Find the length of the larger diagonal of the parallelogram with sides 3√2 cm and 1 cm and an angle of 45 degrees.

Given: ABCD – parallelogram, ے BAD = 45 degrees, AD = 3√2 cm, AB = 1 cm.
Find: large diagonal.
Solution: Since the AC diagonal is opposite to the obtuse angle ABD of the parallelogram, it means that AC is the larger diagonal.
1) Let’s draw the height CH and consider the rectangular ∆АСН. Applying the Pythagorean theorem (in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs), we obtain the equality:
AC ^ 2 = AH ^ 2 + CH ^ 2, where
AH = AD + DH.
2) Consider ∆DCH:
ے CDH = ے BAD = 45 degrees,
ے CDH = 90 degrees because CH – height,
ے DCH = 180 – ے CDH – ے CHD = 180 – 45 – 90 = 45 (degrees).
Hence ∆DCH is rectangular and isosceles.
By the Pythagorean theorem:
DC ^ 2 = DH ^ 2 + CH ^ 2
DC = AB = 1 cm, DH = CH
1 ^ 2 = DH ^ 2 + DH ^ 2
1 = 2 * DH ^ 2
DH ^ 2 = 1/2
DH = 1 / √2
3) Let’s go back to ∆АСН:
AH = AD + DH = 3√2 + 1 / √2 = 7 / √2, CH = DH = 1 / √2
AC ^ 2 = (7 / √2) ^ 2 + (1 / √2) ^ 2 = (49/2) + (1/2) = 50/2 = 25
AC = √25 = 5cm
Answer: AC = 5 cm.