Question

Find the area of a trapezium whose parallel sides are of lengths 33cm12cmand whose non parallel sides are of lengths 17cmand 10cm​

Answer 1

Answer:

$\huge\underline\mathfrak\red{question}$

find the area of a trapezium whose parallel sides are of length 33 cm and 12 cm and whose non parallel sides are of length 17 cm and 10 cm.

$\huge\underline\mathfrak\blue{Answer}$

=> give area of trapezium:-

$= \frac{1}{2} (a + b) \times h$

now;-

=> draw a perpendicular from the vertex A at CD name the point as X and draw another perpendicular at CD from the vertex B and name the point as Y

==> we got two triangle and one rectangle.

=> give:-

=> base of the triangle = difference of the trapezium.

=> base of ∆ = CD-AB

=> base of ∆= 33-12 = 21 cm

find the height of∆

by using Pythagoras theorem

AX²=AD²+DX²

AX²=17²+21²

AX²=730²

AX²=√730=21.01 cm

therefore, height= 27.01 cm.

now:-

=> find the area of trapezium

=> using formula

=½(a+b)×h

= ½(12+33)27.01

= ½(45)27.01

= area of trapezium= 607.91 cm²

Answer 2

$\huge\underline\mathfrak\red{question} </p><p></p><p>$

find the area of a trapezium whose parallel sides are of length 33 cm and 12 cm and whose non parallel sides are of length 17 cm and 10 cm.

$\huge\underline\mathfrak\blue{Answer}$

=> give area of trapezium:-

$\frac{1}{2} (a + b) \times h=$

2

1

(a+b)×h

now;-

=> draw a perpendicular from the vertex A at CD name the point as X and draw another perpendicular at CD from the vertex B and name the point as Y

==> we got two triangle and one rectangle.

=> give:-

=> base of the triangle = difference of the trapezium.

=> base of ∆ = CD-AB

=> base of ∆= 33-12 = 21 cm

find the height of∆

by using Pythagoras theorem

AX²=AD²+DX²

AX²=17²+21²

AX²=730²

AX²=√730=21.01 cm

therefore, height= 27.01 cm.

now:-

=> find the area of trapezium

=> using formula

=½(a+b)×h

= ½(12+33)27.01

= ½(45)27.01

= area of trapezium= 607.91 cm²