Question

∆ABC, angle B= 90 degree BM perpendicular to AC ,BC = 5 cm and AC = 12 CM then find the ratio of areas of ∆ BMC and ∆ABC.

Answer 1

If angle B= 90 degree BM perpendicular to AC ,BC = 5 cm and AC = 12 CM then the ratio of areas of ∆ BMC and ∆ABC 25 : 144.

Step-by-step explanation:

It is given in ∆ABC,

∠B = 90° ….. (i)

BC = 5 cm  

AC = 12 cm

BM ⊥ AC

∴ ∠BMC = ∠BMA = 90° …. (ii)

Now, consider ∆BMC & ∆ABC,

∠C = ∠C ….. [common angle]

∠B = ∠BMC = 90° …… [from (i) & (ii)]

∴ By AA similarity, ∆BMC ~ ∆ABC

Since the ratio of the area of two similar triangles is equal to the square of ratio of their corresponding sides.

∴ [Area of ∆BMC] / [Area of ∆ABC] = [$\frac{BC}{AC}$]² = [$\frac{5}{12}$]² = $\frac{25}{144}$

Thus, the ratio of areas of ∆ BMC and ∆ABC is 25:144.

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Answer 2

Answer:

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