Question

In the figure below, ABCD is a trapezium in
which AE = 3 cm, EB = 5 cm, BC = 5 cm and
3.
AE is one and a half times as long as DC,
Find
(a) the length of DC,
(b) the ratio of the area of AECD to that of
ABCD.​

Answer 1

Given:

ABCD is a trapezium in which AE = 3 cm, EB = 5 cm, BC = 5 cm  

AE is one and a half times as long as DC

To find:

The length of DC

The ratio of the area of AECD to that of ABCD.​

Solution:

Finding the length of DC:

AE is one and a half times as long as DC

i.e., $AE = 1\frac{1}{2}DC$

$\implies AE= \frac{3}{2}DC$

$\implies DC= \frac{2}{3}AE$

$\implies DC= \frac{2}{3}\times 3$

$\implies DC= 2\:cm$

Thus, the length of DC is → 2 cm.

Finding the ratio of the area of AECD to that of the area of ABCD:

The area of trapezium ABCD is,

= $\frac{1}{2}$ × [sum of lengths of parallel sides] × [perpendicular distance between parallel sides]

= $\frac{1}{2} \times (DC + AB) \times CB$

= $\frac{1}{2} \times (2 + 8) \times 5$

= 25 cm²

The area of Δ EBC is,

= $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × EB × CB

= $\frac{1}{2}$ × 5 × 5

= 12.5 cm²

∴ The area of AECD = Ar(trapezium ABCD) - Ar(Δ EBC) = 25 - 12.5 = 12.5 cm²

∴ The ratio of the area of AECD to that of ABCD = $\frac{12.5}{25} = \frac{1}{2}$

Thus, the ratio of the area of AECD to that of ABCD is → 1 : 2.

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In fig ABCD is a trapezium with AB parallel to DC . If triangle AED is similar to triangle BEC. Prove that AD = BC

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