Given: Line A R bisects ∠BAC; AB = AC Triangle A B C is shown. Point R is at the middle of the triangle. Lines are drawn from po
2 answers:
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Your answer is no solution (because a negative square root doesn't exist)
first, you divide all sides by five and then take b, half it then square it
first, you divide all sides by five and then take b, half it then square it
Answer:
Perimeter = (2 + √3)·a
Step-by-step explanation:
Given: ΔABC is equilateral and AB = a
The diagram is given below :
AM is a median , PB ⊥ AB , PM = b
Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.
We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°
Since, AM is perpendicular bisector of BC. So,
Now in ΔAMB , By using Pythagoras theorem
Now, in ΔBMP :
Perimeter of ABM = AB + PB + PM + AM
Hence, Perimeter of ΔABP = (2 + √3)·a units
