Question

the parallel sides of a trapezium are 40cm and 70cm. if the non - parallel sides are equal, each being 25cm , find the area of the trapezium.​

Answer 1

DIAGRAM:

$\setlength{\unitlength}{1.3cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 25 \ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 70 \ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 40 \ cm }\end{picture}$

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• Let, two points E & F are two points making perpendicular AE & DF on the line BC.

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$\frak{Given}\begin{cases}\sf{\;\;\; || \; sides \; of \; the \; trapezium \; are \; \bf{40\; cm} \:\&\;\bf{70\:cm.}}\\\sf{\;\;\;Non\; || \; sides \; of \: the \; trapezium\: \; are \;equal \;to \;\bf{25\: cm.}}\end{cases}$

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Need to find: Area of the trapezium.

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Therefore,

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• AD = 40 cm & BC = 70 cm.
• AB = 25 cm & CD = 25 cm.

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In ∆ABE,

$\underline{\boldsymbol{By\;using\; Hypotenuse\; theorem \: :}}$

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$:\implies\sf (AE)^2 + (BE)^2 = (AB)^2 \\\\\\:\implies\sf (AE)^2 + 15^2 = 25^2 \\\\\\:\implies\sf AE^2 + 225 = 625\\\\\\:\implies\sf AE^2 = 625 - 225\\\\\\:\implies\sf AE^2 = 400\\\\\\:\implies\sf \sqrt{AE^2}= \sqrt{400}\\\\\\:\implies{\underline{\boxed{\frak{\pink{ AE = 20\;cm}}}}}$

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$\therefore{\underline{\sf{Hence, \: value \; of \; AE \; is \; \bf{ 20 \: cm}.}}}$

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⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

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

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

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• Where, a & b are the length of two parallel sides. And, h is the height or distance between two parallel sides.⠀⠀⠀

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Therefore,

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$:\implies\sf Area_{\:(trapezium)} = \dfrac{1}{2} \times AE \times (AD + BC)\\\\\\:\implies\sf Area_{\: trapezium)} = \dfrac{1}{2} \times 20 \times (40 + 70)\\\\\\:\implies\sf Area_{\:(trapezium)} = \dfrac{1}{\cancel{\;2}} \times \cancel{20} \; \times 110\\\\\\:\implies\sf Area_{\:(trapezium)} = 10 \times 110 \\\\\\:\implies{\underline{\boxed{\frak{\purple{Area_{\:(trapezium)} = 1100\;cm^2}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence,\: Area \; of \; the \; trapezium\; is \; \bf{1100\;cm^2 }.}}}$

Answer 2

➻ See the attachment diagram.

Gɪᴠᴇɴ :

• The parallel sides of a trapezium are 40 cm & 70 cm.

$\longmapsto\:\bf{AB\:=\:40\:cm}$

$\longmapsto\:\bf{CD\:=\:70\:cm}$

• Non parallel sides are equal, i.e. 25 cm.

$\longmapsto\:\bf{AC\:=\:BD\:=\:25\:cm}$

Tᴏ Fɪɴᴅ :

• The area of the trapezium.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

✯ As shown in the diagram,

• AX and BY are the altitudes, which are equal to each other.

Let,

➛ $\:\bf{AX\:=\:BY\:=\:h\:cm}$

We know that,

↝ In a right angle triangle,

$\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{(Hypotenuse)^2\:=\:(Height)^2\:+\:(Base)^2\:}}}}}}$

Now,

↝ In ACX rightangle triangle,

➙ $\bf{(AC)^2\:=\:(AX)^2\:+\:(CX)^2\:}$

Where,

• AC = 25 cm

[NOTE : Here, (AB = XY = 40 cm) & (CX = YD)]

• CX = $\sf{\dfrac{70\: - \:40}{2}\:=\:\dfrac{30}{2}}$ = 15 cm

• AX = h cm

➙ $\bf{(25)^2\:=\:h^2\:+\:(15)^2\:}$

➙ $\bf{625\:=\:h^2\:+\:225\:}$

➙ $\bf{h^2\:=\:625\:-\:225\:}$

➙ $\bf{h^2\:=\:400\:}$

➙ $\bf{h\:=\:\sqrt{400}\:}$

➙ $\bf{h\:=\:20\:cm}$

We know that,

↝ Area of a trapezium is,

$\pink\bigstar\:\:{\underline{\green{\boxed{\bf{\blue{Area\:=\:\dfrac{1}{2}\times{h}\times(sum\:of\:parallel\:sides)\:}}}}}}$

➣ $\bf{Area\:=\:\dfrac{1}{2}\times{20}\:(40\:+\:70)\:}$

➣ $\bf{Area\:=\:10\times{110}\:}$

➣ $\bf\purple{Area\:=\:1100\:cm^2}$

$\Large\tt{Therefore,}$

★ The area of the trapezium is 1100 cm².