ΔABC is an isosceles triangle and $$\overline {AB} $$ = $$\overline {AC} $$ = 2a unit, $$\overline {BC} $$ = a unit. Draw $$\overline {AD} $$ ⊥ $$\overline {BC} $$ , and find the length of $$\overline {AD} $$
A. $$\sqrt {15} $$ a unit
B. $$\frac{{\sqrt {15} }}{2}$$ a unit
C. $$\sqrt {17} $$ a unit
D. $$\frac{{\sqrt {17} }}{2}$$ a unit
Answer: Option B
Solution(By Examveda Team)
According to question,Given :
AB = AC = 2a
BC = a
AD ⊥ BC
In isosceles triangle perpendicular sides bisects the opposite side of the length
∴ BD = $$\frac{{BC}}{2}$$
BD = $$\frac{a}{2}$$
In ΔADB using Pythagoras theorem
AB2 = BD2 + AD2
(2a)2 = $${\left( {\frac{a}{2}} \right)^2}$$ + AD2
4a2 = $$\frac{{{a^2}}}{4}$$ + AD2
AD2 = 4a2 - $$\frac{{{a^2}}}{4}$$
AD = $$\frac{{\sqrt {15} }}{2}$$ a units
Related Questions on Triangles
If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is:
A. 50°
B. 70°
C. 60°
D. 80°
In the following figure which of the following statements is true?
A. AB = BD
B. AC = CD
C. BC + CD
D. AD < Cd
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