Question

Find the perimeter of a trapezium PQRS.
SR 20cm
RQ 2cm
PQ 25cm

Answer 1

Answer:

All d sides are given

we have to only find PS.Name the vertex of the triangle formed by joining P and S as O

Step-by-step explanation:

in ∆pso

os=25-rs=5cm

op=rq=2cm

By pgt

h^2=p^2 +b^2

h^2=2^2 + 5^2

h^2=4 + 25

h^2 = 29

h=

$\sqrt{29}$

Therefore...perimeter of trap. pqrs=

$2 + 20 + \sqrt{29} + 25$

=47+√29 cm

Answer 2

Answer:

The perimeter is 52 cm

Hope it helped

Step-by-step explanation:

Consider PST as a right-angled triangle (T is an imaginary point to help find side PS) Also consider TRQP as a rectangle

RQ=TP=2 cm

PQ=TR=25 cm (Due to this step, TS=5 cm)

$a^{2}+b^{2}=c^{2}$ (Pythagoras theorem)

$2^{2} +5^{2} =c^{2}$

$4+25=c^{2}$

$29=c^{2}$

$\sqrt{29}=c$

5cm≈c (This is because the result is 5.385164807)

P=a+b+c+d

P=20+2+5+25

P=52 cm