Respuesta :
Given :
- Base = 20 cm and 14 cm.
- Height = 5 cm.
To find :
- Area of trapezoid.
Solution :
We know,
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}= \dfrac{1}{2 } \times (b_{1} + b_{2}) \times h} }}[/tex]
Now, Substituting the values :
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}= \dfrac{1}{2 } \times (20 + 14) \times 5} }}[/tex]
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}= \dfrac{5}{2 } \times 34} }}[/tex]
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}= \dfrac{5}{ \cancel2 } \times \cancel{34}} }}[/tex]
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}=5 \times 17} }}[/tex]
[tex]{\qquad \dashrightarrow{ \bf{Area_{(Trapezoid)}=85} }}[/tex]
Therefore,
- The area of the trapezoid is 85 cm² .
Answer:
85cm²
Step-by-step 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 :-
[tex]\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} [/tex]
As we know that the area of trapezium is given by ,
[tex]\longrightarrow\rm Area =\dfrac{1}{2}(sum \ of \ || \ sides)\times height [/tex]
- Here 20cm and 14cm are parallel sides .
Substitute the respective values in stated formula,
[tex]\longrightarrow\rm Area =\dfrac{1}{2}(20cm +14cm)5cm \\[/tex]
Solve the parenthesis ,
[tex]\longrightarrow\rm Area =\dfrac{1}{2}\times (34cm)(5cm)[/tex]
Simplify by multiplying ,
[tex]\longrightarrow\rm \underline{\underline{\red{{Area = 85\ cm^2}}}}[/tex]
And we are done !
[tex]\rule{200}4[/tex]
