Relationships in Triangles: practice solutions, Grade 10 – download the PDF

1 Find the midsegment ★★★

\(\overline{MN}\) is a midsegment, so \(MN=\dfrac{1}{2}BC\).

1. \(MN=\dfrac{18}{2}=9\text{ cm}\).
2. \(BC=2\cdot 7.5=15\text{ in}\).

2 Equal distances ★★★

By the Perpendicular Bisector Theorem, \(PA=PB\). So \(3x+4=5x-6\), which gives \(10=2x\) and \(x=5\).

Then \(PA=3(5)+4=19\text{ ft}\) (and \(PB=5(5)-6=19\text{ ft}\)).

3 Splitting a median ★★★

The centroid is \(\dfrac{2}{3}\) of the way from the vertex.

1. \(AG=\dfrac{2}{3}\cdot 15=10\text{ cm}\) and \(GM=15-10=5\text{ cm}\).
2. \(GM=\dfrac{1}{3}AM\), so \(AM=3\cdot 4=12\text{ in}\).

4 Can it be a triangle? ★★★

Add the two smaller lengths and compare with the largest.

1. \(5+8=13<14\): not a triangle.
2. \(6+9=15>14\): a triangle exists.
3. \(7+7=14\), not greater than 14: not a triangle (it would be flat).
4. \(4.5+6.2=10.7>10.5\): a triangle exists.

5 Order the angles ★★★

Each angle faces a side: \(\angle C\) faces \(\overline{AB}=7\), \(\angle B\) faces \(\overline{CA}=9\), and \(\angle A\) faces \(\overline{BC}=12\).

The shorter the opposite side, the smaller the angle, so \(\angle C<\angle B<\angle A\).

6 A right triangle’s circumcenter ★★★

The hypotenuse is \(\sqrt{6^2+8^2}=\sqrt{100}=10\text{ cm}\).

In a right triangle the circumcenter is the midpoint of the hypotenuse, so the circumradius is \(\dfrac{10}{2}=5\text{ cm}\).

7 Name the center ★★★

1. The circumcenter.
2. The incenter.
3. The centroid.
4. The orthocenter.
5. The incenter and the centroid. The circumcenter and orthocenter can lie outside an obtuse triangle.

8 Midsegment with algebra ★★★

Since \(BC=2\cdot DE\): \(6x+10=2(4x-3)=8x-6\). So \(16=2x\) and \(x=8\).

\(DE=4(8)-3=29\) and \(BC=6(8)+10=58\). Check: \(2\cdot 29=58\).

9 Midsegment on a grid ★★★

1. \(D=\left(\dfrac{-2+2}{2},\dfrac{1-5}{2}\right)=(0,-2)\) and \(E=\left(\dfrac{6+2}{2},\dfrac{5-5}{2}\right)=(4,0)\).
2. Slope of \(\overline{DE}\): \(\dfrac{0-(-2)}{4-0}=\dfrac{1}{2}\). Slope of \(\overline{AB}\): \(\dfrac{5-1}{6-(-2)}=\dfrac{4}{8}=\dfrac{1}{2}\). Equal slopes, so the segments are parallel.
3. \(DE=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}\) and \(AB=\sqrt{8^2+4^2}=\sqrt{80}=4\sqrt{5}\). So \(DE=\dfrac{1}{2}AB\).

10 Centroid of a sail ★★★

\(G=\left(\dfrac{1+7+4}{3},\dfrac{2+4+9}{3}\right)=(4,5)\).

The midpoint of \(\overline{BC}\) is \(M=(5.5,6.5)\). From \(A\) to \(G\) we move \((3,3)\); from \(G\) to \(M\) we move \((1.5,1.5)\). The first move is twice the second, so \(AG:GM=2:1\).

11 Inradius of a 5-12-13 triangle ★★★

Area \(K=\dfrac{1}{2}(5)(12)=30\) and semiperimeter \(s=\dfrac{5+12+13}{2}=15\). So \(r=\dfrac{K}{s}=\dfrac{30}{15}=2\text{ in}\).

The circumradius is half the hypotenuse: \(\dfrac{13}{2}=6.5\text{ in}\).

12 Range of the third side ★★★

1. \(14-9
2. The whole numbers from 6 to 22: \(22-6+1=17\) possible lengths.

13 Angle bisector distances ★★★

A point on an angle bisector is equidistant from the sides: \(2x+1=4x-9\), so \(10=2x\) and \(x=5\).

Each distance is \(2(5)+1=11\text{ in}\) (and \(4(5)-9=11\text{ in}\)).

14 Angles to sides ★★★

The angles add up to \(180^\circ\): \((2x+10)+(3x-5)+(x+25)=6x+30=180\), so \(x=25\).

