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LearnQuantitative AptitudeGeometry & Mensuration

04.02 · Geometry & Mensuration

Triangles

Elite lesson14 formulas10 questions

On the syllabus of

Elite lesson28 minEnglish

Formulas

At a glance 14 formulas

  1. 1Third side∣b−c∣<a<b+c|b - c| < a < b + ca,b,c=sides; any two exceed the thirda, b, c = \text{sides; any two exceed the third}
  2. 2Acute, right or obtusec2<a2+b2: acutec^2 < a^2 + b^2:\ \text{acute}c2>a2+b2: obtuse; c=longest sidec^2 > a^2 + b^2:\ \text{obtuse};\ c = \text{longest side}
  3. 3Foot of an altitudeBM=c2+a2−b22aBM = \dfrac{c^2 + a^2 - b^2}{2a}M=foot from A; a,b,c opposite A,B,CM = \text{foot from } A;\ a, b, c \text{ opposite } A, B, C
  4. 4Pythagorean triplets
    3, 4
    5
    5, 12
    13
    8, 15
    17
    7, 24
    25
    20, 21
    29
    9, 40
    41
  5. 5Area12×base×height=12absin⁡C\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2}ab\sin CC=angle between sides a and bC = \text{angle between sides } a \text{ and } b
  6. 6Heron's formulaarea=s(s−a)(s−b)(s−c)\text{area} = \sqrt{s(s - a)(s - b)(s - c)}a,b,c=sides, s=a+b+c2a, b, c = \text{sides},\ s = \tfrac{a + b + c}{2}
  7. 7Equilateral, side aheight=32a, area=34a2\text{height} = \tfrac{\sqrt{3}}{2}a,\ \text{area} = \tfrac{\sqrt{3}}{4}a^2inradius=a23, circumradius=a3\text{inradius} = \tfrac{a}{2\sqrt{3}},\ \text{circumradius} = \tfrac{a}{\sqrt{3}}
  8. 8Inradius r, circumradius Rr=areas, R=abc4×arear = \dfrac{\text{area}}{s},\ R = \dfrac{abc}{4 \times \text{area}}s=a+b+c2; hyp. c: R=c2, r=a+b−c2s = \tfrac{a + b + c}{2};\ \text{hyp. } c:\ R = \tfrac{c}{2},\ r = \tfrac{a + b - c}{2}
  9. 9Altitude to hypotenuseh=abch = \dfrac{ab}{c}a,b=legs, c=hypotenusea, b = \text{legs},\ c = \text{hypotenuse}
  10. 10Similar trianglessides in ratio k⇒areas in k2\text{sides in ratio } k \Rightarrow \text{areas in } k^2DE∥BC: ADDB=AEEC; D,E on AB,ACDE \parallel BC:\ \tfrac{AD}{DB} = \tfrac{AE}{EC};\ D, E \text{ on } AB, AC
  11. 11Same heightTriangles with a common apex and bases on one line have areas in the ratio of their bases, so a median cuts a triangle into two equal halves.
  12. 12Angle bisector theoremBDDC=ABAC\dfrac{BD}{DC} = \dfrac{AB}{AC}AD bisects ∠A, D on BCAD \text{ bisects } \angle A,\ D \text{ on } BC
  13. 13Median AD, centroid GAB2+AC2=2(AD2+BD2)AB^2 + AC^2 = 2(AD^2 + BD^2)D=midpoint of BC; AG:GD=2:1D = \text{midpoint of } BC;\ AG : GD = 2 : 1
  14. 14Median as half sideIf the median from A is half of BC, the midpoint of BC is equally far from A, B and C, so ∠A=90∘\angle A = 90^\circ. The converse holds too.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    Which of these sets of lengths, in cm, cannot be the sides of a triangle?

    Correct answer6,7,146, 7, 14

    Why

    Triangle inequality: the two shorter sides must add to more than the longest.

    4+5>84 + 5 > 8, 5+6>105 + 6 > 10 and 7+8>127 + 8 > 12 all pass, but 6+7=136 + 7 = 13 falls short of 14.

    So 6, 7, 14 cannot form a triangle.

  2. Q2Easy

    A cyclist rides 20 km due east and then 21 km due north. How far is she from her starting point, in a straight line?

    Correct answer29 km

    Why

    Spot the triplet 20, 21, 29.

    The two legs of the ride meet at a right angle, so the distance is 400+441=841\sqrt{400 + 441} = \sqrt{841}; adding the legs gives the trap 41.

    So she is 29 km from the start.

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