Get the answer to your homework problem.
Try Numerade free for 7 days
We don’t have your requested question, but here is a suggested video that might help.
Find the area of the triangle: Round to the nearest tenth: a 18.2 b 17.1 12.3
Find the area of the triangle QRS A) 90 cmB) 80 cmC) 110 cmD) 55 cm E) 45 cm So PQ^2 = QR 132 = QR 169-25 = QR QR = 12 cm
169 = QR^2+ 25Subtract 25 on both sides
144 = QR^212 = QR
• Therefore, the base of triangle QRS = 12 cm. We now need to find the height of the triangle so that we can find the area of it. Here, the height will be the altitude drawn from the vertex S to QR.
• First we draw a perpendicular line from S to QR. Imagining the point of this line on QR as A, we then can see a square formed with points SAQT, with SQ as diagonal as shown in the figure.
Therefore, as the opposite sides of a square are equal, the length of TQ is equal to the length of SA.TQ^2 + TS
^2 = QS^2TQ^2 + 82 = 172
TQ^2 + 64 = 289
Subtracting 64 on both sides
TQ
^2 = 289-64TQ^2 = 225
Applying squaring on both sides
TQ = 15So the length of TQ = 15 cmTherefore SA = 15 cmHence the altitude is 15 cm• Now we have the base and height, So we need to find the area of the triangle and we will reach the target• As Area = ½* base * HeightArea = ½* QR * SAArea = ½* 12 cm * 15 cmArea = 90 cm^2Therefore the area of triangle QRS = 90 cm
^2• By using Pythagoras theorem we have found the area of the triangle QRS
So the Answer will be Option A (90 cm).
About the AuthorDisclaimer: The articles provided by GoGMAT are for informational purposes only and cannot be copied or redistributed without the written permission by GoGMAT. F1GMAT publishes guest posts by authors from all over the world. We do not accept any responsibility or liability for the information contained in the article or claims of copyright violation. However, we take immediate action when any copyright violations are reported. Report copyright violation