QUESTION 1
Let the third side of the right angle triangle with sides [tex]x,6[/tex] be [tex]l[/tex].
Then, from the Pythagoras Theorem;
[tex]l^2=x^2+6^2[/tex]
[tex]l^2=x^2+36[/tex]
Let the hypotenuse of the right angle triangle with sides 2,6 be [tex]m[/tex].
Then;
[tex]m^2=6^2+2^2[/tex]
[tex]m^2=36+4[/tex]
[tex]m^2=40[/tex]
Using the bigger right angle triangle,
[tex](x+2)^2=m^2+l^2[/tex]
[tex]\Rightarrow (x+2)^2=40+x^2+36[/tex]
[tex]\Rightarrow x^2+2x+4=40+x^2+36[/tex]
Group similar terms;
[tex]x^2-x^2+2x=40+36-4[/tex]
[tex]\Rightarrow 2x=72[/tex]
[tex]\Rightarrow x=36[/tex]
QUESTION 2
Let the hypotenuse of the triangle with sides (x+2),4 be [tex]k[/tex].
Then, [tex]k^2=(x+2)^2+4^2[/tex]
[tex]\Rightarrow k^2=(x+2)^2+16[/tex]
Let the hypotenuse of the right triangle with sides 2,4 be [tex]t[/tex].
Then; we have [tex]t^2=2^2+4^2[/tex]
[tex]t^2=4+16[/tex]
[tex]t^2=20[/tex]
We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;
[tex][(x+2)+2]^2=k^2+t^2[/tex]
[tex](x+4)^2=(x+2)^2+16+20[/tex]
[tex]x^2+8x+16=x^2+4x+4+16+20[/tex]
[tex]x^2-x^2+8x-4x=4+16+20-16[/tex]
[tex]4x=24[/tex]
[tex]x=6[/tex]
QUESTION 3
Let the hypotenuse of the triangle with sides (x+8),10 be [tex]p[/tex].
Then, [tex]p^2=(x+8)^2+10^2[/tex]
[tex]\Rightarrow p^2=(x+8)^2+100[/tex]
Let the hypotenuse of the right triangle with sides 5,10 be [tex]q[/tex].
Then; we have [tex]q^2=5^2+10^2[/tex]
[tex]q^2=25+100[/tex]
[tex]q^2=125[/tex]
We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;
[tex][(x+8)+5]^2=p^2+q^2[/tex]
[tex](x+13)^2=(x+8)^2+100+125[/tex]
[tex]x^2+26x+169=x^2+16x+64+225[/tex]
[tex]x^2-x^2+26x-16x=64+225-169[/tex]
[tex]10x=120[/tex]
[tex]x=12[/tex]
QUESTION 4
Let the height of the triangle be H;
Then [tex]H^2+4^2=8^2[/tex]
[tex]H^2=8^2-4^2[/tex]
[tex]H^2=64-16[/tex]
[tex]H^2=48[/tex]
Let the hypotenuse of the triangle with sides H,x be r.
Then;
[tex]r^2=H^2+x^2[/tex]
This implies that;
[tex]r^2=48+x^2[/tex]
We apply Pythagoras Theorem to the bigger triangle to get;
[tex](4+x)^2=8^2+r^2[/tex]
This implies that;
[tex](4+x)^2=8^2+x^2+48[/tex]
[tex]x^2+8x+16=64+x^2+48[/tex]
[tex]x^2-x^2+8x=64+48-16[/tex]
[tex]8x=96[/tex]
[tex]x=12[/tex]
QUESTION 5
Let the height of this triangle be c.
Then; [tex]c^2+9^2=12^2[/tex]
[tex]c^2+81=144[/tex]
[tex]c^2=144-81[/tex]
[tex]c^2=63[/tex]
Let the hypotenuse of the right triangle with sides x,c be j.
Then;
[tex]j^2=c^2+x^2[/tex]
[tex]j^2=63+x^2[/tex]
We apply Pythagoras Theorem to the bigger right triangle to obtain;
[tex](x+9)^2=j^2+12^2[/tex]
[tex](x+9)^2=63+x^2+12^2[/tex]
[tex]x^2+18x+81=63+x^2+144[/tex]
[tex]x^2-x^2+18x=63+144-81[/tex]
[tex]18x=126[/tex]
[tex]x=7[/tex]
QUESTION 6
Let the height be g.
Then;
[tex]g^2+3^2=x^2[/tex]
[tex]g^2=x^2-9[/tex]
Let the hypotenuse of the triangle with sides g,24, be b.
Then
[tex]b^2=24^2+g^2[/tex]
[tex]b^2=24^2+x^2-9[/tex]
[tex]b^2=576+x^2-9[/tex]
[tex]b^2=x^2+567[/tex]
We apply Pythagaoras Theorem to the bigger right triangle to get;
[tex]x^2+b^2=27^2[/tex]
This implies that;
[tex]x^2+x^2+567=27^2[/tex]
[tex]x^2+x^2+567=729[/tex]
[tex]x^2+x^2=729-567[/tex]
[tex]2x^2=162[/tex]
[tex]x^2=81[/tex]
Take the positive square root of both sides.
[tex]x=\sqrt{81}[/tex]
[tex]x=9[/tex]
QUESTION 7
Let the hypotenuse of the smaller right triangle be; n.
Then;
[tex]n^2=x^2+2^2[/tex]
[tex]n^2=x^2+4[/tex]
Let f be the hypotenuse of the right triangle with sides 2,(x+3), be f.
Then;
[tex]f^2=2^2+(x+3)^2[/tex]
[tex]f^2=4+(x+3)^2[/tex]
We apply Pythagoras Theorem to the bigger right triangle to get;
[tex](2x+3)^2=f^2+n^2[/tex]
[tex](2x+3)^2=4+(x+3)^2+x^2+4[/tex]
[tex]4x^2+12x+9=4+x^2+6x+9+x^2+4[/tex]
[tex]4x^2-2x^2+12x-6x=4+9+4-9[/tex]
[tex]2x^2+6x-8=0[/tex]
[tex]x^2+3x-4=0[/tex]
[tex](x-1)(x+4)=0[/tex]
[tex]x=1,x=-4[/tex]
We are dealing with length.
[tex]\therefore x=1[/tex]
QUESTION 8.
We apply the leg theorem to obtain;
[tex]x(x+5)=6^2[/tex]
[tex]x^2+5x=36[/tex]
[tex]x^2+5x-36=0[/tex]
[tex](x+9)(x-4)=0[/tex]
[tex]x=-9,x=4[/tex]
We discard the negative value;
[tex]\therefore x=4[/tex]
QUESTION 9;
We apply the leg theorem again;
[tex]10^2=x(x+15)[/tex]
[tex]100=x^2+15x[/tex]
[tex]x^2+15x-100=0[/tex]
Factor;
[tex](x-5)(x+20)=0[/tex]
[tex]x=5,x=-20[/tex]
Discard the negative value;
[tex]x=5[/tex]
QUESTION 10
According to the leg theorem;
The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the portion of the hypotenuse adjacent to that leg.
We apply the leg theorem to get;
[tex]8^2=16x[/tex]
[tex]64=16x[/tex]
[tex]x=4[/tex] units.
QUESTION 11
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Question 12
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