#### Question

In the given figure ΔABC and ΔAMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

1) Prove ΔABC ~ ΔAMP

2) Find AB and BC.

#### Solution

In ABC and AMP

∠ABC = ∠AMP (each 90°)

∠BAC = ∠PAM (common)

∴ΔABC ~ ΔAMP (By AA similarity)

Since the triangles are similar, we have

`(AB)/(AM) = (BC)/(MP) = (AC)/(AP)`

`=> (AB)/(AM) = (BC)/(MP) = (AC)/(AP)`

`=> (AB)/(AM) = (BC)/12 = 10/15`

Taking `(BC)/12 = 10/15 => BC = (12 xx 10)/15` = 8 cm

Now, using Pythagoras theorem in ΔABC

`(AB)^2 + (BC)^2 = (AC)^2`

`=> (AB)^2 = 10^2 - 8^2 = 36`

=> AB = 6 cm

Hence AB = 6 cm and BC = 8 cm

Is there an error in this question or solution?

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#### APPEARS IN

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In the Given Figure δAbc and δAmp Are Right Angled at B and M Respectively. Given Ac = 10 Cm, Ap = 15 Cm and Pm = 12 Cm 1) Prove δAbc ~ δAmp 2) Find Ab and Bc. Concept: Similarity of Triangles.

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