Question

in the given figure,triangle ABC and triangle AMP right angled at B and M respectively given AC=10cm AP=15cm and PM=12cm prove that triangle ABC is similar to triangle AMP and also find AB and BC​

Answer 1

Answer:

Step-by-step explanation:

In triangle ABC and AMP

Angle ABC = angle AMC (given)

Angle A = angle A (common)

Triangle ABC ~ Triangle AMP

AB/AM=BC/MP=AC/AP

BC/PM=AC/AP

BC/12=10/15

BC=8cm

AB =1

Answer 2

Given :  AC=10 cm  AP=15 cm and PM=12 cm

To prove: $\bigtriangleup ABC \sim \bigtriangleup AMP$

To find : AB and AC

Solution :

in triangle ABC and AMP

∠BAC = ∠PAM (common)

∠ABC = ∠PMA (each 90)

thus

by AA similarity

$\bigtriangleup ABC \sim \bigtriangleup AMP$

NOW ,

$AM = \sqrt{AP^2-PM^2}\\\\AM = \sqrt{15^2-12^2}=9$

since

the triangles are similar

$\frac{AB}{AM}= \frac{BC}{PM}=\frac{AC}{AP}\\\\\frac{AB}{9}= \frac{BC}{12}=\frac{10}{15}\\\\\frac{AB}{9}= \frac{10}{15}\\\\AB = 6 cm\\\\\frac{BC}{12}=\frac{10}{15}\\\\BC = 8 cm$

hence AB = 6 cm and BC = 8 cm

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