A ball is inscribed in a regular quadrangular pyramid, the side of the base of which is 4 cm.

| July 7, 2021 | education

A ball is inscribed in a regular quadrangular pyramid, the side of the base of which is 4 cm. Find the volume of the pyramid, knowing the chateau’s radius of the ball is 1 cm.

The inscribed ball touches the side face at point H and the base at point O.

Segments KH = KO = R = 1 cm.

Segment OM = AD / 2 = 4/2 = 2cm.

Rectangular triangles POM and PKH are similar in acute angle OPM.

Then, the coefficient of similarity is: ОМ / КН = 2/1 = 2.

Let the length of the segment PH = X cm, then PM = 2 * X cm, and the length PO = (X + 1) cm.

In a right-angled triangle POM, according to the Pythagorean theorem: PM ^ 2 = PO ^ 2 + OM ^ 2.

4 * X ^ 2 = (X + 1) 2 + 4 = X ^ 2 + 2 * X + 1 + 4.

3 * X ^ 2 – 2 * X – 5 = 0.

Let’s solve the quadratic equation.

X = 5/3.

Then PO = 5/3 + 1 = 8/3 cm.

The area of ​​the base of the pyramid is equal to: Sbn = AD2 = 16 cm2.

V = Sbase * PO / 3 = 16 * (8/3) / 3 = 128/9 = 14 (2/9) cm3.

Answer: The volume of the pyramid is 14 (2/9) cm3.



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