Photo AI

EHGF is a rectangle - NSC Mathematics - Question 10 - 2024 - Paper 1

Question icon

Question 10

EHGF-is-a-rectangle-NSC Mathematics-Question 10-2024-Paper 1.png

EHGF is a rectangle. HE is produced $x^2$ cm to N and EH is produced $x^2$ cm to P. NF produced intersects PG produced at M to form an isosceles triangle MNP with NM... show full transcript

Worked Solution & Example Answer:EHGF is a rectangle - NSC Mathematics - Question 10 - 2024 - Paper 1

Step 1

Show that the area of EFHG is given by A(x) = 6x^2 - 3x^4

96%

114 rated

Answer

To find the area of rectangle EFHG, we first identify the lengths of sides NE and EH. From the diagram, we can see that:

  • The ratio of NE to EF is given by:

    NE=x23×2NE = \frac{x^2}{3} \times 2

  • Hence, we can express EF as:

    EF=23x2EF = \frac{2}{3} x^2

  • Given EH = 4 - 2 * NE, we have:

    EH=42x23=42x23EH = 4 - 2 \cdot \frac{x^2}{3} = 4 - \frac{2x^2}{3}

The area A of the rectangle EFHG can now be calculated by:

A(x)=EFEH=(23x2)(42x23)A(x) = EF \cdot EH = \left( \frac{2}{3} x^2 \right) \left( 4 - \frac{2x^2}{3} \right)

After simplifying, we will derive:

A(x)=23x2(42x23)=6x23x4A(x) = \frac{2}{3} x^2 (4 - \frac{2x^2}{3}) = 6x^2 - 3x^4

Step 2

Calculate the maximum area of rectangle EFHG

99%

104 rated

Answer

To determine the maximum area, we first set:

A(x)=612x2A'(x) = 6 - 12x^2

Setting the derivative equal to zero to find critical points:

1212x2=0    x2=1    x=112 - 12x^2 = 0\implies x^2 = 1 \implies x = 1

We then evaluate the area at this critical point:

A(1)=6(1)23(1)4=63=3A(1) = 6(1)^2 - 3(1)^4 = 6 - 3 = 3

Thus, the maximum area of rectangle EFHG is 3cm23 cm^2.

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;