At the base of a regular triangular pyramid ABCD lies a triangle ABC with side equal to 6. The lateral edge

| March 22, 2021 | education

At the base of a regular triangular pyramid ABCD lies a triangle ABC with side equal to 6. The lateral edge of the pyramid is 4. Through a point T of the edge AD such that AT: TD = 3: 1, a plane is drawn parallel to lines AC and BD. a) Prove that the section of the pyramid by the indicated plane is a rectangle. b) Find the cross-sectional area.

Let’s build the height BE of the triangle ABC. Since the pyramid is correct, the projection of the BD edge will lie at the height BE, then the BD edge is perpendicular to the AC side.

By condition, НT and KM are parallel to BD, then ВT and KM are perpendicular to AC.

By condition, TC and НM are parallel to AC, then TC and НM are perpendicular to TН and CM, and the section of TCMН is a rectangle, which was required to be proved.

Since AT / TD = 3/1.

AT = 3 * TD.

AT + TD = 4 cm, then 3 * TD + TD = 4.

4 * TD = 4.TD = 1 cm, AT = 3 cm.

Triangles ADB and ATН are similar in two angles, then AD / AT = BD / TH.

TH = AT * BD / AD = 3 * 4/4 = 3 cm.

Triangles ADC and DTC are similar in two angles, then AD / TD = AC / TC.

TK = TD * AC / AD = 1 * 6/4 = 3/2 = 1.5 cm.

Then Ssec = TH * TK = 3 * 1.5 = 4.5 cm2.

Answer: The cross-sectional area is 4.5 cm2.



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