Question

a trapezium shaped field whise parallel sides 42m 30m and other sides are 18m and 18m find its area

Answer 1

Answer:

432 $\sqrt{2}$ cm²

Step-by-step explanation:

Let the 4 sides of Trapezium to be A, B, C and D. Draw perpendiculars from A and B on the side CD of the trapezium at points E and F.

Length of EF would be equal to AB

DE and FC would be 6 cm each.

Apply Pythagoras Theorem in Δ ADE:-

(AD)² = (AE)² + (ED)²

(18)² = (AE)² + (6)²

324 = (AE)² + 36

(AE)² = 288

AE = $\sqrt{288}$

AE = 12 $\sqrt{2}$ cm

So, the height of the triangle is 12 $\sqrt{2}$ cm.

Find the area of both the triangles:-

= $\frac{1}{2}$ x b x h

= $\frac{1}{2}$ x 6 x 12 $\sqrt{2}$

= 36 $\sqrt{2}$ cm²

Area of the rectangle ABFE would be:-

= l x b

= 30 x 12 $\sqrt{2}$

= 360 $\sqrt{2}$ cm²

Area of Trapezium would be:-

= Area of 2 triangles + Area of rectangle

= 2(36 $\sqrt{2}$ ) +  360 $\sqrt{2}$

= 72 $\sqrt{2}$ + 360 $\sqrt{2}$

= 432 $\sqrt{2}$ cm²

So, the area of the Trapezium is 432 $\sqrt{2}$ cm² !

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Answer 2

Answer:

432 $\sqrt{2}$ cm²

Step-by-step explanation:

Let the 4 sides of Trapezium to be A, B, C and D. Draw perpendiculars from A and B on the side CD of the trapezium at points E and F.

Length of EF would be equal to AB

DE and FC would be 6 cm each.

Applying Pythagoras Theorem in Δ ADE:-

(AD)² = (AE)² + (ED)²

(18)² = (AE)² + (6)²

324 = (AE)² + 36

(AE)² = 288

AE = $\sqrt{288}$

AE = 12 $\sqrt{2}$ cm

So, the height of the triangle is 12 $\sqrt{2}$ cm.

Area of both the triangles:-

= $\frac{1}{2}$ x b x h

= $\frac{1}{2}$ x 6 x 12 $\sqrt{2}$

= 36 $\sqrt{2}$ cm²

Area of the rectangle ABFE would be:-

= l x b

= 30 x 12 $\sqrt{2}$

$= 360 \sqrt{2}$ cm²

Area of Trapezium would be:-

= Area of 2 triangles + Area of rectangle

$= 2(36 \sqrt{2}$ $) + 360 \sqrt{2}$

$= 72 \sqrt{2}$ $+ 360 \sqrt{2}$

= 432 $\sqrt{2}$ cm²