Iln triangle ABC, BC=8 , CA=6 , AB=10. A line dividing the triangle ABC into two regions of equal area is perpendicular to AB at point X. Find the value of 
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Answer:
The key to solving this problem is to notice that the given lengths of the 3 sides of triangle ABC, ie, 6, 8 and 10, are a Pythagorean Triple. This tells us that the triangle is right angled (6^2 + 8^2 = 10^2). A couple of other low numbered Pythagorean Triples are 3, 4 and 5, and 5, 12 and 13.
To help with understanding my solution, sketch the following: First draw a horizontal line, AB, to represent the hypotenuse of triangle ABC. Use your judgement to position a point C somewhere above AB so that when you draw the triangle it looks right angled. Label the 3 angles A,B and C and draw a small square in angle C to signify that it is right angled.
We are told that at a point X, somewhere between A and B, a line drawn vertically from X will divide triangle ABC into 2 regions of equal area. It is not necessary to draw the line so that it actually achieves a 50%, 50% division. Placing point X near the centre of line AB is good enough. Now draw a vertical line from X up to line BC. Mark the point where this line meets BC with ‘Y’. This is the line which divides triangle ABC into 2 equal areas. Label the quadrilateral shaped region ‘P’ and the triangle shaped region ‘Q’. There is already a small square marking the right angle in region P. Now mark the right angle in region Q in the same way.
Notice that the 3 angles of triangle ABC are the same as those in triangle BXY; this is because the two triangles are ‘similar’. Draw a small circle in angle A and another in the angle of equal size in triangle BXY. Notice also that just one dot is needed to mark the angles which are the same size in both triangles ABC and BXY!
Make another sketch of triangle ABC but this time with CB as the horizontal base and with vertex (highest point) A. Label the sides with their lengths, 6 cm, 8 cm, and 10 cm and inside the triangle write ‘P+Q’ and place a square in angle C to show that it is a right angle, also a circle in angle A and a dot in angle B. Now sketch triangle BXY with base XB and vertex Y. Mark its angles with a square, circle and dot. Inside this triangle write ‘Q’.
NOW TO FIND BX.
As regions P and Q are of equal area, it is clearly evident that triangle ABC has twice the area of triangle BXY.
It is also evident that triangles ABC and BXY are similar as the three angles in one are equal to the angles in the other, ie, corresponding angles are equal.
[To follow the following you need to know the relationship between lengths and areas of similar figures.]
So, the ratio of AREAS of (P+Q) : (P) = 2 : 1
Therefore, the ratio of corresponding LENGTHS = SQ RT 2 : SQ RT 1 = SQ RT 2 : 1
So, length of CB : length of BX = SQ RT 2 : 1
As CB = 8 cm, BX = 8/ SQ RT 2
BX = 5.65684 . . .
ANSWER: BX = 5.657 cm. (to 4 significant figures/ 3 decimal places)
Answer:
hope it is helpful to you.
