Question

the hypotenuse of a right triangle is 2 1/2 and another side is 1 1/2m calculate its perimeter​

Answer 1

Step-by-step explanation:

given:

Hypotenuse=h=5/2 cm(or m)

another side( perpendicular)= p= 3/2 cm( or m)

let the 3rd side i.e base be b cm.

we know that,

h²=p²+b². [Pythagoras theorem]

A.T.Q,

(5/2)²=(3/2)²+b²

=> 25/4= 9/4+ b²

=>25/4 - 9/4= b²

=>b²=16/4

=>b²=4

=b=√4=2cm.

perimeter= h+b+p

=5/2 +3/2+ 2

=(5+3+4)/2 [ by LCM, 2 becomes 4/2]

=12/2=6cm(or m)

hope it helps.

Answer 2

Answer:

$4(4 + \sqrt{10} )m$

Step-by-step explanation:

$given \: that \\ h = \frac{21}{2}m \: and \\ p = \frac{11}{2} m \: \: \: \: \: \: \: \: \: \\ therefore \: b = \sqrt{ {h}^{2} - {p }^{2} } \\ b = \sqrt{ {( \frac{21}{2}) }^{2} - {( \frac{11}{2} )}^{2} }$

$\: b = \sqrt{ \frac{441}{2} - \frac{121}{2} }$

$= \sqrt{ \frac{441 - 121}{2} } \\ = \sqrt{ \frac{320}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{160} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = 4 \sqrt{10} m \: \: \: \: \: \: \: \: \: \: \: \: \:$

so the perimeter is

$\frac{21}{2} + \frac{11}{2} + 4 \sqrt{10 } \\ = \frac{32}{2} + 4 \sqrt{10} \: \: \: \: \: \: \: \: \\ = 16 + 4 \sqrt{10} \: \: \: \: \: \: \: \: \: \\ = 4(4 + \sqrt{10})m \: \: \: \:$