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
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