B4 (i) The length of each side of an
equilateral triangle is .
Without using a calculator, express
the area of the triangle in the form
, where and are
integers.
(ii) The triangle in part (i) is the base
of a prism. Given that the volume
of the prism is ,
show that the height of the prism
can be expressed in the form
, where and
are fractions in the simplest form.

Ask by Guzman John. in Singapore

Mar 22,2025

Question

B4 (i) The length of each side of an
equilateral triangle is .
Without using a calculator, express
the area of the triangle in the form
, where and are
integers.
(ii) The triangle in part (i) is the base
of a prism. Given that the volume
of the prism is ,
show that the height of the prism
can be expressed in the form
, where and
are fractions in the simplest form.

Ask by Guzman John. in Singapore

Mar 22,2025

UpStudy AI · Accepted Answer

Let the side length be 
$$
s=6(\sqrt{3}-1) \text{ cm}.
$$

**(i) Finding the area of the equilateral triangle**

The area of an equilateral triangle is given by 
$$
A=\frac{\sqrt{3}}{4}s^2.
$$
First, compute $$s^2$$:
$$
s^2=[6(\sqrt{3}-1)]^2=36(\sqrt{3}-1)^2.
$$
Now, calculate $$(\sqrt{3}-1)^2$$:
$$
(\sqrt{3}-1)^2= (\sqrt{3})^2 - 2\sqrt{3}\cdot 1 + 1^2 = 3-2\sqrt{3}+1=4-2\sqrt{3}.
$$
Thus,
$$
s^2=36(4-2\sqrt{3})=144-72\sqrt{3}.
$$
Substitute into the area formula:
$$
A=\frac{\sqrt{3}}{4}(144-72\sqrt{3})=\frac{144\sqrt{3}}{4}-\frac{72\sqrt{3}\cdot\sqrt{3}}{4}.
$$
Since $$\sqrt{3}\cdot\sqrt{3}=3$$,
$$
A=36\sqrt{3}-\frac{216}{4}=36\sqrt{3}-54.
$$
Thus, the area of the triangle is 
$$
A=(-54+36\sqrt{3}) \text{ cm}^2.
$$

**(ii) Finding the height of the prism**

The volume $$V$$ of the prism is given by the product of the base area and its height $$h$$:
$$
V=A\cdot h.
$$
We are given 
$$
V=9(\sqrt{3}-1) \text{ cm}^3,
$$
and from part (i) we have 
$$
A=36\sqrt{3}-54.
$$
Thus,
$$
h=\frac{V}{A}=\frac{9(\sqrt{3}-1)}{36\sqrt{3}-54}.
$$
Factor the denominator:
$$
36\sqrt{3}-54=18(2\sqrt{3}-3).
$$
Then,
$$
h=\frac{9(\sqrt{3}-1)}{18(2\sqrt{3}-3)}=\frac{\sqrt{3}-1}{2(2\sqrt{3}-3)}.
$$
To simplify, multiply the numerator and the denominator by the conjugate $$(2\sqrt{3}+3)$$:
$$
h=\frac{(\sqrt{3}-1)(2\sqrt{3}+3)}{2(2\sqrt{3}-3)(2\sqrt{3}+3)}.
$$
Compute the denominator:
$$
(2\sqrt{3}-3)(2\sqrt{3}+3)=(2\sqrt{3})^2-3^2=4\cdot3-9=12-9=3,
$$
so the full denominator is:
$$
2\cdot 3=6.
$$
Now expand the numerator:
$$
(\sqrt{3}-1)(2\sqrt{3}+3)=\sqrt{3}\cdot2\sqrt{3}+\sqrt{3}\cdot3-1\cdot2\sqrt{3}-1\cdot3,
$$
$$
=2\cdot3+3\sqrt{3}-2\sqrt{3}-3=6+ \sqrt{3}-3=3+\sqrt{3}.
$$
Thus,
$$
h=\frac{3+\sqrt{3}}{6}=\frac{3}{6}+\frac{\sqrt{3}}{6}=\frac{1}{2}+\frac{1}{6}\sqrt{3}.
$$
Therefore, the height of the prism is
$$
h=\left(\frac{1}{2}+\frac{1}{6}\sqrt{3}\right) \text{ cm}.
$$
**(i) Area of the Triangle** The area of the equilateral triangle is $$ (-54 + 36\sqrt{3}) \, \text{cm}^2 $$. **(ii) Height of the Prism** The height of the prism is $$ \left(\frac{1}{2} + \frac{1}{6}\sqrt{3}\right) \, \text{cm} $$.

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