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ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?
- a)1+√5
- b)1−√5
- c)√7
- d)2√7
- e)None of the above
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
ABCD is a rectangle. The points P and Q lie on AD and AB respectively....
First draw the figure as described
let AB = CD = L, and AD =BC= B
now given QB = 2 we have AQ = L-2
if PD = K (let) we have PA = AP = (B- K)
now by given condition we have 3 tTriangleof equal areas
A ( PAQ) = A (QBC) = A (PCD)
all the three are right triangles where arae = 1/2 product of perpendicular sides
1/2 ( L-2) (B-K) = 1/2 (2B) = (1/2)(K)(L)
if we muliply by 2 we get
2B = (K)(L) gives K = 2B/L
substituting K = 2B/L in equality ( L-2) (B-K) = 2B = ( L-2) (B- 2B/L)
or multiply L on both sides gives 2L = (L-2)^2 or L^2 - 6L + 4 = 0
we solve for L we get L = [ - (-6) +_ √ (-6)^2 - 4 (1) (4) ]/ 2 = 3 +_ √5
so we have two values (3 + √5 ) and (3 - √5 ) for L = AB
now length AQ = AB - 2 or we have (3 + √5 - 2) = 1+ √5 which is the positive value allowed for length
The other value is (3 - √5 - 2) = 1- √5 < o so not allowed
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ABCD is a rectangle. The points P and Q lie on AD and AB respectively....
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ABCD is a rectangle. The points P and Q lie on AD and AB respectively....
Let's denote the length of AD as x and the length of AB as y.
Since ABCD is a rectangle, we have AD = BC = x and AB = CD = y.
Let's consider triangle PAQ first. Since the area of triangle PAQ is equal to the area of triangle QBC, and BQ = 2, we can conclude that the height of triangle PAQ (the perpendicular distance from A to PQ) is twice the height of triangle QBC (the perpendicular distance from Q to BC).
Let the height of triangle QBC be h. Then, the height of triangle PAQ is 2h.
Since the triangles PAQ and PCD have the same area, the base of triangle PAQ (PQ) must be twice the base of triangle PCD (CD). The base of triangle PCD is y, so the base of triangle PAQ is 2y.
Therefore, the length of AQ is y - PQ = y - 2y = -y.
However, since AQ is a length, it cannot be negative. Therefore, there is no solution to this problem as stated.
Since ABCD is a rectangle, we have AD = BC = x and AB = CD = y.
Let's consider triangle PAQ first. Since the area of triangle PAQ is equal to the area of triangle QBC, and BQ = 2, we can conclude that the height of triangle PAQ (the perpendicular distance from A to PQ) is twice the height of triangle QBC (the perpendicular distance from Q to BC).
Let the height of triangle QBC be h. Then, the height of triangle PAQ is 2h.
Since the triangles PAQ and PCD have the same area, the base of triangle PAQ (PQ) must be twice the base of triangle PCD (CD). The base of triangle PCD is y, so the base of triangle PAQ is 2y.
Therefore, the length of AQ is y - PQ = y - 2y = -y.
However, since AQ is a length, it cannot be negative. Therefore, there is no solution to this problem as stated.
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ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?a)1+√5b)1−√5c)√7d)2√7e)None of the aboveCorrect answer is option 'A'. Can you explain this answer? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?a)1+√5b)1−√5c)√7d)2√7e)None of the aboveCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?a)1+√5b)1−√5c)√7d)2√7e)None of the aboveCorrect answer is option 'A'. Can you explain this answer?.
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