Figure It Out — Questions & Solutions

Given

Two identical squares are each cut along a diagonal into two triangular pieces. Arrange the four pieces into a square of twice the original area.

To Find

A square made from all four triangular pieces

Reasoning

Each piece is an isosceles right triangle with legs equal to the original side a.

Solution

1. Bring the four right-angle vertices together.

   Place their equal legs along four perpendicular rays, with adjacent triangles sharing a leg.

2. The four hypotenuses form the outer boundary.

   Each has the same length and each boundary angle is 45°+45° = 90°.

3. There is no overlap or gap: four triangles of area $\frac{a^2}{2}$ fill the new square.

Result

The assembled square has area 4 × $\frac{a^2}{2}$ = $2a^2$.

Given

Find the hypotenuse of an isosceles right triangle with equal legs (i) 3, (ii) 4, (iii) 6, (iv) 8, (v) 9, and give decimal bounds.

To Find

Exact hypotenuses and strict one-decimal bounds

Reasoning

$c^2$ = $2a^2$, so c = a$\sqrt{2}$; compare squares of positive decimal endpoints.

Solution

1. (i) c = 3$\sqrt{2}$

   4.$2^2$ = 17.64 < 18 < 18.49 = 4.$3^2$.

2. (ii) c = 4$\sqrt{2}$

   5.$6^2$ = 31.36 < 32 < 32.49 = 5.$7^2$.

3. (iii) c = 6$\sqrt{2}$

   8.$4^2$ = 70.56 < 72 < 72.25 = 8.$5^2$.

4. (iv) c = 8$\sqrt{2}$

   11.$3^2$ = 127.69 < 128 < 129.96 = 11.$4^2$.

5. (v) c = 9$\sqrt{2}$

   12.$7^2$ = 161.29 < 162 < 163.84 = 12.$8^2$.

Result

Respectively: 4.2 < c < 4.3; 5.6 < c < 5.7; 8.4 < c < 8.5; 11.3 < c < 11.4; 12.7 < c < 12.8. Exact values are 3$\sqrt{2}$,4$\sqrt{2}$,6$\sqrt{2}$,8$\sqrt{2}$,9$\sqrt{2}$.

Given

Draw a right triangle with perpendicular sides 5 cm and 12 cm, measure its hypotenuse, then verify its length.

To Find

Hypotenuse length

Reasoning

Baudhāyana–Pythagoras theorem.

Solution

1. Draw AB = 12 cm, erect a perpendicular at A and mark AC = 5 cm.

   Join B to C and measure BC.

2. ${\mathrm{BC}}^2$ = ${\mathrm{AB}}^2$+${\mathrm{AC}}^2$ = 144+25 = 169.

   Since length is positive, BC = 13 cm.

3. Your drawing’s measurement should be close to 13 cm

   measurement precision does not replace the exact calculation.

Result

13 cm.

Given

Draw a right triangle with a leg 8 cm and hypotenuse 17 cm, measure the third side, then verify it.

To Find

The other leg length

Reasoning

The squared leg is hypotenuse squared minus the known leg squared.

Solution

1. Draw AB = 8 cm and a perpendicular ray at A.

   The circle centred at B with radius 17 cm meets the ray at C.

2. ${\mathrm{AC}}^2$ = ${17}^2$−$8^2$ = 289−64 = 225

   AC = 15 cm.

3. A careful drawing should measure approximately 15 cm.

Result

15 cm.

Given

Construct squares whose areas are three times and five times the area of a given square.

To Find

Constructed squares of areas 3A and 5A

Reasoning

A right triangle combines two square areas: $a^2$+$b^2$ = $c^2$.

Solution

1. Let the original side be s and its area A = $s^2$.

   Its diagonal d satisfies $d^2$ = 2A.

2. Construct perpendicular legs s and d

   their hypotenuse h satisfies $h^2$ = A+2A = 3A.

   Construct a square on h.

3. For five times the area, construct perpendicular legs s and 2s.

   Their hypotenuse k satisfies $k^2$ = A+4A = 5A.

   Construct a square on k.

Result

Required square sides are s$\sqrt{3}$ and s$\sqrt{5}$.

Given

In a right triangle with legs a,b and hypotenuse c, find the missing side: (i) a = 5,b = 7; (ii) a = 8,b = 12; (iii) a = 9,c = 15; (iv) a = 7,b = 12; (v) a = 1.5,b = 3.5.

