Question

(2) Length of a diagonal of a square is 15
$\sqrt{2}$
3 cm. Find its perimeter.
(A) 15 cm
(B) 30 cm
(C) 45 cm
(D) 60 cm​

Answer 1

Answer:

I think so its 30

Step-by-step explanation:

don't know but I am not sure pls get answers from others too

Answer 2

Correct Statement

$\sf \: Length \: of \: a \: diagonal \: of \: a \: square \: is \:15 \sqrt{2} \: cm.$

$\sf \: Find \: its \: perimeter. \\ </p><p> \sf \: (A) 15 cm \\ </p><p> \sf \: (B) 30 cm \\ </p><p> \sf \: (C) 45 cm \\ </p><p> \sf \: (D) 60 cm$

$\large\underline{\sf{Solution-}}$

Let ABCD be a square whose side AB = 'x' cm.

It is given that

$\rm :\longmapsto\:Diagonal_{(square)} = \: AC \: = 15 \sqrt{2} \: cm$

In ∆ ABC,

Using, Pythagoras Theorem, we have

$\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\rm :\longmapsto\: {(15 \sqrt{2}) }^{2} = {x}^{2} + {x}^{2}$

$\rm :\longmapsto\:225 \times 2 = 2 {x}^{2}$

$\rm :\longmapsto\: {x}^{2} = 225$

$\rm :\implies\:x = 15 \: cm$

We Know,

Perimeter of square is evaluated as

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times side$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times \: x$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times \: 15$

$\rm :\longmapsto\:Perimeter_{(square)} = 60 \: cm$

$\boxed{ \bf \: Hence, \: Option \: (d) \: is \: correct}$

Additional Information :-

$\boxed{ \sf \: Perimeter_{(rectangle)} = 2(l + b)}$

$\boxed{ \sf \: Area_{(rectangle)} = l \times b}$

$\boxed{ \sf \: Area_{(square)} = {(side)}^{2}}$

$\boxed{ \sf \: Area_{(square)} = \dfrac{1}{2} \times {(Diagonal_{(square)})}^{2}}$

$\boxed{ \sf \: Area_{(right \triangle)} = \dfrac{1}{2} \times b \times h}$

$\boxed{ \sf \: Area_{(equilateral \triangle)} = \dfrac{ \sqrt{3} }{4} {(side)}^{2}}$