Question

CD is altitude of acute triangle if AB=8cm and cd=6cm .find the distance between mid points of AD and BC​

Answer 1

hey mate...

here is your answer...

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

Solution:  

Given: ΔABC is an acute-angled Triangle, CD is the altitude

AB = 8units, CD = 6units.

To Find: The length of the mid-points of AD and BC.

Constructions: Let P and Q be the midpoints of AD and AB

respectively  Draw QR perpendicular to AB.

Working:

QR ⊥ AB; CD ⊥ AB

∴ΔBQR ≈ ΔBCD

∴ R is the midpoint of BD  (∵ Q is the midpint of BC)

∴ RB + PA = RD + DP  (∵ RB = RD and PA = DP)  

∴ PR =1/2×AB = 4

Also, QR = 1/2 ×CD = 3 (∵ ΔBQR ≈ ΔBCD)

∴ Applying pythagorean triplet 3-4-5 PQ must be 5 units.

hope it helps....

Answer 2

SOLUTION

Given,

∆ABC is an acute-angled ∆, CD is the altitude.

AB= 8units & CD= 6units

To find:

The length of the mid points of AD&BC.

Construction:

Let P & Q be the midpoints of AD & AB respectively.Draw QRperpendicular to AB

Working:

QR⊥AB ; CD⊥AB

therefore, ∆BQR = ∆BCD

R is the midpoints of BD.&

Q is the midpoints of BC.

=) RB+PA= RD+DPB (RB= RD & PA= DP)

PR= 1/2× AB= 4

QR= 1/2× CD= 3 (∆BQR ≈∆BCD)

Applying Pythagorean triplet 3-4-5 PQ must be 5 units.

hope it helps ☺️