Question

in the given figure ∆ABC&∆AMP are right angled at B &M respectively given AC=10cm,AP=15&PM=12cm prove that i)∆ABC~∆AMP ii)Find AB & BC.
plz fast​

Answer 1

In ∆ABC and ∆AMP

angle B = angle M (90° given)

angle A = angle A (common angle)

by A. A similarity

∆ABC ~ ∆AMP

corresponding sides of the two similar triangle are in equal proportions

or

ratio of sides of two similar triangle are equal

therefore in ∆ABC and ∆AMP

AC/AP = BC/PM = AB/AM

10/15 = BC/12

BC = (10 × 12)/15

BC = 120/15 = 8

as ∆ABC is right angle triangle

by Pythagoras theorem

AC^2 = AB^2 + BC^2

${10}^{2} = {x}^{2} + {8}^{2} \\ 100 = {x}^{2} + 64 \\ {x}^{2} = 100 - 64 \\ {x}^{2} = 36 \\ x = \sqrt{36} = 6$

AB = 6 cm

hope you get your answer

Answer 2

refer to the images above

hope it helps