Question

Show that ΔABC, where A(-2, 0), B(2, 0), C(O, 2) and ΔPQR where P( -4, 0), Q(4, 0), R(0, 2) are similar triangles.

Answer 1

find distance between to point A.,B. B To C angin similar then all triangles all side equal so its equal

Answer 2

Hi there!
Here's the answer:

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Given,

A(-2, 0), B(2, 0), C(O, 2) are the Vertices of ∆ABC

P( -4, 0), Q(4, 0), R(0, 2) are the Vertices of ∆PQR

¶¶Step-1 :
Find the Sides of ∆ABC viz., AB, BC, CA
using distance between two points formula

• A(-2, 0), B(2, 0)
AB = $\sqrt{(2+2)^{2}+(0-0)^{2}}$

=> AB = 4

• B(2, 0), C(0, 2)
BC= $\sqrt{(0-2)^{2}+(2-0)^{2}}$

=> BC = $\sqrt{8}$
=> BC = $2\sqrt{2}$

• A(-2, 0), C(0,2)
AC = $\sqrt{(0+2)^{2}+(2-0)^{2}}$

=> AC = $\sqrt{8}$
=> AC = $2\sqrt{2}$

¶¶Step 2 :
Find the Sides of ∆PQR viz., PQ, QR, PR
using distance between two points formula

• P(-4, 0), Q(4,0)
PQ = $\sqrt{(4+4)^{2}+(0-0)^{2}}$

=> PQ = 4

• Q(4, 0), R(0, 4)
QR= $\sqrt{(0-4)^{2}+(4-0)^{2}}$

=> QR = $\sqrt{32}$
=> QR = $4\sqrt{2}$

• P(-4, 0), R(0,4)
PR = $\sqrt{(0+4)^{2}+(4-0)^{2}}$

=> PR = $\sqrt{32}$
=> PR = $4\sqrt{2}$

¶¶ Step-3: Check the similarity criteria Using the Basic Proportionality Theorem (Thales Theorem)

i.e., Check whether :
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$

$\frac{AB}{PQ} = \frac{4}{8} = \frac{1}{2}$

$\frac{BC}{QR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$

$\frac{AC}{PR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$

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$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$

Thales theorem satisfied

∆ABC is similiar to ∆PQR.

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