The area of the square formed on the diagonal of a rectangle a... | Mathematics MCQ

MathematicsPrevious Year Paper 2024

The area of the square formed on the diagonal of a rectangle as its side is $108 \frac{1}{3} \%$ more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?

Correct Answer

📐 MA • Math Direct FormulaMathematicsPrevious Year Paper 2024

Option C

Quick Summary:

Express the diagonal area condition as $A_{diag} = 2.0833 \times A_{rect}$ (which is $25/12$ ). Using $d^2 = l^2+b^2$ and $A_{rect} = lb$ , we get $l^2+b^2 = \frac{25}{12}lb$ . Combined with perimeter $2(l+b)=28$ or $l+b=14$ , use $(l-b)^2 = (l+b)^2 - 4lb$ .

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📐MAMath SolutionDirect Formula

📋 Given

Perimeter of the rectangle = 28 units. Area of the square formed on the diagonal is 108 1/3% more than the area of the rectangle.

🔢 Formula Used

(l+b)2−4lb=(l−b)2(l+b)^2 - 4lb = (l-b)^2(l+b)2−4lb=(l−b)2

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Confusing '108 1/3% more than' with '108 1/3% of'. 'More than' means $100\% + 108.33\% = 208.33\% = 25/12$ .

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Setup Equations

Let sides be $l$ and $b$ . Given perimeter $2(l+b) = 28 \implies l+b = 14$ . Area of square on diagonal is $d^2 = l^2+b^2$ . Area of rectangle is $lb$ .

l2+b2=(1+325/3100)lb=(1+1312)lb=2512lbl^2+b^2 = (1 + \frac{325/3}{100})lb = (1 + \frac{13}{12})lb = \frac{25}{12}lbl2+b2=(1+100325/3​)lb=(1+1213​)lb=1225​lb

Solve for Area

Substitute $l^2+b^2$ into $(l+b)^2 = l^2+b^2 + 2lb$ : $14^2 = \frac{25}{12}lb + 2lb = \frac{49}{12}lb$ .

196=4912lb  ⟹  lb=196×1249=48196 = \frac{49}{12}lb \implies lb = \frac{196 \times 12}{49} = 48196=1249​lb⟹lb=49196×12​=48

Calculate Difference

Use $(l-b)^2 = (l+b)^2 - 4lb = 14^2 - 4(48) = 196 - 192 = 4$ .

l−b=4=2l-b = \sqrt{4} = 2l−b=4​=2

✅

C is correct because the difference between the sides of the rectangle is calculated to be 2 units.

Core Concepts Used

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Algebraic Identities Rectangle Geometry Percentage Increase

💡 EXAM TIP

This problem uses the identity $(l-b)^2 = (l+b)^2 - 4lb$ . Similar logic is frequently tested in quadratic roots and area-perimeter coordinate problems.

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