In ΔRST, r = 40 cm, s = 88 cm and t=70 cm. Find the measure of ∠S to the nearest degree.
2 answers:
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Let the equal sides of the isosceles Δ ABC be x.
Given that the perimeter of Δ ABC = 50m.
Therefore, 2x + AC = 50 --- (1)
It is also given that the perimeter of Δ ABD = 40m.
Therefore, x + BD + AD = 40
BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.
So AD = CD.
Or, AD = AC
Therefore,
Multiply both sides by 2.
2x + 2BD + AC = 80
From (1), 2x + AC = 50.
Therefore, 2BD + 50 = 80
2BD = 80 - 50
2BD = 30
BD = 15m.