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1.)

If a convex quadrilateral has sides of lengths 3, 4, 12, and 13, and the two shorter sides are perpendicular to one another as shown, what is the area of the quadrilateral?
2.)

A square and an equilateral triangle have equal perimeters. The area of the triangle is  square inches. What is the number of inches in the length of the diagonal of the square?

3.)

 

The vertex of the right isosceles triangle is the center of the square. What is the area of the overlapping region?
4.)

In triangle $ABC$, $m\angle A = 90^\circ$, $m\angle B = 75^\circ$, and $BC = \sqrt {3}$ units. What is the area of triangle $ABC$? Express your answer as a common fraction.

5.)

What is the number of square centimeters in the area of the triangle shown? (See if you can solve the problem without Heron's Formula!)
6.)

 

In the figure, what is the area of triangle ? Express your answer as a common fraction.
 

Jdaye  Feb 15, 2018
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2+0 Answers

 #1
avatar+1960 
+1

I think I can answer #1

 

 

The diagram below meets all the given criteria. It is a quadrilateral with the given side lengths. Since \(\overline{AC}\) is the hypotenuse of \(\triangle ABC\), we can use the Pythagorean Theorem to find the length of the missing side length. Of course, this triangle satisfies the most famous pythagorean triple, a 3-4-5 right triangle, so \(AC=5\).

 

Yet again, \(\triangle DAC\) is in a similar situation. It is a 5-12-13 triangle, which is another Pythagorean triple. Because of this, \(m\angle DAC=90^\circ\)

 

I can break up the area of the original quadrilateral into two smaller parts: the 3-4-5 right triangle and the 5-12-13 right triangle. 

 

Using the formula \(A_{\triangle}=\frac{1}{2}bh\), one can identify the areas of both triangles. This process is even simpler since the height of both triangles is also the perpendicular height. 

 

\(A_1=\frac{1}{2}*3*4\\ A_1=2*3\\ A_1=6\)\(A_2=\frac{1}{2}*5*12\\ A_2=6*5\\ A_2=30\) 

 

Now, just add the areas together. 

 

\(A_1+A_2=A_{\text{total}}\\ 6+30=36\text{units}^2\)

 

Many of the questions seem to suggest that there is a complementary diagram or some missing information (2, 3, 5 and 6)

TheXSquaredFactor  Feb 15, 2018
edited by TheXSquaredFactor  Feb 15, 2018
 #2
avatar+86548 
+1

4.)

In triangle $ABC$, $m\angle A = 90^\circ$, $m\angle B = 75^\circ$, and $BC = \sqrt {3}$ units. What is the area of triangle $ABC$? Express your answer as a common fraction.

 

 

C

                 √3

 

A   90°           75°   B

 

 

sin 75°  =  AC / √3

AC  = √3* sin (75°)

 

Angle ACB  =  90 - 75  =  15°

sin(15°)  = AB / √3

AB  =  √3 * sin (15°)

 

So   the area  is

 

(1/2) AB * AC   =

 

(1/2) (√3 sin(15°) * √3 sin (75°)   =  (3/2)sin(15°)sin(75°)  =

 

 

(3/2)sin(15”) sin ( 90° - 15°)  =

 

(3/2)sin(15°)  (sin90”cos15°  - cos90°sin15°)  =

 

(3/2)sin(15°) (cos (15°) =        

 

(3/2)(1/2) sin (2 * 15)  =

 

(3/2)(1/2)sin(30°) = 

 

(3/2)(1/2)sin (30°)  =

 

(3/2) (1/2)(1/2)  =  3/8  units^2

 

 

cool cool cool

CPhill  Feb 15, 2018
edited by CPhill  Feb 15, 2018
edited by CPhill  Feb 15, 2018

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