Pythagorean Theorem: Difficult Problems with Solutions

Solution:
Let the sides be [tex]8k[/tex] and [tex]15k[/tex].
By the Pythagorean theorem: [tex](8k)^2+(15k)^2=68^2 \Rightarrow 289k^2=68^2 \Rightarrow 17k=68 \Rightarrow k=4[/tex]
The sides are [tex]32[/tex] and [tex]60[/tex], and the perimeter is
[tex]2(32+60)=184[/tex]

Solution:
Let the shorter leg be [tex]x[/tex]; then the other leg is [tex]x+5[/tex] and the hypotenuse is [tex]x+10[/tex]:
[tex]x^2+(x+5)^2=(x+10)^2[/tex]
[tex]x^2+x^2+10x+25=x^2+20x+100[/tex]
[tex]x^2-10x-75=0 \Rightarrow (x-15)(x+5)=0[/tex]
Since [tex]x\gt 0[/tex], [tex]x=15[/tex]: the sides are [tex]15[/tex], [tex]20[/tex], [tex]25[/tex], and the hypotenuse is [tex]25[/tex].

Solution:
Let [tex]P[/tex] be the foot of the ladder, [tex]R[/tex] and [tex]L[/tex] its top points on the right and the left wall, and [tex]R_0[/tex], [tex]L_0[/tex] the bottoms of the walls. The right triangles [tex]PR_0R[/tex] and [tex]LL_0P[/tex] have equal hypotenuses [tex]PR=PL[/tex] (the ladder).
Since [tex]\angle LPR=90^\circ[/tex], we have [tex]\angle LPL_0+\angle RPR_0=90^\circ[/tex]; also [tex]\angle PRR_0+\angle RPR_0=90^\circ[/tex]. Hence [tex]\angle PRR_0=\angle LPL_0[/tex], and the two triangles are congruent. Therefore [tex]PR_0=LL_0=3[/tex]:
[tex]PR^2=PR_0^2+RR_0^2=3^2+4^2=25 \Rightarrow PR=5[/tex]

Solution:
Let the base be [tex]x[/tex]; then the height is [tex]x+2[/tex] and a leg is [tex]x+3[/tex]. The height bisects the base:
[tex]\left(\dfrac{x}{2}\right)^2+(x+2)^2=(x+3)^2[/tex]
[tex]\dfrac{x^2}{4}+x^2+4x+4=x^2+6x+9 \Rightarrow \dfrac{x^2}{4}=2x+5[/tex]
[tex]x^2-8x-20=0 \Rightarrow (x-10)(x+2)=0 \Rightarrow x=10[/tex]
The base is [tex]10[/tex], the height is [tex]12[/tex], each leg is [tex]13[/tex], and the perimeter is
[tex]10+13+13=36[/tex]

Solution:
Let the roads be [tex]x[/tex] and [tex]y[/tex] long. They are the legs of a right triangle with hypotenuse [tex]13[/tex]:
[tex]x+y=17[/tex], [tex]x^2+y^2=169[/tex]
[tex](x+y)^2=x^2+y^2+2xy \Rightarrow 289=169+2xy \Rightarrow xy=60[/tex]
Two numbers with sum [tex]17[/tex] and product [tex]60[/tex] are [tex]5[/tex] and [tex]12[/tex] (check: [tex]5^2+12^2=169[/tex]). The farther village is [tex]12[/tex] miles away.

Solution:
First find the height to the base (it bisects the base):
[tex]h^2=17^2-8^2=289-64=225 \Rightarrow h=15[/tex]
[tex]S=\dfrac{16\times 15}{2}=120[/tex]
The same area can be computed with a leg and the height [tex]h_1[/tex] to it:
[tex]\dfrac{17\times h_1}{2}=120 \Rightarrow h_1=\dfrac{240}{17}=14\dfrac{2}{17}[/tex]

Solution:
Let the legs be [tex]x[/tex] and [tex]2x[/tex].
By the Pythagorean theorem: [tex]x^2+(2x)^2=10^2 \Rightarrow 5x^2=100 \Rightarrow x^2=20[/tex]
Express the area in two ways (we do not even need [tex]x[/tex] itself):
[tex]\dfrac{x\times 2x}{2}=\dfrac{10\times h}{2} \Rightarrow 10h=2x^2=40 \Rightarrow h=4[/tex]

Solution:
The height from a vertex of the shorter base cuts off from the longer base a segment of [tex]\frac{14-4}{2}=5[/tex]; the remaining part of the longer base is [tex]14-5=9[/tex].
[tex]h^2=13^2-5^2=169-25=144 \Rightarrow h=12[/tex]
The diagonal is the hypotenuse of a right triangle with legs [tex]12[/tex] and [tex]9[/tex]:
[tex]d^2=12^2+9^2=144+81=225 \Rightarrow d=15[/tex]

Solution:
The line through the centers is perpendicular to the common chord and bisects it; half of the chord is [tex]12[/tex]. The distances from the centers to the chord are:
[tex]d_1^2=13^2-12^2=25 \Rightarrow d_1=5[/tex]
[tex]d_2^2=15^2-12^2=81 \Rightarrow d_2=9[/tex]
If the centers are on different sides of the chord, the distance is [tex]5+9=14[/tex]; if they are on the same side, it is [tex]9-5=4[/tex]. Both cases are possible.

