A rectangle is inscribed in a triangle with sides 10, 17 and 21 so that two of its vertices lie on the larger side

A rectangle is inscribed in a triangle with sides 10, 17 and 21 so that two of its vertices lie on the larger side of the triangle, and the other two – on the smaller sides. Find the area of a rectangle if its perimeter is 25

Let us define the semiperimeter of the triangle.

Ravs = (AB + BC + AC) / 2 = (17 + 10 + 21) / 2 = 24 cm.

Then Sav = √24 * (24 – 17) * (24 – 10) * (24 – 21) = √24 * 7 * 14 * 3 = √7056 = 84 cm2.

The area of the triangle ABC is also equal to: Savs = AC * BD / 2.

BH = 2 * Savs / AC = 2 * 84/21 = 8 cm.

Let the side of the PH of the rectangle MPСK be equal to X cm, then MR = (25/2 – X) = 12.5 – X cm.

Triangles MBP and ABM are similar in acute angle. BE = BD – ED = BD – PH = 8 – X cm.

Then, in similar triangles:

MP / AC = BE / BD.

(12.5 – X) / 21 = (8 – X) / 8.

168 – 21 * X = 100 – 8 * X.

13 * X = 68.

X = PH = 68/13 = 5.23 cm.

Then MR = 12.5 – 5.23 = 7.27 cm.

Then Sпр = МР * РН = 7.27 * 5.23 ≈ 38.02 cm2.