In the adjacent figure, ABC is a right triangle at A, such that: AB = (V17 – 1) and AC = 2V5 D (V17 + 1). B. BCD is a triangle, where: BD = 2V5 and CD = 4. All the dimensions are expressed in centimeters. (V17 – 1) 1) Show by calculation, that BC is a natural number. 2) Prove that BCD is a right triangle. 3) Let M be the midpoint of the segment [BC]. Calculate AM. A C. (V17 +1)

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Chapter7: Triangles
Section7.1: The Law Of Sines
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Exercise (7)
In the adjacent figure, ABC is a right triangle at A, such that:
AB = (V17 – 1) and AC =
2V5
D
(V17 + 1).
BCD is a triangle, where: BD = 2/5 and CD = 4.
All the dimensions are expressed in centimeters.
(V17 – 1)
1) Show by calculation, that BC is a natural number.
2) Prove that BCD is a right triangle.
3) Let M be the midpoint of the segment [BC]. Calculate AM.
4) Denote by H the foot of the height drawn from D in the triangle BCD.
a. Calculate an exact value of area of the triangle BCD.
A
(V17 +1)
b. Deduce an exact value of DH.
Transcribed Image Text:Exercise (7) In the adjacent figure, ABC is a right triangle at A, such that: AB = (V17 – 1) and AC = 2V5 D (V17 + 1). BCD is a triangle, where: BD = 2/5 and CD = 4. All the dimensions are expressed in centimeters. (V17 – 1) 1) Show by calculation, that BC is a natural number. 2) Prove that BCD is a right triangle. 3) Let M be the midpoint of the segment [BC]. Calculate AM. 4) Denote by H the foot of the height drawn from D in the triangle BCD. a. Calculate an exact value of area of the triangle BCD. A (V17 +1) b. Deduce an exact value of DH.
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