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Apr 2808:00 AM PDT
-11:00 AM PDT
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18 Sep 2025, 08:57
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amitdgr
This is exactly the same as asking the number of rectangles in a p × q grid
To form a rectangle in a p × q grid, we need to select 2 horizontal and 2 vertical lines, i.e. simply pC2 × qC2
In this question, we have 11 horizontal lines (from y=6 to 16) and 10 vertical lines (from x=-4 to 5)
Thus, the number of rectangles = 11C2 × 10C2 = (11×10/2) × (10×9/2) = 55×45
Now, let us focus on the triangle - the vertex P can be any of the 4 vertices of the above rectangle, hence we can form the triangle in 4 ways for each rectangle
Thus, the number of triangles = 55×45×4 = 110×90 = 9900
Ans C
26 Sep 2025, 02:55
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Let's work through this together.
Understanding What We're Looking For
You need to count right triangles where:
- The right angle is at point P
- Side PR is parallel to the x-axis (horizontal)
- All vertices have integer coordinates within \(-4 \leq x \leq 5\) and \(6 \leq y \leq 16\)
The Critical Insight
Here's what you need to see: If PR is horizontal (parallel to x-axis) and the angle at P is \(90^{\circ}\), then PQ must be vertical (perpendicular to PR).
Think about it this way - if one side from P goes horizontally, and we need a right angle at P, the other side from P must go vertically. This is the key that unlocks the entire problem!
What This Means for Our Coordinates
Let's say P is at position \((x_1, y_1)\). Then:
- Point R must be at \((x_2, y_1)\) where \(x_2 \neq x_1\) (same row, different column)
- Point Q must be at \((x_1, y_2)\) where \(y_2 \neq y_1\) (same column, different row)
Counting the Triangles
Now we can count systematically:
Step 1: How many positions for P?
We have 10 x-values (from -4 to 5) and 11 y-values (from 6 to 16)
So P has \(10 \times 11 = 110\) possible positions
Step 2: For each P, how many positions for R?
R needs the same y-coordinate as P but a different x-coordinate
There are 10 total x-values, minus P's x-coordinate = 9 choices
Step 3: For each P, how many positions for Q?
Q needs the same x-coordinate as P but a different y-coordinate
There are 11 total y-values, minus P's y-coordinate = 10 choices
Step 4: Apply multiplication principle
Total triangles = \(110 \times 9 \times 10 = 9,900\)
Answer: (C) 9,900
Notice how the perpendicularity constraint dramatically simplified our counting - once you realize PQ must be vertical when PR is horizontal, the rest follows naturally!
Want to Master This Question Type?
You can check out the step-by-step solution on Neuron by e-GMAT to master the systematic framework that works for all coordinate geometry counting problems. The full solution shows you three alternative approaches and reveals patterns that help you recognize similar constraints instantly in other problems. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice here.
Understanding What We're Looking For
You need to count right triangles where:
- The right angle is at point P
- Side PR is parallel to the x-axis (horizontal)
- All vertices have integer coordinates within \(-4 \leq x \leq 5\) and \(6 \leq y \leq 16\)
The Critical Insight
Here's what you need to see: If PR is horizontal (parallel to x-axis) and the angle at P is \(90^{\circ}\), then PQ must be vertical (perpendicular to PR).
Think about it this way - if one side from P goes horizontally, and we need a right angle at P, the other side from P must go vertically. This is the key that unlocks the entire problem!
What This Means for Our Coordinates
Let's say P is at position \((x_1, y_1)\). Then:
- Point R must be at \((x_2, y_1)\) where \(x_2 \neq x_1\) (same row, different column)
- Point Q must be at \((x_1, y_2)\) where \(y_2 \neq y_1\) (same column, different row)
Counting the Triangles
Now we can count systematically:
Step 1: How many positions for P?
We have 10 x-values (from -4 to 5) and 11 y-values (from 6 to 16)
So P has \(10 \times 11 = 110\) possible positions
Step 2: For each P, how many positions for R?
R needs the same y-coordinate as P but a different x-coordinate
There are 10 total x-values, minus P's x-coordinate = 9 choices
Step 3: For each P, how many positions for Q?
Q needs the same x-coordinate as P but a different y-coordinate
There are 11 total y-values, minus P's y-coordinate = 10 choices
Step 4: Apply multiplication principle
Total triangles = \(110 \times 9 \times 10 = 9,900\)
Answer: (C) 9,900
Notice how the perpendicularity constraint dramatically simplified our counting - once you realize PQ must be vertical when PR is horizontal, the rest follows naturally!
Want to Master This Question Type?
You can check out the step-by-step solution on Neuron by e-GMAT to master the systematic framework that works for all coordinate geometry counting problems. The full solution shows you three alternative approaches and reveals patterns that help you recognize similar constraints instantly in other problems. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice here.
