Question

find the altitude of trapezium was area 78 and parallel sides are 13cm and 26cm​

Answer 1

Answer :

4cm

Explanation :

In a trapezium,

• Parallel sides - 13cm & 26cm.
• It's area - 78cm.

Find the altitude.

We know that,

Area of a trapezium : ½(a + b)h

Where,

• (a + b) are the sum of parallel sides.
• ‘h’ is the altitude.

On substituting the given values :

→ 78 = ½(13 + 26)h

→ 78 = ½(39)h

→ 78*2/39 = h

→ 4 = h

∴ The altitude of the trapezium is 4cm.

Answer 2

$\frak{Given}\begin{cases} \sf{ The\:Area\:of\:Trapezium \:is\:78cm^{2}}\\\\\sf{Two\:Parallel \:Sides\:of\:Trapezium \:are\:13cm\:and\:26\:cm\:}\end{cases}$

Exigency to find : The Altitude ofTrapezium .

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

❍ Let the altitude of the trapezium be h cm respectively.

$\underline{\frak{Diagram:}}$

$\setlength{\unitlength}{1.1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf B }\put(-0.3,-0.3){ \bf D }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf ?\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 13\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 26\ cm }\end{picture}$ ⠀⠀

⠀⠀⠀$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Area_{\;(trapezium)} = \bigg( \dfrac{1}{2} \times (a + b) \times h\bigg)}}}}$

Where,

• a and b are the two parallel sides and h is distance between two parallel sides or height of trapezium.⠀⠀⠀⠀

$\dag\;{\underline{\frak{Now,\: Substituting\:values\:in\;formula,}}}$

$:\implies\sf 78 = \dfrac{1}{2}(13 + 26) \times h \\\\\\:\implies\sf 78 \times 2 = (13 + 26)\times h \\\\\\:\implies\sf 156 = (13 + 26) \times h \\\\\\:\implies\sf 156 = 39 \times h \\\\\\:\implies\sf 156 = 39h\\\\\\:\implies\sf 156 = 39b \\\\\\:\implies\sf h = \cancel\dfrac{156}{39}\\\\\\:\implies{\underline{\boxed{\frak{\pink{h = 4\;cm}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence, \;the\;Altitude \; of \;Trapezium \;is\;\bf{ 4\;cm}.}}}.$

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