Question

area \of trapezium is 160sq cm .lenghts of parallel sides are in ratio 1:3.If smaller of the parallel sides is 10cm in lenght ,the find the perpendicular disctance between the.

Answer 1

Question :

area \of trapezium is 160sq cm .lenghts of parallel sides are in ratio 1:3.If smaller of the parallel sides is 10cm in lenght ,the find the perpendicular disctance between them.

Answer :

$\sf\pink{given}$

• Area of Trapezium is 160 cm²

• Ratio of length of parallel sides is 1:3

• Smaller of the parallel sides is 10 cm.

$\sf\pink{To\:find}$

• Perpendicular distance or the height of Trapezium = ?

Explanation :

To find this we first need to find the measure of the other parallel side .

• Let the length of other side be "x"

Here ,

$\longrightarrow\bf{\dfrac{1}{3} = \dfrac{10}{x}}$

$\longrightarrow\bf{1\times x = 3\times 10}$

$\longrightarrow\bf\pink{x = 30}$

• Therefore the length of other parallel side is 30 cm

Area of Trapezium = ${\dfrac{1}{2}\times (a+b)\times h}$

here ,

• a and b are the measures of parallel sides that is 10 cm and 30 cm .

• h is the height that we have to find .

$\longrightarrow\bf{160 = \dfrac{1}{2}\times ( 10+30)\times h }$

$\longrightarrow\bf{160 = \dfrac{1}{2}\times 40\times h }$

$\longrightarrow\bf{160 = 20\times h }$

$\longrightarrow\bf{160 = 20h }$

$\longrightarrow\bf{h = \dfrac{160}{20}}$

$\longrightarrow\bf{h = \dfrac{16}{2}}$

$\longrightarrow\bf\pink{h = 8}$

• Therefore the height is 8 cm