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Given: An Isosceles trapezoid EFGH in which EF =GH
To prove: ΔFHE ≅ ΔGEH
Proof: In Isosceles trapezoid EFGH, Considering two triangles ΔFHE and ΔGEH
1. FE ≅ G H → [ Given]
2. ∠H = ∠E
→ Draw GM⊥HE and FN ⊥EH, and In Δ GMH and ΔFNE,
GH=FE [Given]
∠M+∠N=180° so GM║FN and GF║EH, So GFMN is a rectangle.]
∴ GM =FN [opposite sides of rectangle]
∠GMH = ∠FNE [ Each being 90°]
Δ GMH ≅ ΔFNE [ Right hand side congruency]
→∠H =∠E [CPCT]
→ Side EH is common i.e EH ≅ EH .
→ΔFHE ≅ ΔGEH. [SAS]
Answer:
(C)5
Step-by-step explanation:
From the given graph, we can determine the coordinates of the X an Y.
These are:
X(-4,2) and Y(-1,-3)
Since, XY is the hypotenuse, then At point Z, the angle will be 90° and thus, XZ is the altitude.
Join, YZ such that the coordinates of Z is (-1,2).
Then, the length of the segment YZ will be:
Therefore, the length of the YZ is 5.