ABCD is an isoceles trapezium such that AD||BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm 2) of trapezium?
44
In isosceles trapezium ABCD with AD∥BC, drop perpendiculars from A and D to BC at E and F. Then EF=AD=8 and BE=CF.
Step 1 — Find BE:
\[BE=\frac{BC-AD}{2}=\frac{14-8}{2}=3\text{ cm}\]
Step 2 — Height by Pythagoras in ▵ABE:
\[h=\sqrt{AB^2-BE^2}=\sqrt{25-9}=4\text{ cm}\]
Step 3 — Area:
\[\text{Area}=\tfrac{1}{2}(AD+BC)\times h=\tfrac{1}{2}(8+14)(4)=44\text{ cm}^2\]
Therefore the area is 44 cm².
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