Geometry Forum - Problem of the Week
Solutions - Solar Panels, Nov. 28-Dec. 2, 1994
Annie says:What I loved most about this problem is that it proves there is almost always more than one way to solve a problem. Three or four are shown in the solutions below. In fact, the three solutions from Pat Daley's class at Fairfield are all different! When I originally looked at it, I figured all you needed to know was that the sum of the angles of a triangle is 180 degrees and that vertical angles are equal (and a couple other definitions, like isosceles triangles and right angles...).
Megan Guichard, Moira Conway, and Sarah McLaughlin most accurately reproduced exactly the answer I was thinking of - short and sweet. But their answers aren't any more right than anyone else's answer included here! Great job, folks.
Bipin Mujumdar
The wall falling straight from the extension meets the other side of the roof 10 feet from the vertex. No matter what the extension is or at what angle the roof is at, it will always cut off the same amount of the extension. The following is the proof. Draw a altitude straight down from the vertex of the roof. This will form 2 congruent right triangles. If you let the base angles of the roof equal x, the complement angle in each congruent triangle is (90-x). Then draw two vertical lines from the base angles of the roof. The complement of the base angles will also be (90-x). By the alternate interior angle theorem one of the angles in the triangle formed by the roof addition and the dropped wall is now also (90-x). The same thing happens with the upper right angle of the original roof. The altitude which I said to draw first and the wall are now parallel and thus by the alternate interior angle theorem again, the other angle in the triangle formed by the wall and addition is (90-x). Since both the angles of that triangle are 90-x the triangle is isosceles by the Isosceles triangle theorem. Therefore the extension of the left side of the roof must always equal the amount needed to cut of f the right side off the original roof, after the wall is dropped.Farrah Varga and Jessica Fry
Given the triangle as labeled and that triangle ABC has AB=AC because of symmetry, point G is collinear with points A and B, AD = 10 ft.
Solution:
Similar solution by Jennifer Abrahamsen, grade 9
Similar solutions by Samantha Brenner and Betsy McCormick.
Extend segment FD to segment AC at G and draw line BH parallel to
segment FG with H on segment AB. The following angles are
congruent:
Moira Conway (10th) and Sarah McLaughlin (10th) each submitted
Sketchpad sketches in which they did a
general proof.
Introduce perpendicular HBG. By definition triangle ABC is
isosceles, therefore the perpendicular is the perpendicular
bisector of angle ABC. Brad Warner
First, since segment FD falls straight down to the house, it
will intersect at a right angle to segment AC if extended past
point D. Call the point of intersection E. Two right triangles
are formed, triangles EDC and EFA. Since segment AB is congruent
to segment BC, angle A is congruent to angle C because of the
Isosceles Triangle Theorem. Also, angle FEA is congruent to angle
DEC because all right angles are congruent. Using the Third
Angle Theorem, angle F is congruent to angle EDC. Angle EDC is
congruent to angle FDB also because vertical angles are
congruent. Using the Transitive Property of Equality, angle F is
congruent to angle FDB. Segment BF is congruent to segment BD
because if two angles of a triangle are congruent, then the sides
opposite those angles are congruent. So, since we are given that
BF is 10 feet, BD must be 10 feet and the wall will intersect the
roof 10 feet down from the top.
Kim Carbone
To find out how far down the roof the straight wall for solar
panels will go, first of all we know that AB is congruent to BC
and that the extension of AB equals 10 feet. Because two sides of
a triangle are congruent, then the angles opposite those sides
are congruent, angle A and angle C are therefore congruent.
