In an isosceles triangle ABC, the base is BC = 12, the lateral side is 10cm. From vertex A, a segment AD

| April 17, 2021 | education

In an isosceles triangle ABC, the base is BC = 12, the lateral side is 10cm. From vertex A, a segment AD = 15 cm is drawn, perpendicular to the plane of the triangle ABC. Find the distance from point D to the side BC.

Let’s draw the height AH of the isosceles triangle ABC, which in it will also be the bisector and median, then CH = BH = BC / 2 = 12/2 = 6 cm.

In the right-angled triangle ABH, according to the Pythagorean theorem, we determine the length of the leg AH.

AH ^ 2 = AB ^ 2 – BH ^ 2 = 100 – 36 = 64.

AH = 8 cm.

In a right-angled triangle АDН, according to the Pythagorean theorem, we determine the length of the hypotenuse DH, which is the required distance.

DH ^ 2 = AD ^ 2 + AH ^ 2 = 225 + 64 = 289.

DН = 17 cm.

Answer: From point D to straight BC 17 cm.



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