In the figure below, GRAH is a rectangle, RA=7, and GA=10. Determine AH.

Answer:
AH = [tex]\sqrt{51}[/tex]
Step-by-step explanation:
The opposite sides of a rectangle are congruent , so
GH = RA = 7
Using Pythagoras' identity in right Δ GHA
AH² + GH² = GA²
AH² + 7² = 10²
AH² + 49 = 100 ( subtract 49 from both sides )
AH² = 51 ( take the square root of both sides )
AH = [tex]\sqrt{51}[/tex] ≈ 7.14 ( to 2 dec. places )
Answer:
√51 (or in LaTeX $sqrt51$)
Step-by-step explanation:
We know that Triangle GRA is a right triangle because Angle GRA is one of the angles of rectangle GRAH, and all interior angles of a rectangle are right.
We also know the lengths of two sides of GRA, so we can use the Pythagorean Theorem to solve for the third side. Specifically, the Pythagorean Theorem tells us that
(GR)^2 + (RA)^2 = (GA)^2.Plugging in RA=7 and GA=10, we get
(GR)^2 + 49 = 100,so (GR)^2=51. We know that GR is positive, so we conclude that $GR=\sqrt{51}$. Finally, since GRAH is a rectangle, its opposite sides have equal length. Thus AH = GR, and so AH = √51, which cannot be simplified.