Question

show that triangle ABC where A (22 0) B(2 0) C(0 2) and triangle PQR where P(-4,0) Q(4,0) R(0,4) are similar triangle

Answer 1

Answer:

Here, ABC is a triangle having vertices A (22, 0) B(2, 0) and C(0 , 2),

And, PQR is also a triangle having vertices P(-4,0) Q(4,0) and R(0,4),

By the distance formula,

$AB=\sqrt{(2-22)^2+(0-0)^2}=\sqrt{20^2}=20$

Similarly,

$BC=2\sqrt{2}$

$CA=2\sqrt{122}$

$PQ=8$

$QR=4\sqrt{2}$

$RP=4\sqrt{2}$

$\because \frac{20}{8}\neq \frac{2\sqrt{2}}{16\sqrt{2}}\neq \frac{2\sqrt{122}}{4\sqrt{2}}$

$\implies \frac{AB}{P Q}\neq \frac{BC}{QR}\neq \frac{CA}{RP}$

Hence, triangles ABC and PQR are not similar.

Answer 2

Step-by-step explanation:

Here, ABC is a triangle having vertices A (22, 0) B(2, 0) and C(0 , 2),

And, PQR is also a triangle having vertices P(-4,0) Q(4,0) and R(0,4),

By the distance formula,

AB=\sqrt{(2-22)^2+(0-0)^2}=\sqrt{20^2}=20AB=

(2−22)

2

+(0−0)

2

=

20

2

=20

Similarly,

BC=2\sqrt{2}BC=2

2

CA=2\sqrt{122}CA=2

122

PQ=8PQ=8

QR=4\sqrt{2}QR=4

2

RP=4\sqrt{2}RP=4

2

\because \frac{20}{8}\neq \frac{2\sqrt{2}}{16\sqrt{2}}\neq \frac{2\sqrt{122}}{4\sqrt{2}}∵

8

20



=

16

2

2

2



=

4

2

2

122

\implies \frac{AB}{P Q}\neq \frac{BC}{QR}\neq \frac{CA}{RP}⟹

PQ

AB



=

QR

BC



=

RP

CA

Hence, triangles ABC and PQR are not similar.