Question

find the area of trapezium whose parallel sides are 4 cm and 7 cm and distance between parallel lines is 5 cm .(please jaldi batao)​

Answer 1

Answer:

27.5 cm2

Step-by-step explanation:

a = 4cm

b = 7cm

h = 5 cm

Area of trapezium = 1/2*(a+b)*h

Therefore area =

1/2 *(4+7)*5

= 27.5 cm2

Answer 2

Correct Question :-

Find the area of trapezium whose parallel sides are 4cm and 7cm and distance between parallel lines is 5cm.

Answer :-

The area of trapezium is 27.5cm².

Step-by-step explanation :-

To Find,

• The area of trapezium

Solution,

Figure,

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.5mm}\qbezier(0,0)(0,0)(1,3)\qbezier(5,0)(5,0)(4,3)\qbezier(1,3)(1,3)(4,3)\qbezier(3,0)(8.2,0)(0,0)\put(2,-0.5){ \large{\sf 7cm }}\put(2,3.4){ \sf \: 4cm }\put(1,0){\line(0,0){3}}\put(1,1.2){ \sf \: 5cm }\end{picture}$

Given that,

• Parallel sides of trapezium = 4cm and 7cm
• Distance between them = 5cm

As we know that,

$\dag \: \boxed{ \sf{Area \: of \: trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times distance \: between \: them}}$

Therefore,

$\sf{\mapsto \: \dfrac{1}{2} \times ( \: 4 \: + \: 7 \: ) \times 5} \\ \\ \\ \sf{\mapsto \: \dfrac{1}{2} \times 11 \times 5} \\ \\ \\ \sf{\mapsto \: \dfrac{1}{ \cancel{2}} \times \cancel{11} \times 5} \\ \\ \\ \sf{ \mapsto \: 1 \times 5.5 \times 5} \\ \\ \\ \sf{ \mapsto \: 5.5 \times 7} \\ \\ \\ \frak{ \mapsto \: 27.5} \: \: \: \: \: \: \: \bigstar$

Hence, area of trapezium is 27.5cm².