Recall that t<span>wo shapes are said to be congruent if the lengths of the sides and the angles are the same.
For triangle A B C to be congruent to triangle D E F, then A B = D E, A C = D F and B C = E F
Also, the triangle D E F is a result of a rigid body translation of the triangle A B C.
Given that A is at point (1, 0) and D is at point (-1, 0), also given that B is at point (-1, 2) and E is at point (1, -2).
</span><span>It can be seen that points D E is a result of refrecting points A B
across the y-axis and then the x-axis.
</span><span>Thus, given point C as (2, 3),
refrection of point C across the y-axis will result in point (-2,
3) and then reflecting the resulting point across the x-axis will
result to point (-2, -3).
</span>Therefore, the coordinate for 'F' that would make triangle A B C and triangle D E F congruent is (-2, -3).
<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.
As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation
∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm
A possible answer would be 32*0.01
