Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
10.9k views
in Pythagoras Theorem by (47.6k points)
closed by

∆ABC is an equilateral triangle. Point P is on base BC such that PC = 1/3 BC, if AB = 6 cm find AP.

Given: ∆ABC is an equilateral triangle. 

PC = 1/3 BC, AB = 6cm. 

To find: AP 

Consttuction: Draw seg AD ± seg BC, B – D – C.

1 Answer

+1 vote
by (46.9k points)
selected by
 
Best answer

∆ABC is an equilateral triangle. 

∴ AB = BC = AC = 6cm [Sides of an equilateral triangle] 

pc = 1/3 BC [Given]

= 1/3 (6) 

∴ PC = 2cm 

In ∆ADC, 

∠D = 90° [Construction] 

∠C = 60° [Angle of an equilateral triangle] 

∠DAC = 30° [Remaining angle of a triangle] 

∴ ∆ ADC is a 30° – 60° – 90° triangle.

∴ AD = √3/2 AC [Side opposite to 60°] 

∴ AD = √3/2  (6) 

∴ AD = 3√3 cm

∴ CD = 1/2 AC [Side opposite to 30°] 

∴ CD = 1/2 (6) 

∴ CD = 3cm 

Now DP + PC = CD [D – P – C] 

∴ DP + 2 = 3 

∴ DP = 1cm 

In ∆ADP, 

∠ADP = 900

AP2 = AD2 + DP2 [Pythagoras theorem] 

∴ AP2 = (3√3)2 + (1)2 

∴ AP2 = 9 × 3 + 1 = 27 + 1 

∴ AP2 = 28 

∴ AP = √28 

∴ AP = (√4 x 7)

∴ AP = 2√7 cm 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...