in Quantitative Aptitude edited by
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13 votes

Right triangle $PQR$ is to be constructed in the $xy$ - plane so that the right angle is at $P$ and line $PR$ is parallel to the $x$-axis. The $x$ and $y$ coordinates of $P, Q,$ and $R$ are to be integers that satisfy the inequalities: $−4\leq x\leq 5$ and $6 \leq y \leq16.$ How many different triangles could be constructed with these properties?

  1. $110$
  2. $1,100$
  3. $9,900$
  4. $10,000$
in Quantitative Aptitude edited by

2 Answers

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Here $PR$ is parallel to $x-$axis so $y-$ coordinate of vertex $P$ and $R$ would always be the same.

For ex- If $P= (-1, \textbf{7})$ then $R=( 2, \textbf{7})$ or $(5, \textbf{7})$

So total number of possible  $x-$ coordinate of $P$ and $R$ for a particular $y$ $( -4\leq x \leq 5) = {}^{10}P_2 = 90.$

Similarly, for $11$ different $y-$ coordinates, total coordinates of $P$ and $R = 11 \times 90 = 990.$

Now since, its a right angle at $P$ so vertex $Q$ has same $x$-coordinate as $P.$ So total possible coordinates for vertex $Q ( 6\leq y\leq 16) = {}^{10}C_1= 10.$

For ex- If $P= (\textbf{-1}, 7) $ then $Q= ( \textbf{-1}, 8)$ or $( \textbf{-1}, 16)$ but not $(\textbf{-1}, 7).$

Total number of triangles $= 10 * 990 = 9900$

Ans- C
edited by


for more clarity add this also.         there 11  ( 6 to 16) points in y coordinate to choose for Q still we are using 10C1 because we are leaving the one y coordinate point which has been chosen for P and R , otherwise we will not be able to draw triangle ,  it will form a straight line in same P,R,Q      Y coordinate case.
0 votes
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So the question states that how many different right angled triangles can be formed such that one of its edge is parallel to X axis. Thus to select such points  for P- 10*11

                                                            for Q- 10

                                                            for R- 9

Thus total right angles triangles satisfying the condition are= 10*11*10*9= 9900


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