<p>Triangles ABC and PQR are shown below. The given side lengths are in centimeters. The area of triangle ABC is 30 square centimeters. What is the area of triangle PQR in square centimeters?</p>
<p>Diagram of the triangles: <a href="http://img.ctrlv.in/4e235b954dc5d.jpg%5B/url%5D">http://img.ctrlv.in/4e235b954dc5d.jpg</a></p>
<p>a. 15
b. 19
c. 25
d. 30
e. 33</p>
<p>I've been trying to figure out this problem for a while but I just can't get it. Help would be greatly appreciated. :)</p>
<p>Are the triangles isosceles? It says sides are given in cm but no side lengths are given, only x and y. </p>
<p>After some consideration however, this problem requires NO calculation. </p>
<p>Notice that the second triangle QPR MUST be larger than ABC. To demonstrate this, do the following:
Notice the angles: 110 and 70 add up to 180 degrees. Therefore, a straight line. Combine the two triangles into one larger one, with the two “X” sides on one side and the angles that are given being supplementary.</p>
<p>Now, notice that the 110 degree angle triangle, with its larger angle, comprises more than half of the larger triangle. Therefore it must be greater than ABC in area, and the only answer that gives this is </p>
<p>Given that ABC is 30, the answer for QPR must be
e)33.</p>
<p>Mathmatically I’ll have to think a bit more of how to solve this. Probably has something to do with law of cosines with x and y as a variable; either way it would take quite a lot more work to solve.</p>
<p>The answer is actually d, 30.</p>
<p>The question didn’t give any values for x or y and neither did it mention what kind of triangle it was. ):</p>
<p>Haha that makes sense. Basically make a triangle from the diameter of a circle and a point of the circle and split it in half at any point-makes the same result.</p>
<p>Essentially this is rule of sines, combined with area of a triangle.</p>
<p>You get an area formula A=xysin(theta). For proof, see trangle ABC. Draw a line from A to BC at a right angle. Area=Base times height/2.
Base here=BC=y. </p>
<p>Height=the line we drew, lets call T. Sin 70=T/x. Therefore height=T=xsin70. </p>
<p>Base times height=(y times T)/2= (xysin70)/2. </p>
<p>This works with the other triangle as well, if you extend PQ out for a right angle.</p>
<p>…After googling the secondary formula, see [Area</a> of a triangle](<a href=“http://www.clarku.edu/~djoyce/trig/area.html]Area”>Area of a triangle)</p>
<p>Area Triangle A = 30 </p>
<p>so 0.5*xysin70 = 30 xysin70 = 60 xy = 63.85</p>
<p>Then triangle B, Area = 0.5<em>xysin110 = 0.5</em>63.85*sin110 = 30</p>
<p>X and Y are the same in both triangles. The reason why you get the same area
is because sin110 = sin 70.</p>
<p>@js</p>
<p>Right. Thanks for the easier explanation.</p>
<p>The problem is fast and easy to solve with the right idea. Connect the triangles with those (triangles) generated when you BISECT a parallelogram ABCD with its diagonals AC and then BD. each of them has the area of 1/2 ABCD and one can have 70 degrees ( say, the triangle ABD ) . And the other diagonal generates triangle BCD, which MUST then have angle (180-70) = 110 degrees. The opposite ( and parallel and equal ) sides will be X, X and the other pair YY. Hence each of these triangles is THE SAME area !
It takes me MUCH longer to type this than it does to see it in a sketch . Whatever the area of the first one ( given here as 30 ) , the second one MUST be the same. </p>
<p>The problem is nearly trivial, but CONNECTING the idea of using a PARALLELOGRAM to solve this TRIANGLE AREA problem is not trivial and critically depends on your ACT prep.</p>
<p>If you are more of an algebra type of person, this method may prove to be more intuitive.
Consider
A = (1/2)absin(C)</p>
<p>We can set the triangle’s area equal to each other, with some factor Ill call k.</p>
<p>Area of Triangle 1 = k*area of triangle 2.</p>
<p>(1/2)xysin(70) = (1/2)xysin(110)<em>k
simplifying we get
sin(70) = sin(110)</em>k
youll find that k = 1.
Thus the areas of the same, leaving 30 as your answer. For me, this what I instantly did.
I know trying to visualize parrelograms is hard.</p>
