Question

The parallel sides of a trapezium are 25cm and 11cm, while its nonparallel sides are 15cm and 13cm. Find the area of the trapezium​

Answer 1

Answer:

216 cm²

Step-by-step explanation:

let ABCD be the trapezium where AB =11 cm BC=13 cm CD= 25 cm DA= 15 cm

then, drop perpendicular from A and B respectively on CD at points M and N

therefore, MN= 11 cm

let length DM = x cm and NC = y cm and height AM=BN= h cm

now consider two right angled triangles so formed i.e. ΔAMD and ΔBNC

so x +y +11=25 ⇒ x + y = 14

now apply pythagoras theorem on both the right angled triangles,

this will give 15²=h²+x² and 13²= h²+y²

subtract these eqns

x²-y²=15²-13²

now using a²-b²= (a+b)(a-b)

we get, x- y = 4

therefore, y = 5 cm

substitute this to any of the two eqns,

we will get h = 12 cm

Now,

area of trapezium is given by

1/2 (sum of parallel sides)x height

=1/2 x (11+ 25)x 12 cm²

=216 cm²

Answer 2

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

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

• The parallel sides of a trapezium are 25cm and 11cm, while its nonparallel sides are 15cm and 13cm.

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

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

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

Let ABCD be the trapezium is which AB // DC, AB = 25cm, CA = 11cm, AD = 13cm and BC = 15cm

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

AMCD is a parallelogram.

Now,

⟹ MC = AD = 13cm

⟹ AM = DC = 11cm

⟹ MB = (AB - Am)

⟹ MB = (25 - 11)

⟹ MB = 14cm

Thus,

In ∆CMB, we have;

• CM = 13cm
• MB = 14cm
• BC = 15cm

S = 1/2 (13 + 14 + 15)

S = 1/2 × 42

S = 21cm

(s - a) = (21 - 13) = 8cm

(s - b) = (21 - 14) = 7cm

(s - c) = (21 - 15) = 6cm

Area of∆CMB = √s(s - a)(s - b)(s -c)

⟹ √21 × 8 × 7 × 6

⟹ √168 × 42

⟹ √7066

⟹ 84cm²

⟹ 1/2 × MB × CL = 84

⟹ 1/2 × 14 × CL = 84

⟹ 7 × CL = 84

⟹ CL = 84/7

⟹ CL = 12cm

Area of trapezium = 1l2 × (AB = DC) × CL

⟹ 1/2 × (25 + 11) × 12

⟹ 1/2 × 36 × 12

⟹ 18 × 12

⟹ 216cm²

Hence, the area of the trapezium is 216cm².