## Area of Trapezium

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)

**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 cm*^{2}, 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 cm^{2}

Area of trapezium = 1/2 (p_{1} + p_{2}) 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.*

**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,

BC^{2} = BP^{2} + PC^{2}

13^{2} = BP^{2} + 5^{2}

⇒ 169 = BP^{2} + 25

⇒ 169 - 25 = BP^{2}

⇒ 144 = BP^{2}

⇒ 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 cm^{2}

**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. *

**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 cm^{2}

Draw CP ⊥ EB.

Area of ∆EBC = 1/2 × EB × CP

108 = 1/2 × 24 × CP

108/12 = CP

⇒ CP = 9 cmTherefore, 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(p_{1} + p_{2}) × h

= 1/2 × 48 × 9

= 216 cm^{2}

**4.*** The area of a trapezium is 165 cm*^{2} 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 cm^{2}

Height of trapezium = 10 cm

Now, area of trapezium = 1/2 (p_{1} + p_{2}) × h

⇒ 165 = 1/2(x_{1} + 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.

**● Mensuration**

**Area and Perimeter**

**Perimeter and Area of Rectangle**

**Perimeter and Area of Square**

**Area of the Path**

**Area and Perimeter of the Triangle**

**Area and Perimeter of the Parallelogram**

**Area and Perimeter of Rhombus**

**Area of Trapezium**

**Circumference and Area of Circle**

**Units of Area Conversion**

**Practice Test on Area and Perimeter of Rectangle**

**Practice Test on Area and Perimeter of Square**

**● Mensuration - Worksheets****Worksheet on Area and Perimeter of Rectangles**

**Worksheet on Area and Perimeter of Squares**

**Worksheet on Area of the Path**

**Worksheet on Circumference and Area of Circle**

**Worksheet on Area and Perimeter of Triangle**

7th Grade Math Problems

8th Grade Math Practice

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