Question

$\huge\mathfrak\red{Question}$

ln a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm
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Answer 1

Answer:

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL2 = CM2 + LM2 (Pythagoras Theorem)

262 = CM2 + 102

CM2 = 262 – 102

Using algebraic identities, we get; 262 – 102 = (26 – 10) (26 + 10)

hence,

CM2 = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm2

Answer 2

ɢɪᴠᴇɴ:

• ln a trapezium, the parallel sides measure 40 cm and 20 cm.

• Its non-parallel sides are equal having the lengths of 26 cm

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ᴛᴏ ꜰɪɴᴅ:

• Tha area of the trapezium.

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ꜱᴏʟᴜᴛɪᴏɴ:

• Let the Trapezium be ABCD
• Let AB and CD be the parallel sides.

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ꜰɪɢᴜʀᴇ:

• Refer to the attachment !!

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ɴᴏᴡ:

• The distance between the two parallel sides or the height of the trapezium will be CM.

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ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ:

• Area of Trapezium = ½ × Sum of parallel sides × Height

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ʜᴇʀᴇ:

• We have to find the height.

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ɪɴ ᴛʜᴇ ᴅɪᴀɢʀᴀᴍ:

• Draw CL || AD

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ɴᴏᴡ:

• ALCD is a parallelogram → CL = AD = 26 cm

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• *As AD = CB,

• CL = CB → ΔCLB is an isosceles triangle with CB as its height.

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ʜᴇʀᴇ:

$~~~~~~~~~$ $\sf \dashrightarrow$ $\sf BL = AB – AL = (40 – 20) = 20 cm$

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ꜱᴏ:

$~~$ $\sf \dashrightarrow$ $\sf LM = MB = \dfrac{1}{2} BL = \dfrac{1}{2} × 20 = 10 cm$

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ɴᴏᴡ, ɪɴ Δᴄʟᴍ:

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$~$ $\sf \dashrightarrow$ $\sf CL² = CM² + LM² (Pythagoras Theorem)$

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$~~~~~~~~~$ $\sf \dashrightarrow$ $\sf 26² = CM² + 10²$

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$~~~~~~~~~$ $\sf \dashrightarrow$ $\sf CM² = 26² – 10²$

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ᴜꜱɪɴɢ ᴀʟɢᴇʙʀᴀɪᴄ ɪᴅᴇɴᴛɪᴛɪᴇꜱ:

$\sf \dashrightarrow$ $\pmb{\bf{\underline{We \: get, 26² – 10² = (26 – 10) (26 + 10) \: = 16 × 36 = 576}}}$

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$\sf \dashrightarrow$ $\sf CM = √576 = 24 cm$

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ɴᴏᴡ, ᴛʜᴇ ᴀʀᴇᴀ ᴏꜰ ᴛʀᴀᴩᴇᴢɪᴜᴍ ᴄᴀɴ ʙᴇ ᴄᴀʟᴄᴜʟᴀᴛᴇᴅ !!

$\sf \dashrightarrow$ ABCD = $\dfrac{1}{2} × (AB + CD) × CM \:$

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$\sf \dashrightarrow$ $\sf ABCD = \dfrac{1}{2}× (20 + 40) × 24$

Area of Trapezium ABCD $\sf \dashrightarrow$ $\pmb{\bf{\underline{= 720 cm²}}}$

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