Question

In the triangles, QR = DE and SR = FE.

Answer 1

We are given triangles QRS and DEF.

QR = DE and SR = FE.

Also we are given angle < R = 62 degrees and angle < E = 50 degrees.

<R is greater than the < E.

Note: Corresponding side of smaller angle is small.

In triangle QRS, < R is greater angle than the <E in triangle DEF.

Therefore, side DF is smaller side of QS.

Therefore, correct statement is B. DF < QS .

Answer 2

We have been given that in the given triangles QR=DE and SR=FE and we are asked to find which of the given statements is true about the sides of both triangles.

SSS inequality theorem states that if two sides of one triangle are congruent to two sides of another triangle and third side of one triangle is smaller than third side of second triangle, then measure of angle between pair of congruent sides of first triangle will be smaller than corresponding angle between second triangle.

We can see that in our triangles,

$\angle E=50^{o}$

$\angle R=62^{o}$

$50^{o}$

By SSS inequality theorem $DF < QS$, therefore, option B is the correct choice.