Question

In the given figure find the area of trapezium ABCD

Answer 1

Answer:

48 cm²

Step-by-step explanation:

In the given trapezium ABCD, AB=9 cm, BC=AD=5 cm. ...... (1)

AL and BM are the perpendiculars on DC from points A and B respectively.

Given that, AL=BM=4 cm. ....... (2)

Now it is clear that ΔALD and ΔBMC are two right-angled triangles having ∠ALD and ∠BMC as right angles or 90°.

So, Applying Pythagoras's theorem in ΔALD and ΔBMC we get,

AL²+LD²=AD² and BM²+MC²=BC²

⇒LD²=AD²-AL² and MC²=BC²-BM²

Applying equations (1) and (2)

⇒LD²=5²-4²  and  MC²= 5²-4²

⇒LD²=9 and MC²=9

⇒LD=3 and MC=3 ....... (3)

Now, ABML is a rectangle having AB=LM .....(4) and AB║ML

Hence, AB║DC and DC= DL+LM+MC= 3+9+3= 15 cm. {From (3) and (4)}...... (5)

Now, the perpendicular distance from AB to DC is AL= BM= 4 cm.

Therefore, The area of the trapezium ABCD  

=$\frac{1}{2}(AB+DC).AL$

=$\frac{1}{2}(9+15).4$ {From (1), (2), and (5)}

=$24*2$

=48 cm²

(Answer)