Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

In triangleABC , altitude AD = 18, median BE = 9√5 and median CF = 15. Find BC .

 



In



 


triangleABC, altitude AD = 18, median BE = 9√5

 and

median



 


CF = 15. Find BC

.


Grade:10

1 Answers

Latika Leekha
askIITians Faculty 165 Points
7 years ago
In the given triangle ABC, let AH be the median for BC. Let G be the centroid i.e. G is the point of intersection of the medians AD, BE and CF.
Let GH be the height of \Delta GBC and FK be the height of \DeltaFBC relative to the side BC.
Using the property of median that it divides in the ratio 1:2, we get
CG = 2/3 CF = 2/3. 15 = 10.
BG = 2/3 BE = 2/3 . 9√5 = 6√5.
This means \DeltaBFK is ismilar to \Delta BAD.
Hence, FK/AD = BF/BA.
Now BF= ½ BA (Since CF is the median)
Hence, FK = ½ AD = ½ . 18 = 9.
On similar lines we get, \DeltaCGH is similar to \DeltaCFK.
So, CG/CF = GH/FK
Hence, 10/15 = GH/FK
This means, GH = 2/3 FK = 2/3. 9 = 6.
Now, usinh pythagoras theorem in \DeltaGHB, we get
GH2 + HB2 = GB2.
Hence, 62 + HB2 = (6√5)2.
Hence, this gives, HB2 = 180-36 = 144
Hence, HB = 12.
Again using Pythagoras theorem in \DeltaGHC, we get
GH2 + HC2 = GC2.
62 + HC2 = 102.
Hence, HC = 8.
Now, BC = HB + HC = 12 + 8 = 20.
Hence, the length of BC is 20.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free