The height of the rectangular trapezoid KMPT is equal to its smaller base MP. Diagonal KP is perpendicular

| February 11, 2021 | education

The height of the rectangular trapezoid KMPT is equal to its smaller base MP. Diagonal KP is perpendicular to the lateral side of PT. a) Prove that the diagonal KP is the bisector of the angle K of this trapezoid. b) Find the lengths of the bases of the trapezoid if its height is 6 cm.

By condition, KM = MР = PH, therefore, the triangle KMP is isosceles and right-angled, then the angle MKР = MРK = 45.

The TКР angle is equal to the MРC angle as the criss-crossing angles at the intersection of the parallel straight lines CP and MР secant КР. Then the angle TKP = MKP = 45, and therefore KP is the bisector of the MKT angle, which was required to be proved.

If PH = MK = 6 cm, then MR is also 6 cm.

The KРT triangle is rectangular and isosceles, since the angles at the base are equal to 450. Then the height of the PH is the median of the KРT triangle, and KН = TH = MР = 6 cm. KT = 2 * KН = 2 * 6 = 12 cm.

Answer: If the height is 6 cm, then the bases of the trapezoid are 6 cm and 12 cm.



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