CBSE Class 10 Maths Question Paper 2018 Solution | Q25A

CBSE Solved Question Paper 4 Mark | Equilateral Triangles

Question 25A: In an equilateral ∆ABC, D is a point on side BC such BD = $$frac{1}{3}$BC. Prove that 9$AD)2 = 7(AB)2

NCERT Solution to Class 10 Maths

Explanatory Answer | CBSE Class 10 Math Paper 2018 Q25A

ABC is an equilateral triangle.
BD = $$frac{1}{3}$ BC Let the side of equilateral triangle measure 'a' units. Draw AH perpendicular to BC. In right triangle AHC, AH2 + HC2 = AC2 By SAS rule AHB and AHC are congruent, HC = $\frac{1}{2}$ BC = $\frac{1}{2}$ a In right triangle AHB, AH2 = AB2 - HB2 AH2 = a2 -$$$frac{1}{2}$ a)2 = a2 - $\frac{1}{4}$ a2 = $\frac{3}{4}$ a2 In right triangle ADH, AD2 = AH2 + HD2 HD = BH - BD = $\frac{1}{2}$a - $\frac{1}{3}$a = $\frac{1}{6}$a ∴ AD2 = $\frac{3}{4}$a2 +$$$frac{1}{6}$a)2 = $\frac{3}{4}$a2 + $\frac{1}{36}$a2 = $\frac{27a^2 + a^2}{36}$ = $\frac{28}{36}$a2 = $\frac{7}{9}$a2 ∴ AD2 = $\frac{7}{9}$$AB)2

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