5 Marks
32
In a rectangular park measuring $50\text{ m} \times 40\text{ m}$, the gram panchayat plans to construct a rectangular swimming pool in the middle, surrounded by a uniform-width grass strip throughout the park. Find the dimensions of the swimming pool, if the area of the grassy strip is $1184\text{ m}^2$.
Concept:
1
This is a classic quadratic equation word problem. The area of the grass strip is equal to the total area of the park minus the area of the inner swimming pool. Since the strip has a uniform width ($x$) on all four sides, the pool's length is reduced by $2x$ and its breadth is reduced by $2x$.
Hint:
1
Area of Park = $50 \times 40 = 2000\text{ m}^2$.
2
Pool Dimensions: Length = $(50 - 2x)$, Breadth = $(40 - 2x)$.
3
Set up equation: $2000 - (50 - 2x)(40 - 2x) = 1184$.
4
Expand, simplify to a standard quadratic equation, and factorize to find $x$.
Mistakes to Avoid:
!
Subtracting only $1x$: The most common error is defining the pool's length as $50 - x$. The grass strip exists on the left AND right, so you must subtract $2x$ from the length. Same for top and bottom ($2x$ from the breadth).
!
Stopping at $x$: The question asks for the "dimensions of the swimming pool", not the width of the grass strip. Once you find $x$, you must substitute it back into $50 - 2x$ and $40 - 2x$.
Official CBSE Solution:
Let '$x$' be the uniform width of the grass strip.
Then, length of the swimming pool, $l = (50 - 2x)\text{ m}$
and breadth of the swimming pool, $b = (40 - 2x)\text{ m}$
Area of rectangular park $-$ Area of swimming pool = Area of grass strip.
$50 \times 40 - (50 - 2x)(40 - 2x) = 1184$
$2000 - (2000 - 100x - 80x + 4x^2) = 1184$
$2000 - 2000 + 180x - 4x^2 = 1184$
$4x^2 - 180x + 1184 = 0$
Divide the entire equation by 4:
$x^2 - 45x + 296 = 0$
Factorizing by splitting the middle term ($37 \times 8 = 296$):
$(x - 8)(x - 37) = 0$
$x = 8 \text{ or } x = 37$
Rejecting $x = 37$, as length and breadth cannot be negative integers ($40 - 2(37) < 0$).
Therefore, $x = 8$.
Calculate dimensions:
Length of the swimming pool $= 50 - 2(8) = 50 - 16 = 34\text{ m}$.
Breadth of the swimming pool $= 40 - 2(8) = 40 - 16 = 24\text{ m}$.
Ans: Dimensions are 34 m and 24 m.
🎯 CBSE Marking Scheme (5 Marks):
• Identifying length $(50-2x)$ & breadth $(40-2x)$: 1 Mark (½ + ½)
• Forming the correct equation $50 \times 40 - (50-2x)(40-2x) = 1184$: 1 Mark
• Simplifying to quadratic $x^2 - 45x + 296 = 0$: 1 Mark
• Finding roots $x=8, 37$: ½ Mark
• Rejecting $x=37$ and stating $x=8$: 1 Mark
• Calculating final length and breadth: ½ Mark
• Identifying length $(50-2x)$ & breadth $(40-2x)$: 1 Mark (½ + ½)
• Forming the correct equation $50 \times 40 - (50-2x)(40-2x) = 1184$: 1 Mark
• Simplifying to quadratic $x^2 - 45x + 296 = 0$: 1 Mark
• Finding roots $x=8, 37$: ½ Mark
• Rejecting $x=37$ and stating $x=8$: 1 Mark
• Calculating final length and breadth: ½ Mark
5 Marks
33
Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, divides the other two sides in the same ratio.
Using the above theorem solve the following:
PQRS is a trapezium with $PQ \parallel SR$. X and Y are points on non-parallel sides PS and QR respectively such that $XY \parallel PQ$. Show that $\frac{PX}{XS} = \frac{QY}{YR}$.
Using the above theorem solve the following:
PQRS is a trapezium with $PQ \parallel SR$. X and Y are points on non-parallel sides PS and QR respectively such that $XY \parallel PQ$. Show that $\frac{PX}{XS} = \frac{QY}{YR}$.
Concept:
1
Part 1: State and prove the Basic Proportionality Theorem (BPT/Thales Theorem) formally.
2
Part 2: Apply BPT to a trapezium. By drawing a diagonal ($PR$), you split the trapezium into two separate triangles ($\triangle PSR$ and $\triangle PQR$). Because $PQ \parallel XY \parallel SR$, the line $XY$ cuts both triangles parallel to their bases. Apply BPT to both triangles individually and equate the ratios through the shared intersection point $M$.
Hint:
1
Construction: Join PR. Let it intersect XY at M.
2
In $\triangle PSR$, $XM \parallel SR$. Apply BPT to get a ratio equal to $\frac{PM}{MR}$.
3
In $\triangle PQR$, $MY \parallel PQ$. Apply BPT to get a ratio equal to $\frac{PM}{MR}$. Equate them.
Mistakes to Avoid:
!
Skipping the construction: You cannot apply BPT to a 4-sided figure directly. BPT strictly applies to *triangles*. Drawing the diagonal is mandatory to create the triangles required for the proof.
Official CBSE Solution:
Part 1: Proof of BPT
(Standard textbook proof of Basic Proportionality Theorem goes here. Ensure you include the Given, To Prove, Construction, and Figure).
Part 2: Application in Trapezium
Construction: Join diagonal PR intersecting XY at M.
Since $XY \parallel PQ$ and $PQ \parallel SR$, it implies $XY \parallel SR$.
Therefore, the segment $XM \parallel SR$ and $MY \parallel PQ$.
In $\triangle PSR$, $XM \parallel SR$. By BPT:
$\frac{PX}{XS} = \frac{PM}{MR}$ --- (i)
In $\triangle PQR$, $MY \parallel PQ$. By BPT:
$\frac{QY}{YR} = \frac{PM}{MR}$ --- (ii)
From (i) and (ii), both ratios are equal to $\frac{PM}{MR}$. Thus:
$\frac{PX}{XS} = \frac{QY}{YR}$
Ans: Hence Proved.
🎯 CBSE Marking Scheme (5 Marks):
• BPT: Correct figure, given, to prove, construction: ½ Mark
• BPT: Correct mathematical proof: 2 Marks
• Trapezium: Drawing diagonal PR (Construction): ½ Mark
• Trapezium: Applying BPT in $\triangle PSR$ (Eq i): 1 Mark
• Trapezium: Applying BPT in $\triangle PQR$ (Eq ii): ½ Mark
• Equating (i) and (ii) to prove final result: ½ Mark
• BPT: Correct figure, given, to prove, construction: ½ Mark
• BPT: Correct mathematical proof: 2 Marks
• Trapezium: Drawing diagonal PR (Construction): ½ Mark
• Trapezium: Applying BPT in $\triangle PSR$ (Eq i): 1 Mark
• Trapezium: Applying BPT in $\triangle PQR$ (Eq ii): ½ Mark
• Equating (i) and (ii) to prove final result: ½ Mark
