# MTAP Reviewer for Grade 9 Solutions(36-40)

This is 8th MTAP reviewer for grade 9 series for from 2015 MMC Elimination questions. Previous solutions and the PDF file of the problems can be check in this website as well.

This is not the official solution from MMC. This is a solution from the author of this website.

Problem 36:

A point E is chosen inside rectangle ABCD and its distances from A, B and C are 2 cm, 3 cm and 4 cm. How far is E from D?

Solution:

If you draw the figure, it will give you thrill how to solve it because the solution to this problem has never been teach for a regular grade 9 students unless you studied the first tutorial we posted in this website or you love reading olympiad math problems.

This problem can be solved using British Flag Theorem, better check that tutorial first before you go through. This theorem states that if a point is selected inside the rectangle, the sum of the square of the distances from a point from opposite vertices are equal.

Thus, $EB^2+ED^2=EA^2+EC^2$

By substitution, $3^2+ED^2=2^2+4^2$ $9+ED^2=4+16$ $ED^2=4+16-9$ $ED^2=11$ $\sqrt{ED^2}=\sqrt{11}$ $\boxed{ED=\sqrt{11}cm}$

Problem 37:

Find the area of a rectangle with diagonal 10 cm that is twice as long as it is wide.

Solution:

The diagonal, length, and the width of the rectangle will form a right triangle.

Let x be the width and the length will be 2x as given.

Using Pythagorean theorem, $a^2+b^2=c^2$ $x^2+(2x)^2=10^2$ $x^2+4x^2=100$ $5x^2=100$ $x^2=\dfrac{100}{5}$ $x^2=20$ $\sqrt{x^2}=\sqrt{20}$ $x=2\sqrt{5}$

Thus, the length is 4√5

Finding the area: $A=LW$ $A=4\sqrt{5}\cdot 2\sqrt{5}$ $A=8\cdot 5$ $\boxed{A=40cm^2}$

Problem 38:

Triangle ABC has a right angle at C. If sin A=1/3,what is cos A?

Solution:

Draw and label the figure as shown below. Label your angle, label the opposite and the hypotenuse since sine is the ratio of opposite side and hypotenuse. We are asked for the value of cosA which is the ratio of adjacent side to opposite side. Since the adjacent side is unknown, we look for it using Pythagorean theorem. $a^2+b^2=c^2$ $a^2+1^2=3^2$ $a^2+1=9$ $a^2=9-1$ $a^2=8$ $\sqrt{a^2}=\sqrt{8}$ $a=2\sqrt{2}$

Solving for cosA: $cosA=\dfrac{a}{H}$

Where a is the adjacent side to the angle A and H is the hypotenuse. $\boxed{cosA=\dfrac{2\sqrt{2}}{3}}$

Problem 39:

A side of a triangle measures 3 cm. A line segment is drawn parallel to this side, forming a trapezoid whose area is 2/3 of the area of the triangle. How long is the line segment?

Solution:

Draw and label the triangle like shown below. Since the segment is parallel to the side with length 3, two similar triangles will be formed.

We let A be the area of the bigger triangle. Given that the area of the trapezoid is 2/3 of A, the area of smaller traingle is 1-2A/3=A/3.

Using the side-area relationship of two similar polygons, we have $\dfrac{s_2^2}{s_1^2}=\dfrac{A_2}{A_1}$

Where s1 and s2 are the length of the sides of bigger and smaller triangle respectively. A1 and A2 are the corresponding areas. $\dfrac{x^2}{3^2}=\dfrac{\frac{A1}{3}}{A_1}$ $\dfrac{x^2}{3^2}=\dfrac{1}{3}$ $x^2=3^2(\dfrac{1}{3})$ $x^2=3$ $\sqrt{x^2}=\sqrt{3}$ $\boxed{x=\sqrt{3}}$

Problem 40:

A cone has volume 64 m^3 vertex equal to 1/4. If the cone is cut parallel to the base at a distance from the of the height of the cone, what is the volume of the resulting cone?

Solution:

The figure would look like this. The volume of the larger cone can be expressed as, $V=\dfrac{\pi R^2H}{3}$ $64=\dfrac{\pi R^2H}{3}$

The volume of smaller cone can be expressed as, $V=\dfrac{\pi r^2h}{3}$

Since the height is 1/4 of the bigger cone, the radius will follow this proportion since the cones formed here are congruent.(Parallel cut) $V=\dfrac{\pi(\frac{R}{4})^2(\frac{H}{4})}{3}$

Dividing the volumes we have, $\dfrac{V}{64}=\dfrac{\dfrac{\pi(\frac{R}{4})^2(\frac{H}{4})}{3}}{\dfrac{\pi R^2H}{3}}$ $\dfrac{V}{64}=(\dfrac{1}{4})^2(\dfrac{1}{4})$ $\dfrac{V}{64}=\dfrac{1}{64}$ $V=64\cdot (\dfrac{1}{64})$ $\boxed{V=1 m^3}$