Question

Find the area of a trapezium whose parallel sides 25 cm 13cm and other sides are 15 cm 15 cm

Answer 1

Answer:

The area of the trapezium is $57\sqrt{21} cm^{2}$.

Step-by-step explanation:

Given:

$ABCD$ is a trapezium

$AB = 25 cm\\DC = 13 cm$

$AD$ and $BC=15 cm$

Construction:

Draw $CE || AD$

To Find :

Area of trapezium $ABCD$

Solution :

$ADCE$ is a parallelogram $( AD || CE and AE || CD)$

Therefore $AE = DC = 13 cm$ ( Opposite side of parallelogram are equal)

$BE = AB - AE$

$BE = 25 - 13$

$BE = 12 cm$

In triangle $BCE$ :

$S = a + b +\frac{c}{2}$

$S = 15 + 15 + \frac{12}{2}$

$S = 21$

$Area of BCE = \sqrt{s(s-a)(s-b)(s-c)} \\Area of BCE = \sqrt{21(21-15)(21-15)(21-12)} \\Area of BCE = \sqrt{21*6*6*9}\\Area of BCE = 18\sqrt{21}cm^{2}$---Let this be (i)

h is the height of $BCE$

$Area of BCE = \frac{1}{2} ( Base * Height )\\\\=0.5(12)(h)\\=6h$--- Let this be (ii)

From (i) and (ii) :

$6h = 18\sqrt{21}\\ h = 3\sqrt{21} cm$

The height of trapezium $ABCD$ is equal to height of  $BCE$.

Area of trapezium :

$= \frac{1}{2} ( AB + CD ) h\\= 0.5 (25 + 13) 3\sqrt{21} cm^{2} \\= 57\sqrt{21} cm^{2}$

Hence the area of the trapezium is $57\sqrt{21} cm^{2}$.

Answer 2

Answer:

Concept:

The total amount of space a trapezium occupies in a two-dimensional plane is its area. Simply multiplying the total of the bases by the height and dividing the result by two yields the area of a trapezium.

Step-by-step explanation:

Given:

Parallel sides 25 cm 13cm and other sides are 15 cm 15 cm

Find:

The area of a trapezium

Solution:

Let ABCD be the given trapezium in which AB=25cm,CD=13cm,BC=15cm and AD=15cm.

         $Draw C E \| A D and C F \perp B E .$

Now, ADCE is a parallelogram in which $\mathrm{AD} \| \mathrm{CE} and \mathrm{AE} \| \mathrm{CD} .$

$\mathrm{AE}=\mathrm{DC}=13 \mathrm{~cm} and \mathrm{BE}=\mathrm{AB}-\mathrm{AE}=25-13=12 \mathrm{~cm}$

          $In \triangle \mathrm{BCE} , we have\\\\ \mathrm{s}=\frac{15+15+12}{2}=21 \mathrm{~cm} \\\\ \therefore Area of \triangle \mathrm{BCE}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

$\Rightarrow Area of \Delta \mathrm{BCE}$ =

$\sqrt{21(21-15)(21-15)(21-12)} =\sqrt{21 \times 6 \times 6 \times 9}=18 \sqrt{21} sq cm ....(i)$

$Let \ \mathrm{h} \ be \ the \ height \ of \triangle \mathrm{BCE} , thenArea of \triangle \mathrm{BCE}=\frac{1}{2}( Base \times Height ) =\frac{1}{2} \times 12 \times \mathrm{h}=6 \mathrm{~h} .....(ii)$

$From (i) and (ii), we have,\\\\ 6 \mathrm{~h}=18 \sqrt{21} \Rightarrow \mathrm{h}=3 \sqrt{21} \mathrm{~cm} \\\\Clearly, the height of trapezium \mathrm{ABCD} is same as that of \triangle \mathrm{BCE} .$

$\therefore Area of trapezium =\frac{1}{2}(\mathrm{AB}+\mathrm{CD}) \times \mathrm{h} \\\\ \Rightarrow Area of trapezium =\frac{1}{2}(25+13) \times 3 \sqrt{21} \mathrm{~cm}^{2} \\\\ =57 \sqrt{21} \mathrm{~cm}^{2}$

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