send mail to support@abhimanu.com mentioning your email id and mobileno registered with us! if details not recieved
Resend Opt after 60 Sec.
By Loging in you agree to Terms of Services and Privacy Policy
Claim your free MCQ
Please specify
Sorry for the inconvenience but we’re performing some maintenance at the moment. Website can be slow during this phase..
Please verify your mobile number
Login not allowed, Please logout from existing browser
Please update your name
Subscribe to Notifications
Stay updated with the latest Current affairs and other important updates regarding video Lectures, Test Schedules, live sessions etc..
Your Free user account at abhipedia has been created.
Remember, success is a journey, not a destination. Stay motivated and keep moving forward!
Refer & Earn
Enquire Now
My Abhipedia Earning
Kindly Login to view your earning
Support
Type your modal answer and submitt for approval
Find the sum of the medians of isosceles triangle, whose sides are 10, 10 and 12.
8 + 2√97 cm
10 + 2√97 cm
10 + 2√67 cm
None of these
In ΔABC, a = BC = 12 cm, b = AC = 10 cm and c = AB = 10 cm Let AD, BE and CF are the medians of ΔABC. In geometry, Apollonius' theorem is a theorem relating the length of a median of a triangle to the lengths of its side. It states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side". ∴ AB2 + AC2 = 2(AD2 + BD2) ∴ (10)2 + (10)2 = 2(AD2 + 62) ∴ 200 + 2(AD2 + 36) ∴ AD = 8 cm Using AB2 + BC2 = 2(BE2 + AE2 ), we get BE = √97 cm Similaryly, using AC2 + BC2 = 2(CF2 + AF2), we get CF = √97 cm Thus, AD + BE + CF = (8 + 2√97) cm
By: Amit Kumar ProfileResourcesReport error
Access to prime resources
New Courses