Answered by Anonymous
21

GIVEN:-

 \sf•ABCD \: is \: a \: trapezium \: in \: which \: AB \parallel DC , \\

 \sf AB=77cm,BC=25cm,CD =60cm  \\ \sf and  \: DA=26cm \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\

TO FIND OUT:-

• \sf Area  \: of \: trapezium  \: ABCD=? \\  \\

SOLUTION:-

 \sf Draw \: DE \parallel  BC \: and \: DL \perp AB \\

 \sf Then,DE=BE=25cm. \\  \\

 \sf AE=(AB-EB)=(AB-DC)=(77-60)cm \\  \implies \sf \: AE=17cm \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\

 \sf In   \: \triangle  DAE,we \: have: \\  \\

 \sf  \longrightarrow a=AE=17cm \\  \sf \longrightarrow b=DE=25cm \\  \longrightarrow  \sf c=DA=26cm \\  \\

 \sf \therefore s= \frac{1}{2}(17+25+26)cm =34cm\\  \\

 \sf \longrightarrow (s-a)=17cm, \\  \longrightarrow  \sf(s-b)=9cm \: ,  \:  \\  \longrightarrow  \sf(s-c)=8cm \:  \:  \:  \\  \\

 \sf \therefore Area  \: of \: \triangle DAE= \sqrt{s(s-a)(s-b)(s-c)}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:      \sf= \sqrt{34×17×9×8}cm² \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =\sf(17×3×4)cm²  \\  \sf=204cm² \:  \\  \\

 \sf Also \: area \: of \:  \triangle DAE = \frac{1}{2}×AE×DL \:  \:  \: sq.unit \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf= \frac{1}{2}  \times 17 \times DL  \: cm² \\  \\

 \sf \therefore  \frac{1}{2} ×17×DL=204  \:  \:  \:  \:  \: \\   \implies \sf DL= \bigg( \frac{204×2}{17} \bigg)cm \\  \implies \sf DL=24 cm² \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\

 \sf \therefore Area \: of \: trapezium \: ABCD= \frac{1}{2} (AB+DC)×DL \:  \:  \: sq.unit \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = \sf  \bigg[ \frac{1}{2} (77+60)×24 \bigg]cm² \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf =1644cm²

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