19 thoughts on “Triangles – Examples 8

    • Kartik, the first one is a simple weighted average 12%….15%….20% so the ratio of original amounts should be 5:3 i.e. A has 5/8th of 1200 = 50. After spending 12% of this i.e. 90 he should be left with 660. Another way to look at it might be to realise that if he spent 12% he is left with 88% and chances are pretty good the amount left with him should be a multiple of 11….:)

      In the second, you have to note that the base changes for each percentage – so the last one must be measured on 68. Suppose the total marks are 100, then passing marks must be 40. So A gets 10% of 40 i,e, 4 less i.e 36 and B gets 11.11% of 36 i.e 4 less i.e 32. Together they score 68. So C must get not more than 28 less than 68 i.e (28/68) * 100 % i.e. 41 (3/17)% less to pass.

      regards
      J

      • J Sir, Thanks a lot 🙂
        2 . I did in exactly the same way : (1-x%)*68% of M = 40 % of M as i mentioned earlier. now I get x% = 28/68 = 7/17 . but we need to multiply by hundred and then we get the right answer( this is what confuses me ,all the time).
        This is very silly question but please help me when shud we multiply by 100 and when not ?. Instead of x% should i have used x ( as a ratio ) , and then multiply by 100.
        but what if i directly wanted to find the percentage value without multiplication.
        Then it shud be : (100 – x%)*68 of M = 40 % of M still not getting the right answer.

      • I don’t have a rule as to when to multiply by 100…I tend to think of the underlying logic, that a percentage is just a fraction * 100. Here I want the percentage change from 68 to 40 and hence the change is 28/68 as a fraction and 28/68 * 100 as a percentage.

        In your equation it would be (100-x)/100 * (68/100) * M = 40/100 * M (you forgot the % on the 68!) but it would still be painful to do. Equations should be your last resort, not the first….logic and understanding beats mechanical application of equations any day.

        regards
        J

  1. Dear J

    Why is the inradius (base+base-hypotenuse)/2 ? Is this a formula for the right angled triangles? Is there a simple proof for this?

    • This a very nice example.
      Sir I had a different start to this problem.
      I had noted down from one of earlier posts on this topic that for a right angled triangle
      1. Orthocentre lies at the vetex opp to hyp.
      2. Circumcentre is the midpoint of hyp.

      Also for any triangle Orthocentre,circumentre and centroid lie on the same line.
      Using that analogy D should have been the midpoint of side AC?
      Am I missing out something here?

      • Yes I think you are confusing points with lines. The three LINES are not the same. Draw all the relevant lines for a right angled triangle and see for yourself.

        regards
        J

  2. Sir, i will like to add one thing here. let I be the inrcentre of our primary tiangle, R be inradius, R1 inradius of right triangle and R2 of left one, then
    H=R+R1+R2
    BI= EF= R rt2 (directly 10rt2 in this case)
    R1^2+R2^2 = R^2
    aR2+cR1 = bR…. where a, b, c are three sides….

    and happy teachers day to you!!!!

  3. This a very nice example.
    Sir I had a different start to this problem.
    I had noted down from one of earlier posts on this topic that for a right angled triangle
    1. Orthocentre lies at the vetex opp to hyp.
    2. Circumcentre is the midpoint of hyp.

    Also for any triangle Orthocentre,circumentre and centroid lie on the same line.
    Using that analogy D should have been the midpoint of side AC?
    Am I missing out something here?

  4. Hi J,
    could you help me with this question

    Two circles with centres P and Q cut each other at two distinct points A and B.The circles have same radii and neither P nor Q falls within the intersection of the circles.What is the smallest range that includes all possible values of the angle AQP?

    I could see that max is 60 degrees.How can the minimum value be determined?

  5. Hi J,
    could you help me with this question

    Two circles with centres P and Q cut each other at two distinct points A and B.The circles have same radii and neither P nor Q falls within the intersection of the circles.What is the smallest range that includes all possible values of the angle AQP?

    I could see that max is 60 degrees.How can the minimum value be determined?

  6. It would go down to very nearly 0 I guess. So 0-60. But it is really difficult to answer a geometry question without a figure 😛 (That too entirely unrelated to the post above!)

    regards
    J

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