Question

In ∆ABC, angle BCA is a right angle, if Q is the mid point of the side BC, AC= 4 cm and AQ=5 cm, find (AB^2)​

Answer 1

Answer:

in triangle,AQC,

AQ^2 = CQ^2 + AC^2

therefore, CQ^2=9

               CQ=3cm

if q is midpoint of BC therefore BC=2*CQ

                                                        =6cm

therefore, In triangle ABC,

By pythagoras theorem, AB^2= AC^2+ BC^2

                                       AB^2 = 52cm^2

or if u want 2*AB=4√13

Plz mark as brainliest

Step-by-step explanation:

Answer 2

Answer:

triangle ABC the right angle is C, the sides are AC and BC, the hypotenuse is AB. Q is the midpoint of BC, so (CQ) = 1/2 (BC). And so (CQ)^2 = 1/4 (BC)^2 or (BC)^2 = 4(CQ)^2

By Pythagoras in triangle ACQ, (AQ)^2 = (CQ)^2 + (AC)^2

(CQ)^2 = (AQ)^2 - (AC)^2

4(CQ)^2 = 4(AQ)^2 - 4(AC)^2

By Pythagoras in triangle ABC: (AB)^2 = (BC)^2 + (AC)^2 then substituting:

(AB)^2 = 4(CQ)^2 + (AC)^2

(AB)^2 = 4(AQ)^2 - 4(AC)^2 + (AC)^2 = 4(AQ)^2 - 3(AC)^2

Q.E.D.

Step-by-step explanation:

Hope it helps