Question

LA
1. The square of side 12 cm is cut on the diagonal. Find the perimeter of one
(2017)
triangle so formed.
mm 101? + 2)cm (4) 12 (2+ 2) cm​

Answer 1

Answer:

12( 2 + √2 ) cm

Step-by-step explanation:

Suppose of a square of length 12 cm.

Since, adjacent sides of square intersect each other at 90°, using Pythagoras theorem length of the diagonal can be found.

= > diagonal^2 = ( 12 cm )^2 + ( 12 cm )^2

= > diagonal^2 = 12^2 x 2 cm^2

= > diagonal = 12√2 cm

Perimeter of the triangle so formed :

= > side of square + side of square + diagonal of square

= > 12 cm + 12 cm + 12√2 cm

= > 12 cm x 2 + 12√2 cm

= > 12( 2 + √2 ) cm

Answer 2

$\bf{\underline{\underline{Given}}}$

$\mathsf{\bigstar\: Side\:of\:square = 12\:cm}$

$\bf{\underline{\underline{To\:find}}}$

$\mathsf{\bigstar\: perimeter\:of\:one\:triangle}$

$\bf{\underline{\underline{Solution}}}$

$\mathsf{let\:us\:consider\:a\:triangle\:BCD}$

$\mathsf{ from\:the \: given\: square \:ABCD}$

$\mathsf{BC = CD = 12\: cm}$

$\mathsf{And \: \angle C = 90°}$

$\mathsf{By\:pythagoras\:theorem}$

$\mathtt{\implies\: BD^{2} = BC^{2} + CD^{2} }$

$\mathtt{\implies\: BD^{2} = 12^{2} + 12^{2} }$

$\mathtt{\implies\: BD^{2} = 144 + 144 }$

$\mathtt{\implies\: BD^{2} = 288 }$

$\mathtt{\implies\: BD = \sqrt{288}}$

$\mathtt{\implies\: BD = 12\: \sqrt{2}\: cm}$

$\bf{\underline{\underline{We\:know\:that}}}$

$\mathsf{Perimeter\:of\:triangle = Sum\: of \: three\: sides}$

$\mathtt{\implies\: BC + CD + BD}$

$\mathtt{\implies\: 12\:cm + 12\:cm + 12\: \sqrt{2}\: cm }$

$\mathtt{\implies\: 24\:cm + 12\: \sqrt{2}\: cm }$

Taking 12 common

$\fbox{\Large{\mathtt{\green{\implies\: 12(2 + \sqrt{2})\: cm }}}}$