Can you figure this out?!

Geometry Level 3

Given above is a triangle within which there are three smaller triangles of areas 2, 3 and 4 as indicated. Find the area of the remaining quadrilateral.


The answer is 7.8.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
Nov 18, 2018

Labelled the figure as above. Split the quadrilateral into two triangles with area a a and b b . Since △ C M P \triangle CMP and △ C B P \triangle CBP are sharing part of a base and having the same height. The ratio of their areas [ C M P ] [ C B P ] = M P B P = 1 2 \dfrac {[CMP]}{[CBP]} = \dfrac {MP}{BP} = \dfrac 12 . Similarly, [ B N P ] [ C B P ] = N P C P = 3 4 \dfrac {[BNP]}{[CBP]} = \dfrac {NP}{CP} = \dfrac 34 . Also

{ [ A M P ] [ A B P ] = 1 2 ⟹ b a + 3 = 1 2 ⟹ 2 b = a + 3 . . . ( 1 ) [ A N P ] [ A C P ] = 3 4 ⟹ a b + 2 = 3 4 ⟹ 4 a = 3 b + 6 . . . ( 2 ) \begin{cases} \dfrac {[AMP]}{[ABP]} = \dfrac 12 & \implies \dfrac b{a+3} = \dfrac 12 & \implies 2b = a + 3 & ...(1) \\ \dfrac {[ANP]}{[ACP]} = \dfrac 34 & \implies \dfrac a{b+2} = \dfrac 34 & \implies 4a = 3b + 6 & ...(2) \end{cases}

From ( 1 ) : a = 2 b − 3 (1): a = 2b - 3 and ( 2 ) : 4 ( 2 b − 3 ) = 3 b + 6 ⟹ b = 3.6 (2): 4(2b-3) = 3b+6 \implies b = 3.6 and a = 4.2 a=4.2 . Therefore, [ A M P N ] = a + b = 7.8 [AMPN] = a+b = \boxed{7.8} .

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Shashi Kamal
Nov 17, 2018

Using ladder theorem. 1/area of triangle + 1/4=1/(4+3) +1/(4+2). Solving this we get area of ∆=84/5 so unknown area =84/5 - 9=39/5=7.8. Or can also be solved using mpg+area base property.

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@Niraj Sawant the area marked with ? isn't a triangle as you said in the problem

Henry U - 2 years, 6 months ago

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Thanks for pointing that out, I have corrected that.

A Former Brilliant Member - 2 years, 6 months ago

Thank you for your solution. You should use Latex for better presentation of your solution. Also, you should put the link of Wiki Page of the theorems you use in your solutions, so for people who dont know about them, can see the wiki.

A Former Brilliant Member - 2 years, 6 months ago

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