Geometry

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GMAT Quantitative › Geometry

Questions 1 - 10

What is the arc length for a sector with a central angle of $45^{\circ}$ if the radius of the circle is ?

$\frac{3\pi }{4}$

$\frac{\pi }{4}$

$\frac{3\pi }{2}$

$\frac{\pi }{2}$

$\pi$

Explanation

Using the formula for arc length, we can plug in the given angle and radius to calculate the length of the arc that subtends the central angle of the sector. The angle, however, must be in radians, so we make sure to convert degrees accordingly by multiplying the given angle by $\frac{\pi }{180}$ :

$S=r\theta$

$S=3(45^{\circ})(\frac{\pi }{180^{\circ}})=3(\frac{\pi }{4})=\frac{3\pi}{4}$

What is the area of the figure with vertices ?

Explanation

This figure can be seen as a composite of two simple shapes: the rectangle with vertices , and the triangle with vertices .

The rectangle has length and height , so its area is the product of these dimensions, or $9 \times 10 = 90$ .

The triangle has as its base the length of the horizontal segment connecting and , which is ; its height is the vertical distance from the other vertex to this segment, which is . The area of this triangle is half the product of the base and the height, which is $\frac{1}{2} \times 10 \times 2 = 10$ .

Add the areas of the rectangle and the triangle to get the total area:

The chord of a $60^{\circ }$ central angle of a circle with area $40 \pi$ has what length?

$2 \sqrt{10}$

$10\sqrt{2}$

$4 \sqrt{5}$

Explanation

The radius of a circle with area $40 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 40 \pi$

$r^{2} = 40$

$r = \sqrt{40 } = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10}$

The circle, the central angle, and the chord are shown below:

Chord

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so $AB = AO= 2 \sqrt{10}$ , the correct response.

A given sector covers of a circle. What is the corresponding angle of the sector?

$252^{\circ}$

$300^{\circ}$

$222^{\circ}$

$202^{\circ}$

$272^{\circ}$

Explanation

A circle comprises $360^\circ$ , so a sector comprising of the circle will have an angle that is of .

Therefore:

$\angle=0.70(360)=\frac{7}{10}(360)=252^{\circ}$

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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:

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

Determine the equation of the tangent line to the following curve at the point :

Explanation

First find the slope of the tangent line by taking the derivative of the function and plugging in the x value of the given point to find the slope of the curve at that location:

So the slope of the tangent line to the curve at the given point is . The next step is to plug this slope into the formula for a line, along with the coordinates of the given point, to solve for the value of the y intercept of the tangent line:

$-7=1(2)+b\rightarrow b=-9$

We now know the slope and y intercept of the tangent line, so we can write its equation as follows:

Two circles are constructed; one is inscribed inside a given equilateral triangle, and the other is circumscribed about the same triangle.

The circumscribed circle has circumference $20 \pi$ . Give the area of the inscribed circle.

$25 \pi$

$400 \pi$

$50 \pi$

$100 \pi$

$200 \pi$

Explanation

Examine the diagram below, which shows the triangle, its three altitudes, and the two circles.

Thingy

The three altitudes of an equilateral triangle divide one another into two segments each, the longer of which is twice the length of the shorter. The length of each of the longer segments is the radius of the circumscribed circle, and the length of each of the shorter segments is the radius of the inscribed circle. Therefore, the inscribed circle has half the radius of the circumscribed circle.

The circumscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The inscribed circle has radius half this, or 5, so its area is

$A= \pi r^{2} = \pi \cdot 5^{2}= 25 \pi$

What is the area of the quadrilateral on the coordinate plane with vertices ?

Explanation

The quadrilateral is a parallelogram with two vertical bases, each with length . Its height is the distance between the bases, which is the difference of the -coordinates: . The area of the parallelogram is the product of its base and its height: