Question

The parallel sides of a trapezium are 20cm and 10cm. Its nonparallel sides are both equal, each being 13cm. Find the area of the trapezium.

Answer 1

ABCD be the given trapezium in which AB = 20 cm, DC = AE= 10 cm, BC = 13 cm and AD = 13cm.


Through C, draw CE || AD, meeting AB at E.

Draw CF ⊥ AB.

Now, EB = (AB - AE) = (AB - DC)

EB = (20- 10) cm = 10 cm;

CE = AD = 13 cm; AE = DC = 13 cm.

Now, in ∆EBC, we have CE = BC = 13 cm.

It is an isosceles triangle.

Also, CF ⊥ AB

So, F is the midpoint of EB.

Therefore, EF = ¹/₂ × EB = 1/2× 10= 5cm.

Thus, in right-angled ∆CFE, we have CE = 13 cm, EF = 5 cm.

By Pythagoras’ theorem, we have

CF = [√CE² - EF²]

CF = √(13² - 5²)

CF= √169-25= √144 = √12×12

CF= 12cm

Thus, the distance between the parallel sides is 12 cm.

Area of trapezium ABCD = ¹/₂ × (sum of parallel sides) × (distance between them)

Area of trapezium ABCD = ¹/₂ × (20 + 10) × 12 cm²

Area of trapezium ABCD = 1/2×(30)× 12

Area of trapezium ABCD= 30×6 = 180 cm²

Hence, Area of trapezium ABCD= 180 cm²

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Answer 2

$\large\underline\pink{Diagram:-}$

$\large\underline\pink{Given:-}$

• The parallel sides of a trapezium are 20cm and 10cm.
• Its nonparallel sides are both equal, each being 13cm.

$\large\underline\pink{To find:-}$

• Find the area of the trapezium.....?

$\large\underline\pink{Solutions:-}$

AB // DC, AB = 20cm, DC = 10cm, and AD = BC = 13cm.

Draw CL ┻ AB and CM // DA meeting AB at L and M.

AMCD is a parallelogram.

Now,

⟹ AM = DC = 10cm

⟹ MB = (AB - AM)

⟹ MB = (20 - 10)

⟹ MB = 10cm

Also,

CM = DA = 13cm

Therefore, ∆CMB is an Isosceles triangle and CL ┻ MB.

L is the mid - point of B.

ML = LB = 1/2 × MB

⟹ 1/2 × 10

⟹ 5cm

from right ∆CLM, we have:-

⟹ CL² = CM² - ML²

⟹ CL² = (13)² - (5)²

⟹ CL² = 169 - 25

⟹ CL² = 144

⟹ CL = √144

⟹ CL = 12cm

Length of CL = 12cm.

Area of the trapezium = 1/2 × (AB + DC) × CL

⟹ 1/2 × (20 + 10) × 12

⟹ 1/2 × 30 × 12

⟹ 15 × 12

⟹ 180cm²

Hence, the area of the trapezium is 180cm².