Question

11. A trapezium of area 105 cm? has parallel sides of length 5 cm and
9 cm. How far apart are the parallel sides?​

Answer 1

Answer:

Given values

Area of Trapezium = 105 cm²

Parallel sides :-

a = 5 cm

b = 9 cm

Height ( h ) = ?

Area of Trapezium = 1/2 ( a + b ) h = 105

===> 1/2 ( 5 + 9 ) h = 105

===> 1/2 ( 14 ) h = 105

===> 7 × h = 105

===> h = 105 / 7

===> h = 15 cm

Height of the parallel sides is 15 cm

Answer 2

☞ Correct Question :

➝ A trapezium of area 105 cm² , has parallel sides of length 5 cm and 9 cm. How far apart are the parallel sides?

☞ Given :

• Area = $105 cm^{2}$

• Parallel side $\rightarrow P_{1} = 5 cm$

• Parallel side $\rightarrow P_{2} = 9 cm$

☞ To Find :

➝ The Distance between the Parallel sides.

☞ We Know :

➝ Formula for area of a Trapezium .

$\mathtt{A = \dfrac{1}{2} \times sum\:of\:parallel\:sides \times height}$

☞ Concept :

By the given area and the parallel sides , we can find the value of height or distance between the parallel sides.

☞ Solution :

Given :

• Area = $105 cm^{2}$

• Parallel side $\rightarrow P_{1} = 5 cm$

• Parallel side $\rightarrow P_{2} = 9 cm$

From the given Formula :

$\mathtt{A = \dfrac{1}{2} \times (P_{1} + P_{2}) \times h}$

We Get:

$\mathtt{h = \dfrac{2A}{(P_{1} + P_{2})}}$

Putting the value in the formula ,We Get :

$\mathtt{\Rightarrow h = \dfrac{2 \times 105}{5 + 9}}$

$\mathtt{\Rightarrow h = \dfrac{2 \times 105}{14}}$

$\mathtt{\Rightarrow h = \dfrac{\cancel{2} \times 105}{\cancel{14}}}$

$\mathtt{\Rightarrow h = \dfrac{105}{7}}$

$\mathtt{\Rightarrow h = \dfrac{\cancel{105}}{\cancel{7}}}$

$\mathtt{\Rightarrow h = 15 cm}$

Hence , the height or the distance between the Parallel sides is 15 cm.

☞ Extra Information :

• Area of parallelogram = $Height \times base$

• Area of Equilateral triangle = $\dfrac{\sqrt{3}a^{2}}{4}$

• Height of a Equilateral triangle = $\dfrac{\sqrt{3}a}{2}$

• Area of a triangle = $\dfrac{1}{2} \times base \times height$