calculate area of the trapeziums , lengths of whose parallel sides a and b and distance h between them are given below :
(i) a=2.5 , b=1.8 dm ,h= 9 cm
(ii) a = 4 m , b=25 dm,h=60 cm.
Answers
Here we will learn how to use the formula to find the area of trapezium.
Area of trapezium ABCD = Area of ∆ ABD + Area of ∆ CBD
= 1/2 × a × h + 1/2 × b × h
= 1/2 × h × (a + b)
= 1/2 (sum of parallel sides) × (perpendicular distance between them)
Area of trapezium
9Save
Worked-out examples on area of trapezium
1. The length of the parallel sides of a trapezium are in the rat: 3 : 2 and the distance between them is 10 cm. If the area of trapezium is 325 cm², find the length of the parallel sides.
Solution:
Let the common ration be x,
Then the two parallel sides are 3x, 2x
Distance between them = 10 cm
Area of trapezium = 325 cm²
Area of trapezium = 1/2 (p₁ + p₂) h
325 = 1/2 (3x + 2x) 10
⇒ 325 = 5x × 5
⇒ 325 = 25x
⇒ x = 325/25
Therefore, 3x = 3 × 13 = 39 and 2x = 2 × 13 = 26
Therefore, the length of parallel sides area are 26 cm and 39 cm.
2. ABCD is a trapezium in which AB ∥ CD, AD ⊥ DC, AB = 20 cm, BC = 13 cm and DC = 25 cm. Find the area of the trapezium.
find the area of trapezium
9Save
Solution:
From B draw BP perpendicular DC
Therefore, AB = DP = 20 cm
So, PC = DC - DP
= (25 - 20) cm
= 5 cm
Now, area of trapezium ABCD = Area of rectangle ABPD + Area of △ BPC
△BPC is right angled at ∠BPC
Therefore, using Pythagoras theorem,
BC² = BP² + PC²
13² = BP² + 5²
⇒ 169 = BP² + 25
⇒ 169 - 25 = BP²
⇒ 144 = BP²
⇒ BP = 12
Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ∆BPC
= AB × BP + 1/2 × PC × BP
= 20 × 12 + 1/2 × 5 × 12
= 240 + 30
= 270 cm²
3. Find the area of a trapezium whose parallel sides are AB = 12 cm, CD = 36 cm and the non-parallel sides are BC = 15 cm and AG = 15 cm.
examples on area of trapezium
9Save
Solution:
In trapezium ABCD, draw CE ∥ DA.
Now CE = 15 cm
Since, DC = 12 cm so, AE = 12 cm
Also, EB = AB - AE = 36 - 12 = 24 cm
Now, in ∆ EBC
S = (15 + 15 + 24)/2
= 54/2
= 27
= √(27 × 12 × 12 × 3)
= √(3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 3 × 3)
= 3 × 3 × 3 × 2 × 2
= 108 cm²
Draw CP ⊥ EB.
Area of ∆EBC = 1/2 × EB × CP
108 = 1/2 × 24 × CP
108/12 = CP
⇒ CP = 9 cm Therefore, h = 9 cm
Now, area of triangle = √(s(s - a) (s - b) (s - c))
= √(27 (27 - 15) (27 - 15 ) (27 - 24))
Now, area of trapezium = 1/2(p₁ + p₂) × h
= 1/2 × 48 × 9
= 216 cm²
4. The area of a trapezium is 165 cm² and its height is 10 cm. If one of the parallel sides is double of the other, find the two parallel sides.
Solution:
Let one side of trapezium is x, then other side parallel to it = 2x
Area of trapezium = 165 cm²
Height of trapezium = 10 cm
Now, area of trapezium = 1/2 (p₁ + p₂) × h
⇒ 165 = 1/2(x₁ + 2x) × 10
⇒ 165 = 3x × 5
⇒ 165 = 15x
⇒ x = 165/15
⇒ x = 11
Therefore, 2x = 2 × 11 = 22
Therefore, the two parallel sides are of length 11 cm and 22 cm.
These are the above examples explained step by step to calculate the area of trapezium.
plz make me brainliest answer
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
∴ Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
∴ Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m2
Plz mark as brainliest