Question

Two adjacent sides of a parallelogram are 24cm and 18cm. If the distance between longer sides is 12cm, find the distance between shorter sides. ​

Answer 1

Answer :-

Given :-

• Adjacent sides of parallelogram are 24 cm and 18 cm.

• Distance between longer sides is 12 cm.

To Find :-

• Distance between shorter sides.

Solution :-

Area of a parallelogram = Base × Height

For longer sides :-

• Base = 24 cm
• Height = 12 cm

→ Area of parallelogram = 24 × 12

→ Area = 288

Area of parallelogram = 288 cm²

For shorter sides :-

• Base = 18 cm
• Area = 288 cm² ( As both are of same parallelogram, area will be same )

→ Area = Base × Height

→ 288 = 18 × H

→ H = 288 / 18

→ H = 16

Distance between shorter sides = 16 cm

Answer 2

Given:

• Shorter sides = 18cm.
• Longer sides = 24cm.
• Distance between longer sides = 12cm.

To find:

• Distance between shorter sides.

Solution:

As we know that,

Area of parallelogram = base × height

(length of longer side × distance between longer sides)

= 24 × 12

= 288cm².

Thus, Area of parallelogram = 288cm².

Now,

Let the distance between shorter sides be x

• As length of longer side × Distance between longer sides = Area.

• So, Length of shorter side × Distance between shorter sides = Area.

So,

length of shorter side × x

= 18 × x = 288

= x = $\bf\dfrac{288}{18}$

= 16 cm.

Therefore, Distance between the shorter sides is 16cm.

$\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}$