Math, asked by yashkhera4, 3 months ago

8. In a trapezium - shaped field, one of the parallel sides is twice the other. If the area of the
field is 9450 m² and the perpendicular distance between the two parallel sides is 84 m
the length of the longer of the parallel sides.

9. The length of the fence of a trapezium-shaped field ABCD is 130 m
and side AB is perpendicular to each of the parallel sides AD
and BC. If BC = 54 m. CD = 19 m and AD = 42 m, find the area of
the field.

10. In the given figure. ABCD is a trapezium in which AD || BC.
ZABC = 90°, AD = 16 cm. AC = 41 cm and BC = 40 cm. Find the
area of the trapezium.

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

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

Answers

Answered by mvpatagar21
0

Answer:

8.

Let one side be a=x

Another side be b=2x

The height is given as 84m

The area is 9450m ²

1/2(a+b)h=9450m ²

3x(84)=18900

x=18900/3×84

x=75

So the length of longer side is 150 m

9.

Perimeter of trapezium = 130 m.

Perimeter = AB + BC+ CD +DA

= AB+ 54 +19+42

130=AB + 115

AB = 130 - 115

height = 15 m.

Area = 1/2 ×( 96 ) × 15

= 48 × 15

Ans=720 m. Sq.

10.

The area of the given trapezium ABCD = 252 cm²

For better understanding of the solution see the attached figure of the problem :

Given : m∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm

To find : Area of trapezium ABCD

Solution : Since, m∠ABC is 90°

So, AB is the height of the trapezium ABCD

Now, to find AB : By using Pythagoras theorem in ΔABC

AC² = AB² + BC²

41² = AB² + 40²

⇒ AB² = 41² - 40²

⇒ AB = 9 cm

Now, Area of the trapezium is given by the formula :

Area= 1/2 ×Height×(Sum of parallel sides)

⟹ Area= 1/2 ×AB×(AD+BC)

⟹ Area= 1 /2×(9)×(40+16)

⟹ Area= 1/2 ×(9)×56

⟹ Area=9×28

Area=252cm ²

Hence, The area of the given trapezium ABCD = 252 cm²

11.

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

12.

Draw CE such that AECD is a parallelogram

AE = CD = 11 cm

AD = CE = 15 cm

EB = 14 cm

Area of ΔECB = where s = (a+b+c)/2

a = 13 cm, b = 14 cm, c = 15 cm

Here s = (13+14+15)/2 = 21 cm

Area of ΔECB =

=

= 84 sq cm

But Area of ΔECB = 1/2 × base (EB) × height (h)

1/2 × 14 × h = 84

h = 12 cm

Area of trapezium ABCD = 1/2 × h × (AB + CD)

= 1/2 × 12 × (25 + 11)

= 6 × 36

= 216 sq cm

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