Question

the square of side 12 cm is cut on the diagonal. find the perimeter of one triangle so formed.

(1).24√2 cm.
(2).12√2 cm.
(3).10[2+√2] cm.
(4).12[2+√2] cm.​

Answer 1

Question

the square of side 12 cm is cut on the diagonal. find the perimeter of one triangle so formed.

$\tt{Answer= \red{12(2+√2)cm}}$

Solution

Given that

ABCD is a square in which

$\tt{AB=BC=CD=DA=12cm \red{\ All\ sides\ are\ qual\ in\ square}}$

BD=?

${\boxed{\sf{By\ Pythagoras\ Theorem{\pink{\ h²=b²+p²}}}}}$

$\tt{ →(BD)²=(BC)²+(DC)²}$

$\tt{ →(BD)²=(BC)²+(DC)²}$

$\tt{ →(BD)²=(12)²+(12)²}$

$\tt{ →(BD)²=144+144}$

$\tt{ →(BD)²=288 }$

$\tt{ →BD= \sqrt{288}}$

$\tt{ →BD= \sqrt{2⁴×2×3²}}$

$\tt{→BD= \sqrt{2²×2²×3²} × \sqrt{2}}$

$\tt{→BD=2 \times 2 \times 3 \times \sqrt{2}}$

$\tt{→BD= 12 \sqrt{2}}$

Perimeter Of ∆ BCD→BC+DC+BD

⁣           ⁣      →12√2+12+12

⁣           ⁣      →24+12√12

⁣           ⁣      →12(2+√2)cm

$\tt{perimeter\ of\ one\ triangle\ so\ formed= \red{12(2+ \sqrt{2})cm}}$