Solution:
Let the tangent touch the circles at [tex]A[/tex] and [tex]B[/tex]. The radii [tex]O_1A[/tex] and [tex]O_2B[/tex] are perpendicular to [tex]AB[/tex], so [tex]O_1ABO_2[/tex] is a right trapezoid. Draw [tex]O_2K\parallel AB[/tex] ([tex]K[/tex] on [tex]O_1A[/tex]); then [tex]O_2K=AB[/tex] and
[tex]O_1O_2=50+32=82[/tex], [tex]O_1K=50-32=18[/tex]
[tex]AB^2=82^2-18^2=(82-18)(82+18)=64\times 100=6400 \Rightarrow AB=80[/tex]

Solution:
Let the legs be [tex]a[/tex] and [tex]b[/tex]. Each median to a leg is the hypotenuse of a right triangle whose legs are the other leg of the triangle and half of this leg:
[tex]b^2+\dfrac{a^2}{4}=52[/tex], [tex]a^2+\dfrac{b^2}{4}=73[/tex]
Adding the two equalities:
[tex]\dfrac{5}{4}(a^2+b^2)=125 \Rightarrow a^2+b^2=100 \Rightarrow c=10[/tex]

Solution:
Let the altitude be [tex]CH=h[/tex], [tex]AH=9[/tex], [tex]BH=16[/tex]. By the Pythagorean theorem in the two small right triangles:
[tex]AC^2=h^2+81[/tex], [tex]BC^2=h^2+256[/tex]
and in the big triangle [tex]AC^2+BC^2=AB^2=25^2[/tex]:
[tex]2h^2+337=625 \Rightarrow 2h^2=288 \Rightarrow h^2=144 \Rightarrow h=12[/tex]

Solution:
Let [tex]MD=x[/tex]; then [tex]MC=10-x[/tex]. From the right triangles [tex]ADM[/tex] and [tex]BCM[/tex]:
[tex]AM^2=16+x^2[/tex], [tex]BM^2=16+(10-x)^2[/tex]
In the right triangle [tex]AMB[/tex]: [tex]AM^2+BM^2=AB^2[/tex]:
[tex]16+x^2+16+100-20x+x^2=100 \Rightarrow x^2-10x+16=0 \Rightarrow (x-2)(x-8)=0[/tex]
Since [tex]MC\gt MD[/tex], [tex]MD=2[/tex] and [tex]MC=8[/tex].

Solution:
Let the legs be [tex]a[/tex], [tex]b[/tex] and the hypotenuse [tex]c[/tex]. Then
[tex]a+b=60-c[/tex], [tex]ab=2\times 150=300[/tex], [tex]a^2+b^2=c^2[/tex]
[tex](a+b)^2=a^2+b^2+2ab \Rightarrow (60-c)^2=c^2+600[/tex]
[tex]3600-120c+c^2=c^2+600 \Rightarrow 120c=3000 \Rightarrow c=25[/tex]
(The legs are [tex]15[/tex] and [tex]20[/tex].)

Solution:
Let [tex]T[/tex] be the lowest point of the rope. Each part of the rope, from the top of a pole to [tex]T[/tex], is at least as long as the vertical drop [tex]10-2.5=7.5[/tex].
But the two parts together are only [tex]15=7.5+7.5[/tex] long, so each part is exactly [tex]7.5[/tex]. If a pole were at a horizontal distance [tex]x\gt 0[/tex] from [tex]T[/tex], that part would be at least the hypotenuse [tex]\sqrt{(7.5)^2+x^2}\gt 7.5[/tex].
Therefore the distance between the poles is [tex]0[/tex]: they stand right next to each other and the rope hangs down doubled.

Solution:
Ben hears the thunder through the phone at the same moment as Alex, i.e. [tex]15[/tex] s after the lightning, and directly [tex]10[/tex] s later, i.e. [tex]25[/tex] s after the lightning. So the lightning is [tex]15\times \frac{1}{5}=3[/tex] miles from Alex and [tex]25\times \frac{1}{5}=5[/tex] miles from Ben.
The lightning is due east and the hut is due north of Alex, so the angle at Alex is a right angle and the distance from Ben to the lightning is the hypotenuse:
[tex]x^2=5^2-3^2=16 \Rightarrow x=4[/tex]
Alex is [tex]4[/tex] miles from the hut.