Question

8. The perimeter of an isosceles trapezium is 134 cm and the bases are
54 cm and 30 cm in length. Find the length of the nonparallel sides
of the trapezium and its area.​

Answer 1

$\sf{\huge{\underline{\green{Answer :-}}}}$

• The length of the nonparallel sides of the trapezium is 25 cm.

• The area of trapezium is 175 cm² .

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$\sf{\huge{\underline{\pink{Solution :-}}}}$

Given :-

• The perimeter of an isosceles trapezium is 134.

• The bases are 54 cm and 30 cm in length.

To Find :-

• The length of the nonparallel sides of the trapezium.

• The area of trapezium.

Solution :-

Let the length of the nonparallel sides of the trapezium be x.

The perimeter of triangle is = a + b + c + d

➝ x + x + 54 + 30 = 134

➝ 2x + 54 + 30 = 134

➝ 2x + 84 = 134

➝ 2x = 134 - 84

➝ 2x = 50

➝ x = $\cancel{\dfrac{50}{2}}$

➝ x = 25 cm.

Hence, The length of the nonparallel sides of the trapezium is 25 cm.

Then, base = b1 - b2

➝ 54 -30

➝ 24

Hypotaneus = non || side = 25 cm.

Hence,

In right triangle COB, using P.G.T.

H² = P² + B²

➝ 25² = h² + 24²

➝ h² = 25² - 24²

➝ h² = 625 - 576

➝ h² = 49

➝ h = √49

➝ h = 7 cm.

The Area of trapezium = ½ × sum of || sides × Hieght

➝ ½ × ( 25 + 25 ) × 7

➝ ½ × 50 × 7

➝ 25 × 7

➝ 175 cm²

Hence, The area of trapezium 175 cm².

Answer 2

Answer:

25cm and 921.06cm²

Step-by-step explanation:

Let the length of equal parallel sides be x

Perimeter=x+x+(54+30)

134=2x+84

2x=134-84

2x=50

x=25cm

To find the height of trapezium

by Pythagoras theorum

h²+((54-30)/2)= 25²

h²+12²=25²

h²=625-144

h²=√481

h=21.93cm

Area=½×h×sum of parallel sides

=½×21.93×84

= 921.06 cm²

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