Question

Two parallel side of an isosceles trapezium are 10 CM and 20 CM and its non-parallel side are each equal to 13 cm . Find the area of trapezium​

Answer 1

Area of the trapezium is 180 cm².

Step-by-step explanation:

The figure is drawn below,

From the figure, it is clear that the lengths CD and EF are each equal to 5 cm due to symmetry of the figure.

Consider Δ BCD which is a right angled triangle.

Using Pythagoras Theorem,

$BC^2=CD^2+BD^2\\13^2=5^2+BD^2\\169=25+BD^2\\BD^2=169-25\\BD^2=144\\BD=\sqrt{144}=12\ cm$

Now, area of trapezium ABCF is given as:

$Area=\frac{1}{2}\times (\textrm{Sum of parallel sides}\times \textrm{Height}\\Area=\frac{1}{2}\times (10+20)\times 12\\Area=\frac{1}{2}\times 30\times 12=15\times 12=180\ cm^2$

Therefore, the area of the trapezium with parallel sides 10 cm and 20 cm is 180 cm².

Answer 2

Area of trapezium=$180 cm^2$

Step-by-step explanation:

In a trapezium ABCD, where AB=20cm and CD=10 cm are two parallel sides, and the non parallel sides are 13cm each .Also AB=AE+EF+FB=5cm+10cm+5cm=20cm

Construction: Construct an altitude DE and CF  perpendicular to AB.

Now, In right angle ΔAED,

$[/tex]</p><p>By applying Pythagoras theorem,</p><p></p><p>[tex](13)^2=(DE)^2+(5)^2$

$169=(DE)^2+25$

$(DE)^2=169-25$

            $(DE)^2 =144 cm$

DE=12cm

Altitude=12cm

Now, $Area of trapezium=1/2 \times (AB+CD)\times height$

                                           $=1/2\times (20+10) \times 12$

                                           $=180 cm^2$