In the diagram, StartFraction S Q Over O M EndFraction = StartFraction S R Over O N EndFraction = 4.

Triangles M N O and S R Q are shown. The length of side N O is 8 and the length of side R Q is 48. The length of side O M is 15 and the length of Q S is 60. The length of side M N is 12 and the length of S R is 32.

To prove that the triangles are similar by the SSS similarity theorem, which other sides or angles should be used?

MN and SR
MN and QR
∠S ≅ ∠N
∠S ≅ ∠O

MN and QR should be used to prove that the triangles are similar by the SSS similarity theorem ⇒ 2nd answer

Step-by-step explanation:

Let us revise the cases of similarity:

In Δs MNO and QRS

∵ SQ = 60

∵ OM = 15

- Find the ratio between SQ and OM

∴

∵ SR = 32

∵ ON = 8

- Find the ratio of SR and ON

∴  

∵ MN = 12 units

∵ QR = 48

- Find the ratio between QR and MN

∴

- The ratio between the corresponding sides are equal

∴

- By using the third case of similarity SSS

∴ Δ MNO is similar to Δ QRS by SSS similarity theorem

∴ You should use sides MN and QR

The given triangles are similar by Side-Side-Side (SSS), similarity when the ratio of the three corresponding pair of sides are equal.

The other sides that should be used to prove that ΔMNO and ΔSRQ are similar are;

Reasons:

The given parameters are;

The triangles are; ΔMNO and ΔSRQ

NO = 8

RQ = 48

OM = 15

QS = 60

MN = 12

SR = 32

Required:

The additional sides to prove that ΔMNO and ΔSRQ are similar by SSS similarity postulate.

Solution:

Based on the given information, we have;

Where ΔMNO and ΔSRQ are similar, the ratio of the corresponding sides are equal, therefore, given that we have;

SQ and SR from ΔSRQ and OM and ON from ΔMNO, the other information on ΔSRQ and ΔMNO are;

For ΔSRQ and ΔMNO to be similar, the ratio of the remaining sides also gives a value of 4, therefore;

Which gives;

Therefore;

To prove that the triangles are similar by SSS, Side-Side-Side, similarity theorem, the sides that should be used are;

Learn more about SSS similarity postulate here:

link

4932,9180,7625 are divisible by 9

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen