Elementary Mathematics 2019 (I)

Step-by-step solution

1. 1
   a square enclosing the circle touches it on all four sides, so its side is the diameter $2r$

   Why: A square containing a circle of radius $r$ cannot be narrower than $2r$, and the enclosing square is exactly that width.

   The largest circle that fits inside a square
2. 2
   circle piece $=2\pi r$; square piece $=4\times 2r=8r$

   Why: Each piece of wire becomes the perimeter of its shape, so the wire lengths are a circumference and a perimeter.

   Area and circumference of a circlePerimeter and area of a square
3. 3
   $2\pi r+8r=12$

   Why: The two pieces came from one $12$ m wire, so their lengths add to $12$ — that single equation is the whole model.

   Turn a worded condition into a linear equation in one unknown
4. 4
   $r\left(2\pi +8\right)=12\Rightarrow r=\frac{12}{2\left(\pi +4\right)}=\frac{6}{\pi +4}$

   Why: Taking $2$ out of the bracket is what brings the answer to the printed form.

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Step-by-step solution

1. 1
   $x+x+4x=180^{\circ }\Rightarrow x=30^{\circ }$, so the angles are $30^{\circ }$, $30^{\circ }$ and $120^{\circ }$

   Why: A ratio of angles plus the angle sum pins the angles down completely.

   Introduce a common factor $k$ for a ratioAngle sum of a triangle
2. 2
   two angles are equal, so the two sides facing them are equal; take each as $1$, and the side facing $120^{\circ }$ is then the largest

   Why: Only the angles are given, so the triangle is fixed up to size — choosing the equal sides as $1$ costs nothing because $k$ is a ratio.

   Base angles of an isosceles triangle
3. 3
   the altitude from the $120^{\circ }$ vertex $A$ bisects $BC$ and splits the $120^{\circ }$ into two $60^{\circ }$ angles, so $\frac{BC/2}{1}=\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ and $BC=\sqrt{3}$

   Why: The altitude in an isosceles triangle lands on the midpoint of the base, turning the problem into one $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle whose ratios are standard.

   Sine, cosine and tangent in a right triangleExact trigonometric ratios of the standard angles
4. 4
   $k=\frac{1+1+\sqrt{3}}{\sqrt{3}}=\frac{2+\sqrt{3}}{\sqrt{3}}=\frac{2}{\sqrt{3}}+1$

   Why: Splitting the fraction term by term is quicker here than rationalising, and it lands straight on the printed form.

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Step-by-step solution

1. 1
   legs $a$ and $b$: $\frac{1}{2}ab=24\Rightarrow ab=48$, and $a^2+b^2=10^2=100$

   Why: In a right-angled triangle the two legs are base and height, so the area gives their product and Pythagoras gives the sum of their squares.

   Area of a triangle from base and heightPythagoras' theorem
2. 2
   $(a+b{)}^2=a^2+b^2+2ab=100+96=196\Rightarrow a+b=14$; $(a-b{)}^2=100-96=4\Rightarrow a-b=2$

   Why: These two identities turn a sum of squares and a product into a sum and a difference, avoiding a quadratic altogether.

   Square of a sum and of a difference
3. 3
   $a=8$, $b=6$: the shorter side is $6$ cm and the longer is $8$ cm

   Why: Adding and subtracting the two results gives the legs individually.

4. 4
   new legs $\frac{6}{2}=3$ and $2\times 8=16$, so the new hypotenuse $=\sqrt{3^2+16^2}=\sqrt{9+256}=\sqrt{265}$ cm

   Why: The right angle sits between the two legs, so changing their lengths leaves it a right triangle and Pythagoras still applies.

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Step-by-step solution

1. 1
   $AB=AC$, so $A$ lies on the perpendicular bisector of $BC$; the centre $O$ does too, since $OB=OC=8$. Hence $A$, $O$ and the midpoint $M$ of $BC$ are collinear and $AM\perp BC$

   Why: Two distinct points equidistant from $B$ and $C$ determine the perpendicular bisector of $BC$, which is what makes one right triangle serve both the chord and the radius.

   The perpendicular from the centre bisects a chordThe circumcentre is the point equidistant from the vertices
2. 2
   write $AM=h$ and $BM=m$; in right triangle $ABM$, $h^2+m^2=12^2=144$

   Why: $AB$ is the hypotenuse of the right triangle cut off by the altitude $AM$.

   Pythagoras' theorem
3. 3
   $O$ lies on $AM$ with $AO=8$, so $OM=h-8$; in right triangle $OBM$, $(h-8{)}^2+m^2=8^2=64$

   Why: The second right triangle uses the radius $OB$ as hypotenuse on the same foot $M$, which is what lets $m$ be eliminated.

4. 4
   subtract: $(h-8{)}^2-h^2=64-144\Rightarrow -16h+64=-80\Rightarrow h=9$

   Why: Subtracting removes $m^2$ and the $h^2$ terms together, leaving one linear equation.

5. 5
   $m^2=144-81=63\Rightarrow m=3\sqrt{7}$, so $BC=2m=6\sqrt{7}$ cm

   Why: The perpendicular from the centre bisects the chord, so the chord is twice $BM$, not $BM$.

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Step-by-step solution

1. 1
   Statement 1. In an isosceles trapezium $ABCD$ with $AB\parallel DC$, the leg $AD$ is a transversal of the parallel sides, so $\angle A+\angle D=180^{\circ }$; and the equal legs give equal base angles, $\angle A=\angle B$

   Why: The parallel sides give one supplementary pair immediately, and the 'isosceles' condition is exactly what supplies the second equality.

   Angles made by a transversal of two parallel lines
2. 2
   $\angle B+\angle C=180^{\circ }$ as well, so $\angle A+\angle C=\angle B+\angle C=180^{\circ }$: the opposite angles $A$ and $C$ are supplementary, and the trapezium is cyclic

   Why: Opposite angles summing to $180^{\circ }$ is the test for a quadrilateral to be inscribable in a circle, so statement 1 holds for every isosceles trapezium.

   Opposite angles of a cyclic quadrilateral are supplementary
3. 3
   Statement 2. In a parallelogram opposite angles are equal: $\angle A=\angle C$. If it is also cyclic then $\angle A+\angle C=180^{\circ }$, so $2\angle A=180^{\circ }$ and $\angle A=90^{\circ }$

   Why: The two facts pull in opposite directions — equal and supplementary — and the only angle that is both is $90^{\circ }$.

   Angles of a parallelogram
4. 4
   a parallelogram with one right angle has all four right, so it is a rectangle: statement 2 holds too, and both statements are correct

   Why: Consecutive angles of a parallelogram are supplementary, so one right angle forces the rest.

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Step-by-step solution

1. 1
   let the ladder be $L$ and the height it reaches $h$: $h^2+2.5^2=L^2$

   Why: The wall is vertical, so wall, ground and ladder make a right triangle with the ladder as hypotenuse.

   Pythagoras' theorem
2. 2
   after the slip the top is at $h-0.8$ and the foot at $2.5+1.4=3.9$: $(h-0.8{)}^2+3.9^2=L^2$

   Why: The ladder's length does not change, so both right triangles share the same hypotenuse $L$ — that is what makes two equations in two unknowns solvable.

3. 3
   $h^2+6.25=h^2-1.6h+0.64+15.21$

   Why: Equating the two expressions for $L^2$ removes $L$, and expanding $(h-0.8{)}^2$ lets $h^2$ cancel as well.

   Square of a sum and of a difference
4. 4
   $6.25=15.85-1.6h\Rightarrow 1.6h=9.6\Rightarrow h=6$

   Why: What is left is linear in $h$, so no quadratic is needed.

5. 5
   $L=\sqrt{6^2+2.5^2}=\sqrt{42.25}=6.5$ m

   Why: Put $h$ back into either triangle; the first is the simpler one.

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Step-by-step solution

1. 1
   let the radius be $r$ with centres $O_1$, $O_2$; $O_2$ lies on the first circle, so $O_1{O}_2=r$

   Why: A circle is the set of points at distance $r$ from its centre, so 'passes through the centre of the other' says precisely that the centre distance is $r$.

   A circle is the locus of points at a fixed distance from a fixed point
2. 2
   the common chord $PQ$ is perpendicular to $O_1{O}_2$ and bisected by it; the circles being equal, it is bisected at the midpoint $M$ of $O_1{O}_2$, so $O_{1}M=\frac{r}{2}$

   Why: $P$ and $Q$ are each $r$ from both centres, so $O_1{O}_2$ is the perpendicular bisector of $PQ$ and, by the same symmetry, $PQ$ bisects $O_1{O}_2$.

   The perpendicular from the centre bisects a chord
3. 3
   in right triangle $O_{1}MP$: $MP=\sqrt{r^2-{\left(\frac{r}{2}\right)}^2}=\frac{r\sqrt{3}}{2}$, so $PQ=r\sqrt{3}$

   Why: Half the chord, the distance from the centre and the radius form the usual right triangle.

   Pythagoras' theorem
4. 4
   $r\sqrt{3}=10\sqrt{3}\Rightarrow r=10$, so the diameter is $20$ cm

   Why: The $\sqrt{3}$ cancels, which is why the question printed the chord with a $\sqrt{3}$ in it.

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Step-by-step solution

1. 1
   Statement 1. A circle through $A$, $B$, $C$ has its centre equidistant from all three, so that centre is the circumcentre — the one point where the perpendicular bisectors of $AB$ and $BC$ meet. Non-collinear points make those bisectors non-parallel, so they meet once and there is exactly one such circle

   Why: Uniqueness comes from the bisectors crossing at a single point; 'infinitely many' is what happens with two points, not three.

   The circumcentre is the point equidistant from the vertices
2. 2
   Statement 2. Let the chord $BC$ subtend $\theta <180^{\circ }$ at the centre on the minor-arc side. A point $P$ in the minor segment lies on the minor arc, so the central angle on the far side of $BC$ from $P$ is the reflex angle $360^{\circ }-\theta >180^{\circ }$, and $\angle BPC=\frac{360^{\circ }-\theta }{2}=180^{\circ }-\frac{\theta }{2}>90^{\circ }$

   Why: The angle-at-centre theorem halves whichever central angle lies on the opposite side of the chord from the vertex, and for a vertex on the minor arc that is the reflex one.

   The angle at the centre is twice the angle at the circumference
3. 3
   the angle in a minor segment is therefore obtuse, not acute, so statement 2 fails as well: neither statement is correct

   Why: Both claims have been refuted, so the answer is the 'neither' option.

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Step-by-step solution

1. 1
   the triangle inequality on $△ABC$ gives $AB+BC>AC$, $AB+AC>BC$ and $BC+AC>AB$

   Why: These three are the whole content of the theorem, so any always-true inequality among the sides must be one of them rearranged.

   Triangle inequality
2. 2
   from $AB+BC>AC$, subtract $AB$: $BC>AC-AB$, i.e. $AC-AB<BC$ — statement 1

   Why: Subtracting the same length from both sides preserves the inequality.

3. 3
   from $AB+AC>BC$, subtract $AC$: $AB>BC-AC$, i.e. $BC-AC<AB$ — statement 2

   Why: The same move on the second of the three.

4. 4
   from $BC+AC>AB$, subtract $BC$: $AC>AB-BC$, i.e. $AB-BC<AC$ — statement 3

   Why: The same move on the third, so all three statements are the theorem in disguise.

5. 5
   all three are correct

   Why: Each of the three printed statements has been derived, so none can be dropped.

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Step-by-step solution

1. 1
   Statement 1. Doubling the median $AD$ past $D$ makes $AB{A}^{\prime }C$ a parallelogram, and the triangle inequality in $△AB{A}^{\prime }$ gives $2m_a<AB+AC$; writing the same for the other two medians and adding gives $2\left(m_a+m_b+m_c\right)<2(a+b+c)$

   Why: The doubled median is a diagonal of a parallelogram whose sides are two sides of the triangle, which is what converts a median into an ordinary triangle inequality.

   The sum of the medians lies between $\frac{3}{4}$ of the perimeter and the perimeter
2. 2
   so $m_a+m_b+m_c<a+b+c$: the perimeter exceeds the sum of the medians, and statement 1 is correct

   Why: Dividing by $2$ leaves exactly the claim made.

3. 3
   Statement 2. For $D$ on $BC$, $△ABD$ gives $AB+BD>AD$ and $△ACD$ gives $AC+DC>AD$

   Why: $D$ on $BC$ splits the triangle into two triangles sharing the segment $AD$, so the inequality can be written twice.

   Triangle inequality
4. 4
   add them: $AB+AC+(BD+DC)>2AD$, and $BD+DC=BC$, so $AB+BC+CA>2AD$ — statement 2 is correct too, and both statements hold

   Why: Adding is what makes the two halves of $BC$ recombine into the whole side, giving the perimeter on the left.

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Step-by-step solution

1. 1
   let the cube's edge be $a$; inscribed in the sphere, all eight vertices lie on it, so the sphere's diameter is the space diagonal $a\sqrt{3}$ and $R=\frac{a\sqrt{3}}{2}$

   Why: The longest line inside a cube is the space diagonal, so that is what the sphere must accommodate.

   Space diagonal of a cube and a cuboid
2. 2
   the cylinder touches all four vertical faces, so its diameter is the edge $a$ and $r=\frac{a}{2}$; standing in the cube, its height is $a$

   Why: A circle touching all four sides of a square of side $a$ has diameter $a$ — the horizontal cross-section of the cube is that square.

   The largest circle that fits inside a square
3. 3
   the cone shares the cylinder's height and diameter, so $r=\frac{a}{2}$ and $h=a$ for it as well

   Why: The stem says the heights are the same and the diameters equal, which fixes the cone completely.

4. 4
   $V_{\text{sphere}}=\frac{4}{3}\pi {\left(\frac{a\sqrt{3}}{2}\right)}^3=\frac{4}{3}\pi \cdot \frac{3\sqrt{3}{a}^3}{8}=\frac{\sqrt{3}\pi a^3}{2}$

   Why: Cubing $\frac{\sqrt{3}}{2}$ gives $\frac{3\sqrt{3}}{8}$ — the step where a $\sqrt{3}$ is most easily lost.

   Volume and surface area of a sphere
5. 5
   $V_{\text{cone}}=\frac{1}{3}\pi {\left(\frac{a}{2}\right)}^{2}a=\frac{\pi a^3}{12}$

   Why: A cone is one third of the cylinder on the same base with the same height.

   Volume of a right circular cone
6. 6
   $\frac{V_{\text{sphere}}}{V_{\text{cone}}}=\frac{\sqrt{3}\pi a^3}{2}\times \frac{12}{\pi a^3}=6\sqrt{3}$, so the ratio is $6\sqrt{3}:1$

   Why: Both $\pi$ and $a^3$ cancel, which is why no value of $\pi$ is needed here.

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Step-by-step solution

1. 1
   cube volume $=a^3$

   Why: Everything will be written in the one length the configuration offers.

   Volume and wall area of a cuboid
2. 2
   the cylinder touches all four vertical faces, so $r=\frac{a}{2}$, and it stands the full height of the cube, so $h=a$

   Why: The horizontal cross-section of the cube is a square of side $a$, and a circle touching all four of its sides has diameter $a$.

   The largest circle that fits inside a square
3. 3
   cylinder volume $=\pi {\left(\frac{a}{2}\right)}^{2}a=\frac{\pi a^3}{4}$

   Why: Base area times height, with the radius squared — not the diameter.

   Volume of a right circular cylinder
4. 4
   ratio $=a^3:\frac{\pi a^3}{4}=4:\pi$

   Why: $a^3$ cancels, so the ratio does not depend on the size of the cube.

5. 5
   with $\pi =\frac{22}{7}$: $4:\frac{22}{7}=28:22=14:11$

   Why: The printed options are ratios of whole numbers, so $\pi$ has to be taken as $\frac{22}{7}$ to land on one.

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Step-by-step solution

1. 1
   with the cube's edge $a$: $R=\frac{a\sqrt{3}}{2}$ for the sphere, and $r=\frac{a}{2}$, $h=a$ for both the cylinder and the cone

   Why: The sphere takes the space diagonal as its diameter; the cylinder's diameter is the edge, because it touches all four vertical faces.

   Space diagonal of a cube and a cuboidThe largest circle that fits inside a square
2. 2
   sphere surface $=4\pi R^2=4\pi \cdot \frac{3a^2}{4}=3\pi a^2$

   Why: Squaring $\frac{\sqrt{3}}{2}$ gives $\frac{3}{4}$, which cancels the $4$ neatly.

