Question

find the hieght of a trapezium in which parallel sides are 25cm adn 77cm and nonparallel side are 25cmand 60cm. given the area of trapezium as 1644cm^2

Answer 1

Answer:

CF=DF−CD

=(77−60)cm

=17cm

For △BCF

Permiter of triangle △BCF=BC+BF+CF

=(25+26+17)cm

=68cm

Semiperimeter of the triangle (s)=

2

P

=

2

68

=34cm

By Heron's formula

Area of triangle △BCF=

S(s−BC)(s−BF)(s−CF)

=

34(34−25)(34−26)(34−17)

=

34×9×8×17

=

41616

=204cm

2

let h cm be the length of perpendicular to CF

Area of BCF=

2

1

×CF×BE

⇒ 204=

2

1

×17×h

⇒ h=

14

204×2

=24cm

Area trapezium of ABCD=

2

1

[(60+77)×24]=

2

1

(137×29)=

2

1

×3288=1644

solution