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3 years ago

15

In △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP . Find the area of △ABC if the area of △BMP

is equal to 21m2.

2 answers:

ddd [48]3 years ago

5 0

M is mid point of CP. M will divide the $\Delta$ BPC in two equal parts $\Delta$ BMC and $\Delta$ BMP.

Area of $\Delta$ BMP is equal to 21m^2

Since, $\Delta$ BMC = $\Delta$ BMP

Area of $\Delta$ BPC = Area of $\Delta$ BMC +Area of $\Delta$ BMP = 21 + 21 = 42m^2

and since ratio of AP:BP =1:3 so the area of $\Delta$ BMP will be 1/3 of Area of $\Delta$ ABC

hence, Area of $\Delta$ ABC = 63m^2

OLga [1]3 years ago

3 0

Answer: 56 m^2

Step-by-step explanation:

The areas of triangle BMP and CMB are equal because the base and height are the same. Therefore, <em>A</em>BMP=<em>A</em>CMB=21m^2 which is due to transitivity.
<em>A</em>PCB=<em>A</em>BMP+<em>A</em>CMB=21+21=42m^2
The ratio of the areas of triangles ACP to PCB is the same as their bases AP and BP which is 1:3. This is due to area of triangles.
After that, we use Substitution and Algebra to calculate the area of: <em>A</em>ACP= $42 / 3 * 1 = 14 * 1 = 14$
Finally, we add all of the areas together. (Or you could've just multiplied in the last equation by 4 rather than 3 to get the <em>A</em>ABC sooner.) <em>A</em>ABC=<em>A</em>ACP+<em>A</em>PCB=14+42=56m^2

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$P(B) = 0.77$

$P(A\ n\ B) = 0.22$

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Given

See attachment for proper data presentation

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

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