Question

Find the area of trapezium whose height is 6 and the sum of parallel sides are 10 and 5

Answer 1

Answer:

A=45

A=45a=Base

A=45a=Base 10

A=45a=Base 10b=Base

A=45a=Base 10b=Base 5

A=45a=Base 10b=Base 5h=Height

A=45a=Base 10b=Base 5h=Height 6

A=45a=Base 10b=Base 5h=Height 6Unit Conversion:

A=45a=Base 10b=Base 5h=Height 6Unit Conversion:Solution

A=45a=Base 10b=Base 5h=Height 6Unit Conversion:SolutionA=a+b

A=45a=Base 10b=Base 5h=Height 6Unit Conversion:SolutionA=a+b2h=10+5

A=45a=Base 10b=Base 5h=Height 6Unit Conversion:SolutionA=a+b2h=10+52·6=45

Answer 2

Answer:

Hence, value of area of the trapezium will be 45 sq. units

Step-by-step explanation:

In context to question asked,

We have to determine the one parallel side of trapezium

As per question,

It is given that,

Parallel sides are = 10 and 5 units

Height of trapezium = 6 units

Sum of parallel sides will be = 10 + 5 = 15 units

As we know that,

Area of trapezium

$A = \frac{a+b}{2}\times h\\Where,\:A = Area\\a,b = Parallel\:sides\:of\:trapezium\\h =Height$

So, for determining the value of area of trapezium we will put the value given in question in above formula,

Thus we will get,

$A = \frac{a+b}{2}\times h\\A = \frac{10+5}{2}\times 6 \\=>A = \frac{15}{2}\times 6\\=>A = 15 \times 3 = 45 \:unit^2$