Question

The parallel sides of a trapezium are 40cm and 16cm. Each of its non - parallel sides is 13cm ,find the area of the trapezium

Answer 1

Answer:

Let ABCD be the trapezium.

In which AD||BC, and AB=DCBC=40cm and AD= 16cm.

let us draw a line DE parallel to AB which will form the parallelogram ABED with DE = 13cm

Now, in the triangle formed DEC if we draw a perpendicular DF, it will bisect the rest (24cm) base in two halves as DE=AB=DC.

In the right angled triangle DFC formed, we know that (DF)^2+(FC)^2=(DC)^2.

=>(DF)^2=13cm^2- 12cm^2(1/2 of 24cm)

=>DF^2 =169cm^2-144cm^2

=>DF^2=25cm^2

=>DF=5cm

And we know that area of a trapezium is 1/2 ×sum if ||sides ×h

=1/2×(16+24)cm×5cm

=40cm/2×5cm

= 100cm^2.

Hope it will help you

Answer 2

$\huge\mathbb{SOLUTION:-}$

$\bf{Given:-}\begin{cases}\sf{Parallel\:sides\: of\: trapezium=40cm\:\:and\:16cm}\\ \sf{Non\: parallel\:sides=13cm}\end{cases}$

$\huge\star{\sf{\red{To\:Find:-}}}$

• The area of trapezium

$\bf\underline{According\:To\: Question:-}$

• First find the height of trapezium

• Check the attachment ! !
• The parallel sides of trapezium becomes 16 cm and rest is 26 cm and the both sides of trapezium are equal then
• it becomes 12 cm
• Check the attachment

$\sf \leadsto height =\sqrt{13{}^{2}-12{}^{2}}\\ \\ \sf\leadsto height=\sqrt{169-144}\\ \\ \sf\leadsto height=\sqrt{25}\\ \\ \sf\leadsto height=5cm$

$\boxed{\star{\sf{\purple{Area\: of\: trapezium=\frac{1}{2}\times (sum\: of\: parallel\:sides)\times height}}}}$

$\sf\implies Area=\frac{1}{2}\times (40+16)\times 5\\ \\ \sf\implies Area=\frac{1}{\cancel2}\times \cancel{56}\times 5\\ \\ \sf\implies Area=28\times 5\\ \\ \sf\implies Area=140cm{}^{2}$

$\underline{\boxed{\sf{\purple{Area\:of\: trapezium=140cm{}^{2}}}}}$

#BAL

#answerwithquality