   Volume and surface area of a sphere
3. 3
   cone slant height $l=\sqrt{{\left(\frac{a}{2}\right)}^2+a^2}=\frac{a\sqrt{5}}{2}$, so its curved surface $=\pi rl=\pi \cdot \frac{a}{2}\cdot \frac{a\sqrt{5}}{2}=\frac{\sqrt{5}\pi a^2}{4}$

   Why: Curved surface needs the slant height, not the vertical height — and that is where the $\sqrt{5}$ in statement 1 comes from.

   Slant height and curved surface of a cone
4. 4
   $3\pi a^2÷\frac{\sqrt{5}\pi a^2}{4}=\frac{12}{\sqrt{5}}\approx 5.37$, not $\sqrt{5}\approx 2.24$: statement 1 is wrong

   Why: The $\sqrt{5}$ sits in the denominator with a $12$ over it, so the factor is far larger than $\sqrt{5}$ itself.

5. 5
   cube surface $=6a^2$; cylinder curved surface $=2\pi \cdot \frac{a}{2}\cdot a=\pi a^2\approx 3.14a^2$ — not equal, so statement 2 is wrong as well and neither statement is correct

   Why: A cube's six faces total $6a^2$, nearly twice the unrolled cylinder wall, so no arithmetic slip could make them agree.

   Volume and wall area of a cuboidCurved and total surface of a cylinder

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Step-by-step solution

1. 1
   $\angle ABC=90^{\circ }$, so $AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{81+1600}=\sqrt{1681}=41$ cm

   Why: $AC$ is the only link between the right-angled half of the quadrilateral and the half the question asks about.

   Pythagoras' theorem
2. 2
   $△ADC$ has $DA=15$, $DC=28$, $CA=41$, so $s=\frac{15+28+41}{2}=42$

   Why: Three sides and no height is exactly the situation Heron's formula is for.

   Heron's formula
3. 3
   area $=\sqrt{42(42-15)(42-28)(42-41)}=\sqrt{42\times 27\times 14\times 1}$

   Why: The $s-a$ values are $27$, $14$ and $1$ — the $1$ is a warning that the triangle is nearly flat, not an error.

4. 4
   $42\times 27\times 14=\left(2\times 3\times 7\right)\left(3^3\right)\left(2\times 7\right)=2^2\times 3^4\times 7^2$, so the area is $2\times 9\times 7=126$ cm${}^2$

   Why: Factorising before taking the root avoids multiplying out to $15876$ and then hunting for its square root.

   Prime factorisation

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Step-by-step solution

1. 1
   $AC=\sqrt{9^2+40^2}=\sqrt{1681}=41$ cm, from the right angle at $B$

   Why: The diagonal is the cut line, so it has to be found before either piece can be measured.

   Pythagoras' theorem
2. 2
   $△ABC$ is right-angled at $B$, so its two legs are base and height: area $=\frac{1}{2}\times 9\times 40=180$ cm${}^2$

   Why: A right angle hands over the height for free; no Heron is needed on this half.

   Area of a triangle from base and height
3. 3
   $△ADC$ has sides $15$, $28$, $41$ and $s=42$: area $=\sqrt{42\times 27\times 14\times 1}=126$ cm${}^2$

   Why: Three sides and no angle is Heron's case.

   Heron's formula
4. 4
   $AC$ is a diagonal, so the two triangles tile the quadrilateral: area $=180+126=306$ cm${}^2$

   Why: A diagonal of a convex quadrilateral splits it into exactly two triangles with no overlap and nothing left over.

   Cut a compound plane figure into standard pieces

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Step-by-step solution

1. 1
   $(AB+BC+CA)-(AD+DC+CA)=(AB+BC)-(AD+DC)$

   Why: $CA$ appears once in each perimeter, so it drops out — which means the right angle is not even needed for this item.

2. 2
   $(9+40)-(15+28)=49-43=6$ cm

   Why: Four small additions, no square roots.

3. 3
   the long way round: $AC=\sqrt{81+1600}=41$, so the perimeters are $9+40+41=90$ cm and $15+28+41=84$ cm, differing by $6$ cm

   Why: Computing both perimeters confirms the cancellation rather than relying on it.

