Question 972914
1. Adjacent sides of parallelogram are 7cm and 9cm. Its 1 diagonal is 8cm. Find its other diagonal.
Two adjacent sides and that diagonal form a triangle,
with the angle between those two sides opposite that diagonal.
 
The law of cosines says that for a triangle with vertices A, B, and C,
opposite sides a, b, and c respectively,
the measures of angles and sides are related by
{{{a^2=b^2+c^2-2bc*cos(A)}}} .
 
There is a triangle formed by
two sides measuring {{{b=7}}}cm and {{{c=9}}}cm flanking angle {{{A}}} ,
and a diagonal measuring {{{a=8}}}cm, connected to both of those sides.
That triangle, with sides so similar in length, must be an acute triangle.
Applying law of cosines to that triangle, we write
{{{8^2=7^2+9^2-2*7*9*cos(A)}}}
{{{64=49+81-2*7*9*cos(A)}}}
{{{2*7*9*cos(A)=49+81-64}}}
{{{126cos(A)=66}}}
{{{cos(A)=66/126=11/21}}} .
The cosine is positive, meaning that {{{A}}} is an acute angle.
In fact, {{{A=about58.4^o}}} , but we do not need to calculate that.
The parallelogram looks like this:
{{{drawing(350,200,-4.5,9.5,-1,7,
triangle(-3.667,5.963,5.333,5.963,0,0),
triangle(0,0,5.333,5.963,9,0),
locate(-0.5,0,D),locate(9,0,A),
locate(-4,6.5,C),locate(5.67,6.5,B),
locate(4.3,0,9cm),locate(2.66,2.98,8cm),
locate(6.5,2.98,7cm)
)}}}
 
The other diagonal {{{d}}} is opposite the other angle of the parallelogram, the angle at D,
which is supplementary to angle A, so that {{{cos(D)=-cos(A)=-11/21}}} ,
and is also flanked by sides measuring 7cm and 9cm.
Those sides and the other diagonal form another triangle.
{{{drawing(350,200,-4.5,9.5,-1,7,
triangle(-3.667,5.963,5.333,5.963,9,0),
triangle(-3.667,5.963,0,0,9,0),
locate(-0.5,0,D),locate(9,0,A),
locate(-4,6.5,C),locate(5.67,6.5,B),
locate(4.3,0,9cm),locate(2.66,2.9,d),
locate(-1.6,2.98,7cm)
)}}}
Applying law of cosines to that other triangle, we write
{{{d^2=7^2+9^2-2*7*9*cos(D)=49+81-126*(-11/21)=49+81+66=196}}}--->{{{d=sqrt(196)=14}}}
The other diagonal measures {{{highlight(14cm)}}} .
 
2. Diagonals of parallelogram are 12cm and 26cm. One of its side is 17cm.Yhe {{{x}}}= length of the other side in cm.
Since the diagonals of a parallelogram bisect each other,
the longer diagonal will be split into two congruent segments,
each one measuring {{{26cm/2=13cm}}} ,
and the shorter diagonal will be split into two congruent segments,
each one measuring {{{12cm/2=6cm}}} .
Then, the parallelogram, divided into two pairs of congruent triangles, will look like this:
{{{drawing(550,176,-1,26.5,-1.4,8.4,
triangle(0,0,17,0,12.41,3.464),triangle(24.84,6.928,17,0,12.41,3.464),
triangle(0,0,17,0,7.84,6.928),triangle(24.84,6.928,7.84,6.928,12.41,3.464),
locate(15.8,6.9,17),locate(6.2,1.8,6),locate(18.5,5.15,6),
locate(9,5.6,13),locate(3.9,3.7,x),
locate(11.1,4.15,green(X)),locate(12.4,4.5,red(Y)),
red(arc(12.41,3.464,2.1,2.1,-143,-16)),
green(arc(12.41,3.464,3,3,164,217))
)}}}
 
The law of cosines says that for a triangle with vertices A, B, and C,
opposite sides a, b, and c respectively,
the measures of angles and sides are related by
{{{a^2=b^2+c^2-2bc*cos(A)}}} .
 
To find {{{x}}} , we can apply law of cosines to the left side triangle and write
{{{x^2=6^2+13^2-2*6*13*cos(green(X))}}} ,
but we do not know {{{green(X)}}} , or {{{cos(green(X))}}} .
However, we know that {{{red(Y)}}} and {{{green(X)}}} are supplementary,
and that because of that {{{cos(red(Y))=-cos(green(X))}}} .
 
We can find about {{{cos(red(Y))}}} and {{{cos(green(X))}}} by applying law of cosines to the top triangle:
{{{17^2=6^2+13^2-2*6*13*cos(red(Y))}}}--->{{{17^2=6^2+13^2+2*6*13*cos(green(X))}}}--->{{{289=36+169+156*cos(green(X))}}}--->{{{cos(green(X))=(289-36-169)/156}}}--->{{{cos(green(X))=(289-36-169)/156=84/156=7/13}}} .
 
Now we get back to the left triangle to find {{{x}}} :
{{{x^2=6^2+13^2-2*6*13*cos(green(X))}}}--->{{{x^2=36+169-156cos(green(X))}}}--->{{{x^2=36+169-156cos(green(X))}}}--->{{{x^2=36+169-156(7/13)}}}--->{{{x^2=121}}}--->{{{x=sqrt(121)}}}--->{{{x=11}}}
The other side measures {{{highlight(11cm)}}} .