Question

3. Find the area of a trapezoid with bases 20 cm and 14 cm and height 5 cm.

Answer 1

Given :

• Base = 20 cm and 14 cm.
• Height = 5 cm.

To find :

• Area of trapezoid.

Solution :

We know,

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))= (1)/(2 ) * (b_(1) + b_(2)) * h} }}`

Now, Substituting the values :

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))= (1)/(2 ) * (20 + 14) * 5} }}`

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))= (5)/(2 ) * 34} }}`

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))= (5)/( \cancel2 ) * \cancel{34}} }}`

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))=5 * 17} }}`

`{\qquad \dashrightarrow{ \bf{Area_((Trapezoid))=85} }}`

Therefore,

• The area of the trapezoid is 85 cm² .

Answer 2

85cm²

Explanation:

Here we are given a trapezium with base = 20cm and 14cm and height = 5cm . And we are interested in finding the area of the trapezium .

Figure :-

`\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){$\bf A $}\put(3,2.4){$\bf D $}\put(-0.3,-0.3){$\bf B$}\put(4,-0.3){$\bf C$}\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){$\bf 5\ cm$}\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){$\bf 20\ cm $}\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){$\bf 14\ cm $}\end{picture}`

As we know that the area of trapezium is given by ,

`\longrightarrow\rm Area =(1)/(2)(sum \ of \ || \ sides)* height`

• Here 20cm and 14cm are parallel sides .

Substitute the respective values in stated formula,

`\longrightarrow\rm Area =(1)/(2)(20cm +14cm)5cm \\`

Solve the parenthesis ,

`\longrightarrow\rm Area =(1)/(2)* (34cm)(5cm)`

Simplify by multiplying ,

`\longrightarrow\rm \underline{\underline{\red{{Area = 85\ cm^2}}}}`

And we are done !

`\rule{200}4`