   Pythagoras' theorem

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Step-by-step solution

1. 1
   in an equilateral triangle the circumcentre is also the centroid, so it lies at $\frac{R}{2}$ from each side; half a side and $\frac{R}{2}$ are then the legs of a right triangle with hypotenuse $R$

   Why: The centroid cuts each median $2:1$, so the short part — the distance to the side — is half the long part, which is $R$.

   The circumcentre is the point equidistant from the verticesSide of a regular polygon inscribed in a circle
2. 2
   ${\left(\frac{a}{2}\right)}^2+{\left(\frac{R}{2}\right)}^2=R^2\Rightarrow \frac{a^2}{4}=\frac{3R^2}{4}\Rightarrow a=R\sqrt{3}$

   Why: The radius drawn to a vertex and the perpendicular dropped to a side make the right triangle that ties $a$ to $R$.

   Pythagoras' theorem
3. 3
   $a=20\sqrt{3}\times \sqrt{3}=20\times 3=60$ cm

   Why: The two $\sqrt{3}$ factors multiply to $3$, which is why the radius was printed with a $\sqrt{3}$ in it.

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Step-by-step solution

1. 1
   the side is $a=R\sqrt{3}=20\sqrt{3}\times \sqrt{3}=60$ cm

   Why: The inscribed equilateral triangle has side $R\sqrt{3}$, which the previous item establishes.

   Side of a regular polygon inscribed in a circle
2. 2
   the median from $A$ is also the altitude: $\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 60=30\sqrt{3}$ cm

   Why: In an equilateral triangle the median, altitude, angle bisector and perpendicular bisector from a vertex are one line.

   Area and altitude of an equilateral triangle
3. 3
   the centroid divides each median $2:1$ from the vertex, so $d=\frac{2}{3}\times 30\sqrt{3}=20\sqrt{3}$ cm

   Why: The question asks for the distance from the vertex, which is the long part of the $2:1$ split.

   The centroid divides each median in the ratio $2:1$

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Step-by-step solution

1. 1
   $l+b+h=22$, and the space diagonal gives $l^2+b^2+h^2=14^2=196$

   Why: The diagonal of a cuboid is $\sqrt{l^2+b^2+h^2}$, so the second given is a statement about the sum of squares and nothing else.

   Space diagonal of a cube and a cuboid
2. 2
   $(l+b+h{)}^2=l^2+b^2+h^2+2(lb+bh+hl)$

   Why: This identity is the only bridge between the two givens and the cross-products the surface area is built from.

   Square of a sum of several terms
3. 3
   $22^2=484=196+2(lb+bh+hl)\Rightarrow 2(lb+bh+hl)=288$

   Why: The bracket is left doubled on purpose — that is already the quantity wanted.

4. 4
   the total surface area of a cuboid is exactly $2(lb+bh+hl)$, so it is $288$ cm${}^2$

   Why: The three dimensions are never needed separately; the identity delivers the surface area whole.

   Volume and wall area of a cuboid

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Step-by-step solution

1. 1
   $S-3V=l^3+b^3+h^3-3lbh=(l+b+h)\left(l^2+b^2+h^2-lb-bh-hl\right)$

   Why: This is the whole question: the expression asked for is precisely the one that factorises, which is why it can be evaluated without knowing the edges.

   $a^3+b^3+c^3-3abc$, and the case $a+b+c=0$
2. 2
   $l+b+h=22$, and the diagonal gives $l^2+b^2+h^2=14^2=196$

   Why: The first factor is a given outright; the second needs one more quantity.

   Space diagonal of a cube and a cuboid
3. 3
   $(l+b+h{)}^2=l^2+b^2+h^2+2(lb+bh+hl)\Rightarrow 484=196+2(lb+bh+hl)\Rightarrow lb+bh+hl=144$

   Why: Squaring the edge sum is the standard way to reach the cross-products from the sum and the sum of squares.

   Square of a sum of several terms
4. 4
   $S-3V=22\times (196-144)=22\times 52=1144$ cm${}^3$

   Why: Both factors are now numbers, so the volume itself — which the data do not determine — never has to be found.