Triangles

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ISEE Upper Level Quantitative Reasoning › Triangles

Questions 1 - 10

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle with sidelength $s = 8 \sqrt{2}$ . Its perimeter is three times this, or $3 \cdot 8 \sqrt{2} = 24 \sqrt{2}$

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

In a right triangle, two sides have length . Give the length of the hypotenuse in terms of .

$2\sqrt{2}t$

$\sqrt{2}t$

$2\sqrt{3}t$

$\sqrt{3}t$

Explanation

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=(2t)^2+(2t)^2\Rightarrow c^2=8t^2\Rightarrow c=\sqrt{8t^2}=2\sqrt{2}t$

The legs of a right triangle measure $2 \frac{2}{5}$ and $3 \frac{1}{5}$ . What is its perimeter?

$9 \frac{3}{5}$

$8 \frac{3}{5}$

$9 \frac{1}{5}$

$10\frac{1}{5}$

Explanation

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ :

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

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

The legs of a right triangle measure $2 \frac{2}{5}$ and $3 \frac{1}{5}$ . What is its perimeter?

$9 \frac{3}{5}$

$8 \frac{3}{5}$

$9 \frac{1}{5}$

$10\frac{1}{5}$

Explanation

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ :

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

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

In a right triangle, two sides have length . Give the length of the hypotenuse in terms of .

$2\sqrt{2}t$

$\sqrt{2}t$

$2\sqrt{3}t$

$\sqrt{3}t$

Explanation

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=(2t)^2+(2t)^2\Rightarrow c^2=8t^2\Rightarrow c=\sqrt{8t^2}=2\sqrt{2}t$

The legs of a right triangle measure $2 \frac{2}{5}$ and $3 \frac{1}{5}$ . What is its perimeter?

$9 \frac{3}{5}$

$8 \frac{3}{5}$

$9 \frac{1}{5}$

$10\frac{1}{5}$

Explanation

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ :

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

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

In a right triangle, two sides have length . Give the length of the hypotenuse in terms of .

$2\sqrt{2}t$

$\sqrt{2}t$

$2\sqrt{3}t$

$\sqrt{3}t$