Next, the top of the extension of AB is F and from there, a wall
is dropped straight down meeting the BC side of the roof at D. If
you extend FD down to AC, the two lines will meet at G. Because
of the Exterior Angle Theorem, angle FGC equals the measure of
angle F + angle A and also angle FGA equals the measure of angle
GDC + angle C. Because the measures of angles FGC and FGA both
equal 90 degrees, then by using the substitution property, 90 =
mF + mA and 90 = mGDC + mA. Using substitution again, substitute
angle A for angle C. 90 = mF + mA and 90 = mGDC + mA. By the
subtraction property, the measure of angle F is equal to the
measure of angle GDC. Since angle GDC is congruent to angle BDF
because vertical angles are congruent, then angle F is also
congruent to angle BDF. Finally, since angle F is congruent to
angle BDF, this proves that BF is congruent to BD since if two
angles of a triangle are congruent, then the sides opposite
those angles are congruent. Knowing that BF equals 10 feet, we
now know that BD equals 10 feet as well. Therefore, the straight
wall FD will hit BC at 10 feet from the present point of the
roof.
Julie Black
The wall that goes straight down should be 10 feet. This is
because triangle ABC is isosceles, therefore angle A is
congruent to angle C and segment AB is congruent to segment CB.
Then if you draw an altitude in triangle ABC from point B, it
forms two right angles. If you call the point where the altitude
meets AC, E, this means that triangle ABE is congruent to CBE
because of AAS or HA or HL. Therefore, angle CBE is congruent to
angle ABE because of CPCTC. Then, extend FD through AC. FD is
parallel to BE because in a plane, if two lines are perpendicular
to the same line, then they are parallel. Angle BDF is congruent
to angle EBD because if two parallel lines are cut by a
transversal, the alternate interior angles are congruent. Angle
DFB is congruent to EBA because they are corresponding angles of
parallel lines. Therefore, angle BDF is congruent to angle BFD
because of substitution. Because of this, segment BD is
congruent to segment BF (if two angles of a triangle are
congruent then the sides opposite those angles are congruent).
BF is 10 feet because it was given. Therefore, BD is 10 feet
because of substitution.
Susan Quan
10 ft.
angles GDC and HBC (corresponding angles) angles HBC and HBA
(Segment BH bisects the vertex angle of isosceles triangle ABC)
angles HBA and DFB (corresponding angles) angles GDC and DFB
(substitution)
angles FDB and GDC (vertical angles)
angles FDB and DFB (substitution or transitive) So triangle DBF
is isosceles and BD = BF = 10 feet.
Megan Guichard
(I have included a Sketchpad file, which I refer to in the proof.
A=A,G=B,B=C, J=F)
It can easily be seen that triangle AJH is similar to triangle
BKH -- angle A is congruent to angle B because triangle AGB is
isosceles, and both angle JHA and angle KHB are 90, and so the
third angles (angle J and angle HKB) must be congruent. Angle
HKB is congruent to angle JKG because they are vertical angles,
and therefore, angle J is congruent to angle JKG, making triangle
KGJ isosceles with JG=KG. Since JG is 10, then so is KG.
Anita Lau
1) first, draw a diagram of the roof, the 10 feet to be extended,
and the vertical drop for the wall.
2) by extending imaginary lines, extending from the existing roof, and
because the roof is symmetrical, the opposite angles of the imaginary
"X" created are equal.
E=imaginary extension
E * * F F=extension to be made
* * ABC=roof
* B *
* *** extensions of roof
+ + +++ original roof
+ +
A + + C
3) since the roof is symmetrical, when you drop a wall straight
down, then the extended part of the house/roof will also be an
isosceles triangle, thus making the length of roof to be ripped
out be the same as the extended portion, in this case, 10 feet.
Nancy Chen
What I see in the diagram, if AB and BC can't
F be change, and so does angle B, then, whatever
* *' the distance to BF is, BD should be the same
* 10 * ' distance.
* * '
* * ' But later I found out that because I don't know
* * ' what is angle B, angle B can be at any angle and
B * ' FD will change to longer or shorter
* * ' determine to angle B.
* * '
* * '
* * '
* *' D
* *
* *
* *
* *
* A *
--------------------- C
Mt. St. Joseph's
Christina Carnevale (9th grade), Sarah Egner (9th grade),
Marianne Ganster (10th), Meghann Keppard (10th) and Siobhan
O'Brien (10th) all figured out that the straight wall would hit
10 feet down from the existing wall. They all did this by using
specific angle measures to come to the conclusion that BFD was
isosceles.
Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page