Dr Rick

@GonetoBlueSky8 @codek_tv No, it gives m²+n²+2mn=16, and if course mn=1 by definition. Hope that helps.

@rfbooth @codek_tv Squaring m+n gives m^2+n^2+2m =16…..

Can you solve This?

@hope_dies_again @Ginger_Tucci @CJ_7171 I assume you are referring here to the Palestine Action activists, who have not in fact been found guilty of any offence. No doubt you know more about all that than the jury did, though.

@Ginger_Tucci @CJ_7171 He has a point when their unelected leader celebrates smashing a policewoman's back with a sledgehammer and people still voted for them. What sort of crazy mass hallucination is that?

We knew this country has been in huge, huge danger the last 10 yrs or so but this is a dark day for everyone when this lot win an election. If you voted for these sectarian, anti-British lunatics you’re a treacherous disease to this country. Manchester is a disgrace

@SilverBali7 Nice.

@MatildaSpicer I think we need to consider some lengths.

ABCD is a square. AC = CE, ∠ACE = 30° Find the measure of ∠AED. #mathpuzzle #Geometry #GeometricProblem

@MatildaSpicer A=25 (segment) So alpha=55 (triangle with exterior 80) So x+55=100 So x=45.

Find the measure of ∠ACB. #mathpuzzle #Geometry #GeometricProblem

@MatildaSpicer 5cos(arccos(⅗)-45°)=(3+4)/√2 =7/√2, so 49/2.

Find the area of square ABCD. #mathpuzzle #Geometry #GeometricProblem

@ybgoi Mass points for this configuration are A2, B4, C1, giving D6 and E3. Thus DP:PC=1:6, and [ADP]=⅔×⅙×[ABC]=⅑[ABC]. Now let vector BC=p and BA=q. So DC is p-⅓q, PA=-⅐(p-⅓q)+⅔q. So directions are 3p-q, -p+5q. The dot is -3p.p+16p.q-5q.q=0, so perpendicular.

i) Prove that AP ⊥ DC ii) Compute the shaded fraction of equilateral ∆ABC?

@viterick Thirty.

Find "a+b" (By Thiago Durado)

@sonukg4india Thirty.

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@SilverBali7 Three tenths.

@ai_meh_s @Asadthegeek @BiancaDante27 @viterick @il1123 @ohmohm I don't mean to be rude, and I appreciate the compliment inherent, but I now have about sixty notifications for these threads. Please don't tag me in problems (unless I wrote them).

α=?

@mnd233445 Shoelace Theorem for the easy win.

Shaded area = ?

@MatildaSpicer Lovely.

∠ABC = 45° ∠ADB = ∠BEC = 90° AE = 3, CE = 7 BF = ? #mathpuzzle #Geometry #GeometricProblem

@sonukg4india Masses A(1), B(1), C(2) give F at centre. So CD=FD and AF=3EF. So [DAF]=[CAF]=3[CEF], which means the answer is ⅓. The equal angles are irrelevant.

#morning with #math Given a triangle as shown. Find the ratio of the area of triangle CEF to the area of triangle DAF.

@MatildaSpicer Sixty-three.

AC || DE BD is the bisector of angle ∠ADC. ∠BFC = ? #mathpuzzle #Geometry #GeometricProblem

@viterick Nice.

Find shaded area

@sonukg4india 121.

Given a rectangle as shown. Find the value of 6xy. #Geometry #Rectangle

@ai_meh_s @BiancaDante27 @izzettinbingol @OkanAtalay1970 @HarishMamania @Bolsotri1 @mwns999 @Abdurrahma35914 @cnhankeli1 @mnk68636175 Twenty.

I: Incenter DC=?

@sonukg4india Beautiful. Took me a while!

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