# MTAP Reviewer for Grade 9 Solution Series(31-35)

Here is the part 7 of the solution series in 2015 MTAP(MMC) reviewer for grade 9 elimination level. Other solutions can be found in this website as well as the PDF copy of all the problems.

Problem 31:

An equilateral triangle and a square have the same perimeter. What is the ratio of the length of a side of the triangle to the length of a side of the square?

Solution:

The perimeter of equilateral triangle can be found using the formula,

$P=3t$

where t is the length of the side

The perimeter of square can be found using the formula,

$P=4s$

where s is the length of the side

Since the perimeter is equal, we equate their perimeters,

$P=P$

$3t=4s$

Since we are asked to find the ratio of the side of triangle to the length of the side of the square we find t/s.

$\dfrac{t}{s}=\dfrac{4}{3}$

Thus, the ratio is 4:3

Problem 32:

John cuts an equilateral triangular paper whose sides measure 2 cm into pieces. He then rearranges the pieces to form a square without overlapping. How long is the side of the square formed?

After cutting, the area of the figure is still the same. To find the area of an equilateral triangle, we use the following formula,

$A=\frac{a^2\sqrt{3}}{4}$

Where a is the length of the side of an equilateral triangle.

Finding the area we have,

$A=\frac{2^2\sqrt{3}}{4}$

$A=\frac{4\sqrt{3}}{4}$

$A=\sqrt{3}$

Now, since the area of this triangle is equal to the area of the square, we can use the formula of the area of the square to find the length of the side.

$A=s^2$

Now, we equate the area of the triangle,

$\sqrt{3}=s^2$

Since we are solving for s, we take the square root of both sides,

$\sqrt{\sqrt{3}}=\sqrt{s^2}$

$\sqrt[4]{3}=s$

Or,

$\boxed{s=\sqrt[4]{3}}$

Problem 23:

The sides of a triangle are of lengths 5, 12 and 13 cm. What is the length of its shortest altitude?

Solution:

Draw the triangle and label like shown below.

Since the triangle formed is a right triangle, 12 and 5 are also altitudes. If we draw another altitude to the hypotenuse and label it x. Both 5 and 12 will become the hypotenuse of the two new triangles formed. Thus, x is the shortest altitude.

Now, take note that $\triangle ABC\sim\triangle BDC\sim\triangle ADB$

By similar triangles we can solve for the value of x.

$\dfrac{BD}{BC}=\dfrac{AB}{AC}$

$\dfrac{x}{5}=\dfrac{12}{13}$

$x=5(\dfrac{12}{13})$

$\boxed{x=\dfrac{60}{13}}$

Problem 34:

Each side of triangle ABC measures 8 cm. If D is the foot of the altitude drawn from A to the side BC and E is the midpoint of AD, how long is segment BE?

Solution:

Draw the figure as shown below,

Since this is an equilateral triangle, triangle ADB is a 36-60-90 triangle. Using the property of this triangle, the side of AD is square root of 3 times the length of BD. Thus,

$AD=\sqrt{3}BD$

$AD=\sqrt{3}(4)$

$AD=4\sqrt{3}$

Since E is the midpoint of AD, $AE=ED=2\sqrt{3}$

Now, triangle BDE also forms a right triangle and we can solve for BE using Pythagorean theorem.

$BE^2=BD^2+ED^2$

$BE^2=4^2+(2\sqrt{3})^2$

$BE^2=16+12$

$BE^2=28$

$\sqrt{BE^2}=\sqrt{28}$

$\boxed{BE=2\sqrt{7}}$

Problem 35:

A point E is chosen inside a square of side 8 cm such that it is equidistant from two adjacent vertices of the square and the side opposite these vertices. Find the common distance.

Solution:

Draw the figure like shown below,

The only way that a point will be equidistant from two adjacent vertices, it should lie in the symmetrical axis of the square. Label the figure accordingly. If your figure is not correct, your solution will not be correct as well.

To find the value of x, we use the right triangle formed in the upper left side of upper right side. Either of the two.

Using Pythagorean theorem we have,

$c^2=a^2+b^2$

$x^2=4^2+(8-x)^2$

$x^2=16+64-16x+x^2$

$16x=80$

$x=\dfrac{80}{16}$

$\boxed{x=5}$

We would like to thank MMC 2013 finalist Daniel James Molina for helping us out solving this problem.