Question

the parallel sides of a trapezium are 9 CM and 7 CM and it's area is 80 cm2 find it's altitude​

Answer 1

Given :

• Parallel Sides of Trapezium are 9 cm and 7 cm .
• Area of Trapezium is 80 cm² .

To Find :

• Altitude/Height of Trapezium .

Solution :

$\longmapsto\tt{Parallel\:Sides=9\:cm\:and\:7\:cm}$

Using Formula :

$\longmapsto\tt\boxed{Area\:of\:Trapezium=\dfrac{1}{2}\times{(Sum\:of\:parallel\:sides)}\times{h}}$

Putting Values :

$\longmapsto\tt{80=\dfrac{1}{2}\times{(9+7)}\times{h}}$

$\longmapsto\tt{80=\dfrac{1}{{\cancel{2}}}\times{{\cancel{16}}}\times{h}}$

$\longmapsto\tt{80=8\:h}$

$\longmapsto\tt{h=\cancel\dfrac{80}{8}}$

$\longmapsto\tt\bf{h=10\:cm}$

So , The Height of Trapezium is 10 cm .

________________________

VERIFICATION :

$\longmapsto\tt{80=\dfrac{1}{{\cancel{2}}}\times{{\cancel{16}}}\times{10}}$

$\longmapsto\tt{80=8\times{10}}$

$\longmapsto\tt\bf{80=80}$

HENCE VERIFIED

Answer 2

Given :-

Trapezium

• Parallel sides = 9 cm,  7 cm
• Area = 80 cm²

To find :-

➟ Altitude = ?

Concept used :-

$\dag \: \boxed{\mathfrak{Area \: of \: a \: trapezium = \dfrac{1}{2} \times (a + b) \times Height}}\\\\\\\bf{Where;}\\\\\star \sf{a \: and \: b \:are \: parallel\: sides}\\\\\star \sf{height = altitude}$

Solution :-

$\sf{Area = \dfrac{1}{2} \times (9 + 7) \times Height}\\\\\\\leadsto \: \sf{80 = \dfrac{1}{2} \times 16 \times Height}$

$\longrightarrow \: \sf{80 = 8 \times Height}\\\\\\\dashrightarrow \: \sf{Height = \dfrac{80}{8}}\\\\\\\therefore \: \boxed{\bf{Height = 10 \: cm^{2} }}$

Know more :-

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$