Answer:
The proof is given below.
Step-by-step explanation:
Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of parallelogram bisect each other.
In ΔACD and ΔBEC
AD=BC (∵Opposite sides of parallelogram are equal)
∠DAC=∠BCE (∵Alternate angles)
∠ADC=∠CBE (∵Alternate angles)
By ASA rule, ΔACD≅ΔBEC
By CPCT(Corresponding Parts of Congruent triangles)
AE=EC and DE=EB
Hence, AE is conruent to CE and BE is congruent to DE
The area of ΔIDK is 32.69 square units.
Solution:
Given data:
ID = 7 and Hypotenuse, KI = 11.67.
Let us first find KD:
Using Pythagoras theorem,
Subtract 49 from both sides.
Taking square root on both sides, we get
KD = 9.34
Area of the triangle =
= 32.69
The area of ΔIDK is 32.69 square units.
Answer:
The image will be congruent to ΔMNP.
Line EG will be perpendicular to the line segments connecting the corresponding vertices.
The line segments connecting corresponding vertices will all be parallel to each other.
Step-by-step explanation:
The reflected image will be the mirror image of ΔMNP, so it will be congruent. The lines connecting each vertex to the corresponding vertex will be parallel, and all three lines are perpendicular to EG.
