Question

Hloo♡
*solve this*
In 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.​

Answer 1

Your answer is given below-:::::

The parallel sides of an isosceles trapezium = 40 cm and 20 cm.

The equal sides = 26 cm.

The distance between the parallel sides = [26^2-{(40–20)/2}^2]^0.5

= [676–10^2]^0.5

= (676–100)^0.5

= 576^0.5

= 24 cm.

Area of trapezium = (40+20)*24/2 = 720 sq cm.

Your answer will be 720 sq . cm

Answer 2

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$\huge \tt {SOLUTION:}$

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From the question statement draw the diagram.

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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.

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We know,

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

(✓)So, height has to be found.

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In the diagram, draw CL || AD

Now, ALCD is a parallelogram

⇒ AL = CD = 20 cm and CL = AD = 26 cm

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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

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

hence,

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

➡CM = √576 = 24 cm

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Now, the area of trapezium can be calculated.

➡Area of trapezium, ABCD = ½ × (AB + CD) × CM×½ × (20 + 40) × 24

Area of trapezium ABCD = 720 cm²

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$\tt {Answer=720~cm^2}$

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