Question

Find the perpendicular distance between two parellel lines of trapezium whose parellel sides are 30 cm and 24 cm , and area is 405 cm square.
please solve it. it is very important question. tomorrow is my exam...​

Answer 1

Answer:

• Perpendicular distance between parallel lines of trapezium is 15 cm.

Step-by-step explanation:

Given :-

• Parallel sides of trapezium are 30 cm and 24 cm.
• Area of trapezium is 405 cm².

To find :-

• Perpendicular distance between two parallel lines.

Solution :-

Let, Distance between parallel lines be x cm.

We know,

Area of trapezium = (a + b)/2 × h

Where,

• a and b are sides of trapezium.
• h is height or we can say perpendicular distance between two parallel lines of trapezium.

Put all values in formula :

⇒405 = (a + b)/2 × h

⇒405 = (30 + 24)/2 × x

⇒405 = 54/2 × x

⇒405 × 2 = 54 × x

⇒810 = 54 × x

⇒810/54 = x

⇒x = 15

We take x be distance between parallel lines.

So,

Perpendicular distance between parallel lines is 15 cm.

Answer 2

Answer:

Diagram:-

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\put (1,2.2){\line (0,-1){2.2}}\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf h\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 24\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 30\ cm }\end{picture}$

Given:-

The length of two parallel lines of a trapezium are 30cm and 24cm

Area of the trapezium =405 sq.cm

To find:-

perpendicular distance between two parallel lines of the trapezium =Height (h)

Solution:-

If the parallel lines are taken as a and b then

• a=30cm
• b=24cm

Let

Height =h cm

As we know that in a trapezium

$\boxed{\sf Area=\dfrac {1}{2}(a+b)\times Height}$

$\\\qquad\quad\displaystyle\sf{:}\rightarrowtail \dfrac{1}{2}(30+24)\times h =405$

$\\\qquad\quad\displaystyle\sf{:}\rightarrowtail \dfrac {1}{2}(54)\times h=405$

$\\\qquad\quad\displaystyle\sf{:}\rightarrowtail 27h=405$

$\\\qquad\quad\displaystyle\sf{:}\rightarrowtail h=\dfrac {405}{27}$

$\\\qquad\quad\displaystyle\sf{:}\rightarrowtail h=15$

$\\\\\therefore\sf Perpendicular\:distance\:between\:the\:parrelal\:lines\:of\:the\:trapezium\:is\:15cm.$