Question

the parallel sides of the trapezium are 20cm and 10cm . its non parallel side are both equal , each being 13cm . find the area of the trapezium.

Answer 1

 ❤Here is your solutions ❤

Given :-

The parallel sides of the trapezium are AD and BE ( 20cm and 10cm)

its non parallel side are both equal , each being 13cm .

Now

Using Pythagoras theorem we will find height of trapezium.(height = DE = AB)

    H^2  = p^2 + b^2

in right angle triangle  FED.

DF^2 = EF^2 + DE^2

13^2 = 5^2 + DE^2

169 - 25 = DE^2

144 = DE^2

12 = DE

so,

height of trapezium  DE = AB  = 12 cm.

Area of trapezium  = 1/2 × (sum of parallel sides ) × height

=>1/2 × (20+10) × 12

=>1/2 × 30×12

=>1/2 × 360

=>180cm^2

Hence

Area of trapezium  = 180 cm^2

Hope it helps you

Answer 2

Solutions :-


Given :
ABCD is the given trapezium.
AB = 20 cm
DC = AE = 10 cm
BC = 13 cm
AD = 13 cm


Draw CF ⊥ AB.

Now,
EB = (AB - AE) = (AB - DC)
= (20- 10) cm = 10 cm


In ∆EBC, we have
CE = BC = 13 cm

EF = ½ × EB
= ½ × 10 = 5 cm

In right-angled ∆CFE, we have
CE = 13 cm, EF = 5 cm

By Pythagoras theorem, we have

CF = √(CE² - EF²)
= √(13² - 5²)
= √(169 - 25)
= √144 = 12 cm


Find the area of trapezium ABCD :-

Area of trapezium = ½ × (sum of parallel sides) × height square unit
= ½ × (20 + 10) × 12 cm²
= ½ × 30 × 12 cm²
= 15 × 12 cm²
= 180 cm²


Hence,
Area of trapezium = 180 cm²