Question

The parallel sides of a Trapezium are 20cm and 13 cm its non parallel sides are 10 cm each find the area of trapezium​

Answer 1

parallel sides of a trapezium are 20cm and 13cm

non- parallel sides are 10cm

Two parallel sides =AB =13cm and DC= 20cm

Non parallel side = AD= BC = 10cm

construction:- draw parallel line to BC

BC=AE=10cm

Draw another line which perpendicular to DC

Now,

AB=EC=13cm

DE=20-13=7cm

as AF is perpendicular to DC it divide DFand FE equally

so, DF= FE =7/2

DF =3.5cm and FE = 3.5cm

AFE triangle and AFD triangle are right angled triangles

triangle AFE

(AF)^2 +(FE)^2 = (AE)^2

(AF)^2+ (3.5)^2 =(10)^2

(AF)^2 +12.25=100

(AF)^2= 10-12.25

(AF)^2 =87.75

$af = \sqrt{87.55}$

AF= 9.36

height (AF) = 9.3 cm

Area of trapezium = 1/2×(sum of all parallel sides ×height

= 1/2× (13+20)×9.3

= 1/2 × 33 × 9.3

= 33.93 /2

= 306.9/2

= 153.45.......answer

HOPE IT HELP U ^^

THANKS❤

Answer 2

Explanation:

parallel sides of a trapezium are 20cm and 13cm

non- parallel sides are 10cm

Two parallel sides =AB =13cm and DC= 20cm

Non parallel side = AD= BC = 10cm

construction:- draw parallel line to BC

BC=AE=10cm

Draw another line which perpendicular to DC

Now,

AB=EC=13cm

DE=20-13=7cm

as AF is perpendicular to DC it divide DFand FE equally

so, DF= FE =7/2

DF =3.5cm and FE = 3.5cm

AFE triangle and AFD triangle are right angled triangles

triangle AFE

(AF)^2 +(FE)^2 = (AE)^2

(AF)^2+ (3.5)^2 =(10)^2

(AF)^2 +12.25=100

(AF)^2= 10-12.25

(AF)^2 =87.75

af = \sqrt{87.55} af=

87.55

AF= 9.36

height (AF) = 9.3 cm

Area of trapezium = 1/2×(sum of all parallel sides ×height

= 1/2× (13+20)×9.3

= 1/2 × 33 × 9.3

= 33.93 /2

= 306.9/2

= 153.45.......answer

HOPE IT HELP U ^^

THANKS❤