Question

a string of length 12is bent into square PQRS and then into an isosceles triangle PQT by keeping the side PQ of the square at the base then the area of square PQRS :area of triangle PQT​

Answer 1

Answer:

Solution:

Length of string = 12

First bent into square, then perimeter 4×side=12

⇒ side = 3

Then bent into isosceles triangle, with base as square side

perimeter = 2x+3=12

⇒x=4.5

Where x is the equal side of triangle

Now height of Ah

2

=

(4.5)

2

−(3/2)

2

=18

⇒h=3

2

Areaoftriangle

Areaofsquare

=

2

1

×9×

2

3×3

=

2

2

=

1

2

=

2

:1

Answer 2

Question :

a string of length 12is bent into square PQRS and then into an isosceles triangle PQT by keeping the side PQ of the square at the base then the area of square PQRS :area of triangle PQT

step by step

side PQ of the square PQRS.

$= \frac{12}{4}$

$= 3$

Hence ,

side PT and QT

$= \frac{9}{2}$

$= 4.5$

height

$\: {4.5}^{2} - {1.5}^{2} = {h}^{2}$

$h = \sqrt{18}$

area of triangle

$= \frac{ \sqrt{18 } \times 4.5 }{2}$

area of square = 9

ratio

$= \frac{ \sqrt{1.8} }{45}$

$\frac{2}{ \sqrt{4.5} }$

answer ;

$\frac{2}{ \sqrt{4.5} }$