In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm,

Question:

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD) = ?

(a) $24 \mathrm{~cm}^{2}$

(b) $40 \mathrm{~cm}^{2}$

(c) $55 \mathrm{~cm}^{2}$

(d) $27.5 \mathrm{~cm}^{2}$

 

Solution:

(b) $40 \mathrm{~cm}^{2}$

In right angled triangle MBC, we have:

$M C=\sqrt{5^{2}-4^{2}}=\sqrt{9}=3 \mathrm{~cm}$

In right angled triangle ADL, we have:

$D L=\sqrt{5^{2}-4^{2}}=\sqrt{9}=3 \mathrm{~cm}$

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm

$\therefore$ Area of the trapezium $=\frac{1}{2} \times$ (sum of parallel sides) $\times$ distance between them

$=\frac{1}{2} \times(13+7) \times 4$

$=40 \mathrm{~cm}^{2}$

 

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