In triangle ABC AB = 11 cm, BC = 7 cm, BD – height. which of the segments is larger – AD or DC?

Consider two right-angled triangles ABD and BCD, which have a common leg BD.

Let the length of the leg BD = X cm.

Then, by the Pythagorean theorem, AB ^ 2 = AD ^ 2 + BD ^ 2.

AD ^ 2 = 11 ^ 2 – X ^ 2. (one).

ВС ^ 2 = СD ^ 2 + ВD ^ 2.

CD ^ 2 = 7 ^ 2 – X ^ 2. (2).

Subtract equation 2 from equation 1.

AD ^ 2 – CD ^ 2 = 121 – X ^ 2 – 49 + X ^ 2 = 72.

The square of AD is greater than the square of CD by 72, therefore, AD is more than CD.

AD = √ (72 + CD ^ 2).

Answer: AD> CD.