Question

The area of a trapezium is 262.5sq.cm and the perpendicular distance between its parallel sides is 15 cm.what is the sum of its parallel sides?

Answer 1

Solution :-

Area of the trapezium = 262.5 cm²

Perpendicular distance between two parallel sides h = 15 cm

Sum of parallel sides (a + b) = ?

By using area of the trapezium formula

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

⇒ 262.5 = 1/2 * h * (a + b)

⇒ 262.5 * 2 = 15 * (a + b)

⇒ 525 = 15 * (a + b)

⇒ 525/15 = a + b

⇒ 35 = a + b

⇒ a + b = 35

Therefore the sum of parallel sides is 35 cm .

Answer 2

Answer:

$\bold\green{Sum\:of\:parallel\:sides}$ = $\bold{35\: cm}$

Step-by-step explanation:

$\frak{Given}\begin{cases}\tt\green{Area\: of \: trapezium= 262.5 \:{cm}^{2}}\\\tt\red{Distance \: b/w \: parallel\: sides = 15\: cm }\\\tt\purple{Sum \: of \: parallel \: sides = ?}\end{cases}$

$\rule{300}{1}$

We know that,

${\boxed{\bold\green{Area\: of \: trapezium = \dfrac{1}{2}(a+b) \times h }}}$

Where, (a + b) = Sum of parallel sides.

h = distance between them.

$\tt Substituting \: values :-$

$\Rightarrow\sf 262.5 = \dfrac{1}{2}(a + b) \times 15 \\\\\\\Rightarrow\sf 262.5 = \dfrac{15(a + b)}{2} \\\\\\\Rightarrow\sf 15(a + b) = 262.5 \times 2 \\\\\\\Rightarrow\sf 15(a + b) = 525 \\\\\\\Rightarrow\sf (a + b) = \dfrac{\cancel{525}}{\cancel{15}} \\\\\\\Rightarrow\huge{\boxed{\sf\purple{(a + b) = 35 \: cm }}}$

${\boxed{\therefore{\bold\green{Sum\:of\:its\: parallel\: sides = 35 \: cm }}}}$