GMAT Quantitative - Right Triangles

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

Explanation

Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ =\sqrt{2^2+2^2}=\sqrt{8}$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $\sqrt{ (2^2-(\sqrt2)^2)}=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)*\sqrt2*\sqrt2=1$

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Triangle is a right triangle with . What is the length of its height ?

$\frac{12}{5}$

$\frac{5}{2}$

$\frac{36}{20}$

$\frac{52}{20}$

Explanation

The height AE, divides the triangle ABC, in two triangles AEC and AEB with same proportions as the original triangle ABC, this property holds true for any right triangle.

In other words, $\frac{AE}{AB}=\frac{AC}{BC}$ .

Therefore, we can calculate, the length of AE:

$AE= \frac{AC\cdot AB}{BC}= \frac{12}{5}$ .

A right triangle has a height of and a base of . In order for another triangle to be congruent, what must be the length of its hypotenuse?

$\sqrt{17}$

Explanation

In order for two triangles to be congruent, they must be identical. That is, the lengths of the corresponding sides of two congruent triangles must be equal. This means that in order for a triangle to be congruent to one with a height of and a base of , its hypotenuse must be the same length as the hypotenuse of that triangle, which we can find using the Pythagorean Theorem:

$c^2=169\rightarrow c=\sqrt{169}=13$

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Triangle is a right triangle with sides . What is the size of the height ?

$\frac{24}{5}$

$\frac{24}{10}$

$\frac{36}{5}$

$\frac{11}{2}$

Explanation

As we have previously seen, the height of a right triangle divides a it into two similar triangles with sides of same proportion.

Therefore, we can set up the following equality: $\frac{AE}{AB}=\frac{AC}{BC}$ or $AE=\frac{AC\cdot AB}{BC}$ .

By plugging in the numbers, we get, $\frac{48}{10}$ or $\frac{24}{5}$ .

Which of the following right triangles is similar to one with a height of and a base of ?

Explanation

In order for two right triangles to be similar, the ratio of their dimensions must be equal. First we can check the ratio of the height to the base for the given triangle, and then we can check each answer choice for the triangle with the same ratio:

$Given:\frac{h}{b}=\frac{6}{9}=\frac{2}{3}$

So now we can check the ratio of the height to the base for each answer option, in no particular order, and the one with the same ratio as the given triangle will be a triangle that is similar:

$Option1:\frac{h}{b}=\frac{5}{10}=\frac{1}{2}$

$Option2:\frac{h}{b}=\frac{7}{12}$

$Option3:\frac{h}{b}=\frac{8}{14}=\frac{4}{7}$

$Option4:\frac{h}{b}=\frac{12}{20}=\frac{3}{5}$

$Option5:\frac{h}{b}=\frac{10}{15}=\frac{2}{3}$

The triangle with a height of and a base of has the same ratio as the given triangle, so this one is similar.

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Triangle is a right triangle, with . What is the size of angle $\widehat{BCA}$ ?

$45^{\circ}$

$40^{\circ}$

$43^{\circ}$

$44^{\circ}$

$46^{\circ}$

Explanation

Triangle ABC is an isosceles right triangle. Therefore, its angles at the basis BC will always be $45^{\circ}$ .

This stems from the fact that the sums of the angles of a triangle are $180^{\circ}$ and in our case with ABC a right and isosceles triangle, $180-90=90^{\circ}$ , therefore for the two remaining angles are equal.

There are 90 degrees left, therefore to find the measure of each angle we do the following,

$\frac{90}{2}=45^{\circ}$ .

A right triangle has a base of 4 and a height of 3. What is the perimeter of the triangle?

Explanation

We are given two sides of the right triangle, so in order to calculate the perimeter we must first find the length of the third side, the hypotenuse, using the Pythagorean theorem:

$c^2=25\rightarrow c=5$

Now that we know the length of the third side, we can add the lengths of the three sides to calculate the perimeter of the right triangle:

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is a right isosceles triangle, with height . , what is the length of the height ?

$\frac{5\sqrt{2}}{2}$

$5\sqrt{2}$

$\sqrt{2}$

$\frac{5\sqrt{2}}{4}$

$\frac{5}{2}$

Explanation

Since here ABC is a isosceles right triangle, its height is half the size of the hypotenuse.

We just need to apply the Pythagorean Theorem to get the length of BC, and divide this length by two.

$BC^{2}=AB^{2}+AC^{2}$ , so $BC=\sqrt{50}$ .

Therefore, $BC=5\sqrt{2}$ and the final answer is $\frac{5\sqrt{2}}{2}$ .

Of the two acute angles of a right triangle, one measures fifteen degrees less than twice the other. What is the measure of the smaller of the two angles?

$35^{\circ }$

$25^{\circ }$

$30^{\circ }$

$40^{\circ }$

$20^{\circ }$

Explanation

Let one of the angles measure ; then the other angle measures . The sum of the measures of the acute angles of a triangle is $90^{\circ }$ , so we can set up and solve this equation: