Question

find the area of the trapezium PART whose sides PA=12cm PT=AR=10cm and TR=22 CM ALSO PA ||TR​

Answer 1

Answer:

area is 128cm² hope it will help you

Answer 2

The area of the trapezium PART is $85\sqrt{3} cm^2$.

Step-by-step explanation:

Given:

Here PART is a trapezium.

PA = 12cm, PT = AR = 10cm and TR = 22.

Also PA║TR

Let PC and AD be the perpendiculars drawn to the side TR

Now CT as shown in the figure below,

Here CT = DR = TR - CD

                       = 22 - 12                  [shown in figure]

                       = 10 cm

∴ TC = DR = 5 cm

Using pythogrs theorem,

In ΔPCT,

$PT^2=PC^2+CT^2$

$PC^2= 10^2-5^2$

$PC^2=100-25$

$PC^2=75$

$PC=\sqrt{75}$

$PC= 5\sqrt{3} =AD$                 [shown in fugue]

Area of the trapezium PART = Area of ΔPTC + area of ΔADR + area of PACD

                                               

= $\frac{1}{2} \times base \times height + \frac{1}{2} \times base \times height +base \times height$

= $\frac{1}{2} \times CT \times PC + \frac{1}{2} \times DR \times AD +PA\times AD$

= $\frac{1}{2} \times 5 \times 5\sqrt{3} + \frac{1}{2} \times 5 \times 5\sqrt{3} + 12\times 5\sqrt{3}$

= $5\sqrt{3} [\frac{5}{2}+\frac{5}{2}+12 ]$

= $5\sqrt{3}[\frac{10}{2}+12 ]$

= $5\sqrt{3}[5+12]$

= $5\sqrt{3} \times 17$

= $85\sqrt{3} cm^2$

Therefore the area of the trapezium PART is $85\sqrt{3} cm^2$.