Question

ꜰɪɴᴅ ᴛʜᴇ ᴀʀᴇᴀ ᴏꜰ ᴛʀᴀᴘᴇᴢɪᴜᴍ ᴀʙᴄᴅ ɢɪᴠᴇɴ ɪɴ ᴛʜᴇ ꜰɪɢᴜʀᴇ ʙᴇʟᴏᴡ.​

Answer 1

Answer:

72cm^2

Step-by-step explanation:

Let BE=6cm

In triangle(BEC),

BE^2+EC^2=BC^2

Therefore,

6^2+EC^2=10^2

=>EC^2=100-36

=>EC^2=64

=>EC=8cm (because EC cannot be -8cm)

Area of trapezium= 1/2X(sum of parallel sides)X(height)

                              =1/2X(12+6)X8

                              =1/2X18X8

                              =9X8

                              =72cm^2

Answer 2

GiveN :-

• 1st parallel side ( a ) = 6 cm

• 2nd parallel side ( b ) = 12 cm

To FinD :-

• Area of the trapezium

SolutioN :-

Firstly we have to find perpendicular height of the trapezium.

In triangle hypotenuse and base is given , therefore

$\longrightarrow \boxed{ \sf \red{ H^{2} = P^{2} + B^{2}}} \\ \\ \longrightarrow \sf {(10)}^{2} = P^{2} + {(6)}^{2} \\ \\\longrightarrow \sf 100 =P^{2} + 36 \\ \\\longrightarrow \sf P^{2} = 100 - 36 \\ \\\longrightarrow \sf P = \sqrt{64} \\ \\\longrightarrow \boxed{\sf P = 8 \: cm}$

Height of the trapezium is 8 cm

Now area of the trapezium is :

$: \implies\boxed{ \sf \blue{ Area = \frac{a + b}{2}h}} \\ \\: \implies \sf Area = \frac{6 + 12}{2} \times 8 \\ \\: \implies \sf Area = \frac{18}{2} \times 8 \\ \\: \implies \sf Area =9 \times 8 \\ \\ : \implies \boxed{\sf Area =72 \: {cm}^{2}}$

$\therefore \large \underline{ \bf \green{ Area \: of \: trapezium \: is \: 72 \: {cm}^{2} }}$