Use the given information to prove that PQR = TSR.

10 answers
Question:

Use the given information to prove that PQR = TSR.

Answers

c

Step-by-step explanation:

List the angles in order from largest to smallest.

A)∠R, ∠P, ∠Q

B)∠R, ∠Q, ∠P

C)∠Q, ∠P, ∠R

D)∠Q, ∠R, ∠P

D. [tex]\frac{DE}{PQ} = \frac{3}{2}[/tex]

Step-by-step explanation:

Given:

[tex]\frac{DE}{PR}=\frac{FE}{RQ} = \frac{3}{2}[/tex]

SSS Similarity theorem: Triangles are similar if all three sides in one triangle are in the same proportion to the corresponding sides in the other.

Applying SSS theorem to ΔDEF and Δ PQR :

[tex]\frac{DE}{PQ} = \frac{EF}{QR} = \frac{DF}{PR}[/tex]

But

[tex]\frac{DF}{PR} = \frac{EF}{RQ} = \frac{3}{2}[/tex]

Therefore,

[tex]\frac{DE}{PQ} = \frac{3}{2}[/tex]

This is the additional information needed to show the triangles are similar as per SSS similarity theorem.

D. DE/PQ= 3/2

Step-by-step explanation:

2020 edg

I answered then realized my answer was incorrect. I edited it out to make sure that it isn't read.

Step-by-step explanation:

Option D.

Step-by-step explanation:

Given information: [tex]\frac{DF}{PR}=\frac{FE}{RQ}=\frac{3}{2}[/tex].

We need to find the additional information which is needed to prove △DEF ~ △PQR using the SSS similarity theorem.

According to the SSS similarity theorem, two triangles are similar if their corresponding sides are proportional.

Using  SSS similarity theorem, both △DEF and △PQR are similar if

[tex]\frac{DE}{PQ}=\frac{E F}{Q R}=\frac{DF}{PR}[/tex]

If can be written as

[tex]\frac{DE}{PQ}=\frac{FE}{RQ}=\frac{DF}{PR}[/tex]

It is given that [tex]\frac{DF}{PR}=\frac{FE}{RQ}=\frac{3}{2}[/tex].

So, the additional information which is needed to prove △DEF ~ △PQR using the SSS similarity theorem is [tex]\frac{DE}{PQ}=\frac{3}{2}[/tex].

Therefore, the correct option is D.

The additional information required to prove ΔDEF ~ ΔPQR is the value of the ratio DE/PQ which has to be equal to three-halves for ΔDEF to be similar to ΔPQR

Step-by-step explanation:

Given DF/PR = FE/RQ = 3/2

The Side Side Side,  SSS, similarity theorem states that where there are two triangles that have corresponding sides that are proportional to each other, the two triangles are said to be similar

Given ΔDEF and ΔPQR, have sides DF/PR = FE/RQ, to prove that ΔDEF and ΔPQR, then the additional information required is the ratio of the third sides of the triangles which is DE/PQ.

If DE/PQ = Three-halves, the two triangles ΔDEF and ΔPQR are similar, if not, that is DE/PQ ≠ Three-halves, then the two triangles ΔDEF and ΔPQR are not similar.

The answer is D on Edge

The correct option is;

DE/PQ = 3/2 which is StartFraction DE Over PQ Endfraction = three-halves

Step-by-step explanation:

The information given bout the triangles are;

DF/PR = FE/RQ = 3/2

Therefore since the given sides of triangle DEF are 3/2 times the sides of triangle PQR, the given sides of triangle DEF have been scaled 3/2 times to get the given sides of triangle PQR

Therefore, to prove that ΔDEF ~ ΔPQR, the ratio of the third side of triangle DEF, that is DE, to the third side of triangle PQR, which is PQ must be three-halves.

I know 17. The formula for INTERIOR angles is 180(n-2) divided by n. therefore 180(6-2) / 6 = 180 x 4 / 6 = 720 / 6 = 120.

and EXTERIOR angles is 360 divided by n, therefore 360/6 = 60.

3. Given: R is the midpoint of both PT and QS and TSR. ∆ are right triangles. Prove: PQR. ∆. ~ TSR. ∆. 5. Given: AB = CB = 9 cm.

Step-by-step explanation:

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