\(\angle A=60^\circ\), \(\angle B=70^\circ\), \(\angle C=50^\circ\).

\(\overline{AB}\) faces \(\angle C\) (smallest), \(\overline{BC}\) faces \(\angle A\), and \(\overline{AC}\) faces \(\angle B\) (largest). So \(AB

15 A perpendicular bisector equation ★★★

1. The midpoint is \((2,4)\). The slope of \(\overline{AB}\) is \(\dfrac{6-2}{5+1}=\dfrac{2}{3}\), so the bisector has slope \(-\dfrac{3}{2}\). Then \(y-4=-\dfrac{3}{2}(x-2)\), which gives \(y=-1.5x+7\).
2. Distance squared to \(A\): \(1^2+5^2=26\). Distance squared to \(B\): \(5^2+1^2=26\). They are equal, so \((0,7)\) is equidistant from \(A\) and \(B\); it is also on the line since \(-1.5(0)+7=7\).

16 Circumcenter on the grid ★★★

The perpendicular bisector of \(\overline{AB}\) is \(x=4\). So \(O=(4,y)\) with \(OA=OC\):

\(16+y^2=(2-4)^2+(6-y)^2=4+36-12y+y^2\), so \(16=40-12y\) and \(y=2\).

\(O=(4,2)\). The radius is \(OA=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx 4.47\). Check with \(C\): \(\sqrt{4+16}=\sqrt{20}\).

17 Orthocenter on the grid ★★★

The altitude from \(C\) is perpendicular to the x-axis: \(x=2\).

The slope of \(\overline{BC}\) is \(\dfrac{6-0}{2-8}=-1\), so the altitude from \(A\) has slope \(1\): \(y=x\). At \(x=2\), \(y=2\), so \(H=(2,2)\).

Check: the slope of \(\overline{AC}\) is \(3\), so the altitude from \(B\) is \(y=-\dfrac{1}{3}(x-8)\); at \(x=2\) it gives \(y=2\). It passes through \(H\).

18 The Euler line ★★★

1. \(G=\left(\dfrac{0+8+2}{3},\dfrac{0+0+6}{3}\right)=\left(\dfrac{10}{3},2\right)\).
2. \(H\), \(G\) and \(O\) all have \(y=2\), so they lie on the horizontal line \(y=2\).
3. \(HG=\dfrac{10}{3}-2=\dfrac{4}{3}\) and \(GO=4-\dfrac{10}{3}=\dfrac{2}{3}\). So \(HG=2\cdot GO\): the centroid is twice as far from \(H\) as from \(O\).

19 Midpoints of a quadrilateral ★★★

1. In \(\triangle ABC\), \(\overline{PQ}\) is a midsegment, so \(PQ\parallel AC\) and \(PQ=\dfrac{1}{2}AC\). In \(\triangle ADC\), \(\overline{SR}\) is a midsegment, so \(SR\parallel AC\) and \(SR=\dfrac{1}{2}AC\). Hence \(PQ\parallel SR\) and \(PQ=SR\): one pair of opposite sides is parallel and congruent, so \(PQRS\) is a parallelogram.
2. \(PQ=SR=\dfrac{12}{2}=6\text{ cm}\). In the same way (using triangles \(ABD\) and \(CBD\)), \(QR=PS=\dfrac{16}{2}=8\text{ cm}\). The perimeter is \(2(6+8)=28\text{ cm}\).

20 Where to put the cell tower ★★★

1. A point equidistant from three vertices is the circumcenter.
2. The triangle is right-angled at \(P\), so the circumcenter is the midpoint of the hypotenuse \(\overline{QR}\): \(O=\left(\dfrac{10+0}{2},\dfrac{0+24}{2}\right)=(5,12)\). The hypotenuse is \(\sqrt{100+576}=26\), so the distance is \(13\text{ mi}\). Check: \(OP=\sqrt{25+144}=13\).

The tower stands at \((5,12)\), 13 miles (about 20.9 km) from each town.

21 Trail lengths ★★★

1. \(6.5-4.2
2. The whole numbers in that interval go from 3 to 10. The shortest is 3 km and the longest is 10 km.

22 Triangle inequality with algebra ★★★

All three lengths must be positive, and each sum of two sides must exceed the third.

\(2x+(x+5)>3x-4\) gives \(5>-4\): always true.

\(2x+(3x-4)>x+5\) gives \(4x>9\), so \(x>2.25\).

\((x+5)+(3x-4)>2x\) gives \(2x>-1\): true when \(x>2.25\).

Positivity: \(3x-4>0\) needs \(x>\dfrac{4}{3}\), also satisfied. So the triangle exists exactly when \(x>\dfrac{9}{4}\). Test: \(x=3\) gives sides 6, 8, 5, and \(5+6>8\).