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Areas of Parallelograms and Triangles — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsAreas of Parallelograms and Triangles1 Mark
MCQ
In the given figure ABCD is a trapezium such that $\text{AL}\perp\text{DC}$ and $\text{BM}\perp\text{DC}.$ If AB = 7cm, BC =AD = 5cm and AL = BM = 4cm, then ar(trapezium ABCD) = ?

  • A
    $24cm^2$
     
  • ✓
    $40cm^2$
     
  • C
    $55cm^2$
     
  • D
    $27.5cm^2$
✓

Answer

Correct option: B.
$40cm^2$
 
$40cm^2$
In right $\triangle BMC$,
By pythagoras theorem,
$BC^2=BM^2+C CM^2$
$\Rightarrow CM^2=BC^2-BM^2$
$\Rightarrow CM^2=5^2-4^2$
$\Rightarrow CM^2=25-16$
$\Rightarrow CM^2=9$
$\Rightarrow CM=3 cm$
In right $\triangle ALD$,
By pythagoras theorem,
$A D^2=A L^2+D L^2$
$\Rightarrow D L^2=A D^2-A L^2$
$\Rightarrow D L^2=5^2-4^2$
$\Rightarrow D L^2=25-16$
$\Rightarrow D L^2=9$
$\Rightarrow D L=3 cm$
Since AMML forms a rectangle, the opposite sides are equal.
So, $A B=L M=7 cm$
Area of a trapezium $=\frac{1}{2} \times$ (Sum of the parallel sides) $\times$ Distance between the parallel sides
$=\frac{1}{2} \times(AB \times CD) \times AL$
$=\frac{1}{2} \times(3+7+3) \times 4$
$=40 cm^2$

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