Question

the parallel side of a trapezium are 25cm and 13cm it's non parallel sides are equal each being 10cm find the area of the trapezium

Answer 1

Let AB = 25 cm , CD = 13 cm and BC = CD = AD cm

Draw perpendiculars from C and D to AB meeting AB at F and E respectively

Now A E = F B = 6 cm

In triangle AED,

AE2+DE2=AD2AE2+DE2=AD2

DE2=AD2−AE2DE2=AD2−AE2

=102−62=64=102−62=64

DE=8cmDE=8cm

Area of the Trapezium = 1/2*h(a+b) sq cm

Area of the Trapezium = 1/2*8(25+13)sq cm =152 sq cm

Answer 2

${\large{\bold{\rm{\underline{Let's \; understand \; the \; question}}}}}$

★ This question says that the non parallel sides of trapizum are equal and measure 10 cm. The parallel sides measure 13 cm and 25 cm. We have to find out the area of trapizum..! Let's solve this problem..!

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ The non parallel sides of trapizum are equal and measure 10 cm.

★ The parallel sides measure 13 cm and 25 cm.

${\large{\bold{\rm{\underline{To \; find}}}}}$

★ Area of the trapizum

${\large{\bold{\rm{\underline{Solution}}}}}$

★ Area of the trapizum = 152 cm²

${\large{\bold{\rm{\underline{Using \; concepts}}}}}$

★ Formula to find area of the trapizum

★ Phythagoras Theorm

${\large{\bold{\rm{\underline{Using \; formulas}}}}}$

★ Area of the trapizum = 1/2 = Sum of parallel sides × Height

★ Phythagoras Theorm = Perpendicular² = Base² + Height²

${\large{\bold{\rm{\underline{Assumptions}}}}}$

★ ABHI is a trapizum

★ Parallel base AB = 25 cm

★ Parallel base HI = 13 cm

★ Equal sides AI = BH = 10 cm

★ Let distance between two sides (parallel) = IT = HM = h (where height is written)

★ Between AT and MB let point be a

★ AB = 25 cm

${\large{\bold{\rm{\underline{Full \: Solution}}}}}$

____________________________

~ To solve this problem we have to use our all the taken assumptions let's see how to use, where and how to do too..!

★ ABHI is a trapizum

★ Parallel base AB = 25 cm

★ Parallel base HI = 13 cm

★ Equal sides AI = BH = 10 cm

★ Let distance between two sides (parallel) = IT = HM = h (where height is written)

★ Between AT and MB let point be a

★ AB = 25 cm

$\; \; \; \; \; \; \; \; \; \; \;{\tt{Henceforth,}}$

${\rm{:\implies AT \: + \: TM \: + \: MB \: = 25 \: cm}}$

${\rm{:\implies a \: + \: a \: + \: 13 \: = 25 \: cm}}$

${\rm{:\implies 2a \: + \: 13 \: = 25 \: cm}}$

${\rm{:\implies 2a \: = 25 - 13}}$

${\rm{:\implies 2a \: = 12}}$

${\rm{:\implies a \: = 12/2}}$

${\rm{:\implies a = \: 6 \: cm}}$

${\small{\boxed{\bf{Between \: AT \: and \: MB \: let \: point \: be\: a \: or \: 6 \: cm}}}}$

____________________________

~ Now as we are able to see that there are two triangle's formed in this trapizum so let's carry on...! And let's find for just one triangle because it is already cleared that both are equal so no need to find for both again' saying that's because both the triangle are equal..!

____________________________

~ For triangle ATI..!

${\rm{:\implies \angle ATI \: measures \: 90 \degree}}$

${\rm{:\implies AI \: = \: 10 \: cm}}$

${\rm{:\implies TI \: = \: h \: cm}}$

${\rm{:\implies AT \: = a \: or \: 6 \: cm}}$

____________________________

~ Now let's use phythagoras theorm formula and let's put the values..!

${\rm{:\implies IT^{2} \: = AI^{2} \: - \: AT^{2}}}$

${\rm{:\implies h^{2} \: = 10^{2} \: - \: 6^{2}}}$

${\rm{:\implies h^{2} \: = 100 \: - 36}}$

${\rm{:\implies h^{2} \: = 64}}$

${\rm{:\implies h \: = \sqrt{64}}}$

${\rm{:\implies h \: = 8 \: cm}}$

${\small{\boxed{\bf{Distance \: or \: h \: is \: 8 \: cm}}}}$

____________________________

~ Now let's find the area of trapizum..!

${\rm{:\implies \dfrac{1}{2} \: = \: (sum \: of \: parallel \: side) \times height}}$

${\rm{:\implies \dfrac{(25+13) \times 8}{2}}}$

${\rm{:\implies \dfrac{38 \times 8}{2}}}$

${\rm{:\implies \dfrac{304}{2}}}$

${\rm{:\implies 152 \: cm^{2}}}$

${\small{\boxed{\bf{Area \: of \: trapizum \: is \: 152 \: cm^{2}}}}}$

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