ABC is an acute angled triangle and CD be the altitude through C. If AB = 6 cm and CD = 8 cm, then find the distance between mid-points of AD and BC.

Asked by a.behera67 | 2nd Jan, 2016, 09:19: PM

Expert Answer:

 
Let space straight P space and space straight Q space be space the space midpoints space of space AD space and space BC. Draw space QR space perpendicular space to space AB. In space increment BDC space and space increment BRQ angle straight D space equals space angle straight R space equals 90 degree angle straight B equals angle straight B therefore increment BDC space almost equal to space increment BRQ space left parenthesis AA space test space of space similarity right parenthesis Using space converse space of space BPT space theorem space we space can space say space that comma space since space straight Q space is space midpoint space of space BC space so space straight R space is space midpoint space of space BD.  straight P space is space midpoint space of space AD space and space straight R space is space midpoint space of space DB AP plus RB equals PD plus DR space space left parenthesis because RB space equals space DR space and space AP space equals space PD right parenthesis therefore PR space equals 1 half AB equals 4  Also comma QR space equals 1 half CD equals 3 space open parentheses because increment BDC space almost equal to space increment BRQ space and space using space BPT space theorem space close parentheses therefore Applting space pythagorean space triplet space 3 minus 4 minus 5 space PQ space must space be space space 5 space unnits.

Answered by Vijaykumar Wani | 4th Jan, 2016, 10:48: AM