To Find

Five exact missing lengths

Reasoning

Use $c^2$ = $a^2$+$b^2$ or $b^2$ = $c^2$−$a^2$ as appropriate.

Solution

1. (i) $c^2$ = 25+49 = 74, so c = $\sqrt{74}$.

2. (ii) $c^2$ = 64+144 = 208, so c = 4$\sqrt{13}$.

3. (iii) $b^2$ = 225−81 = 144, so b = 12.

4. (iv) $c^2$ = 49+144 = 193, so c = $\sqrt{193}$.

5. (v) $c^2$ = 2.25+12.25 = 14.5 = $\frac{58}{4}$, so c = $\frac{\sqrt{58}}{2}$.

Result

$\sqrt{74}$; 4$\sqrt{13}$; 12; $\sqrt{193}$; $\frac{\sqrt{58}}{2}$, respectively.

Given

Using odd squares as in the source, find five further Baudhāyana triples after (3,4,5) and (5,12,13).

To Find

Five further triples

Reasoning

For odd m ≥ 3, n = $\frac{\left(m^2+1\right)}{2}$ gives (n−1)${}^2$+$m^2$ = $n^2$.

Solution

1. Use m = 7,9,11,13,15, giving n = 25,41,61,85,113.

2. The triples are (24,7,25), (40,9,41), (60,11,61), (84,13,85), (112,15,113).

3. Checks: 576+49 = 625

   1600+81 = 1681

   3600+121 = 3721

   7056+169 = 7225

   12544+225 = 12769.

Result

All five triples satisfy the squared equality.

Given

Can the odd-square generator produce a non-primitive Baudhāyana triple?

To Find

Whether any generated triple has a common divisor greater than 1

Reasoning

A divisor common to consecutive integers divides their difference 1.

Solution

1. The generated triple contains n−1 and n.

2. Any common divisor of all three entries would divide both n−1 and n, hence divide 1.

3. Therefore the greatest common divisor is 1 for every generated positive triple.

Result

No. Every triple from this generator is primitive.

Given

Are there primitive triples not obtained by this odd-square method? Give an example.

To Find

A primitive triple outside the generator

Reasoning

Every generated triple has a leg exactly one less than its hypotenuse.

Solution

1. $8^2$+${15}^2$ = 64+225 = 289 = ${17}^2$

   gcd(8,15,17) = 1.

2. Neither leg is 16, so (8,15,17) cannot be of the form (n−1,m,n).

Result

Yes: (8,15,17).

Given

Find the diagonal of a square of side 5 cm.

To Find

Square diagonal length

Reasoning

The diagonal is the hypotenuse of a right triangle with two legs of 5 cm.

Solution

1. $d^2$ = $5^2$+$5^2$ = 50.

2. d = $\sqrt{50}$ = 5$\sqrt{2}$ cm.

Result

5$\sqrt{2}$ cm.

Given

Find the six missing side lengths in the right-triangle figures continuing on page 53, in reading order.

To Find

The six positive missing side lengths

Reasoning

Use the right-angle marks to identify legs and hypotenuses.

Solution

1. (i) Legs 7 and 9: $c^2$ = 49+81 = 130, so c = $\sqrt{130}$.

2. (ii) Legs 4 and 10 meet at the marked right angle: $c^2$ = 16+100 = 116, so c = 2$\sqrt{29}$.

3. (iii) Hypotenuse 41, leg 40: $b^2$ = 1681−1600 = 81, so b = 9.

4. (iv) Hypotenuse $\sqrt{200}$, leg 10: $b^2$ = 200−100 = 100, so b = 10.

5. (v) Legs 10 and $\sqrt{150}$: $c^2$ = 100+150 = 250, so c = 5$\sqrt{10}$.

6. (vi) Hypotenuse 45, leg 27: $b^2$ = 2025−729 = 1296, so b = 36.

Result

$\sqrt{130}$; 2$\sqrt{29}$; 9; 10; 5$\sqrt{10}$; 36.

Given

Find the side of a rhombus with diagonals 24 units and 70 units.

To Find

Length of each rhombus side

Reasoning

Rhombus diagonals bisect one another at right angles.

Solution

1. Half-diagonals are 12 and 35.

   Each side is a hypotenuse with these legs.

2. $s^2$ = ${12}^2$+${35}^2$ = 144+1225 = 1369

   s = 37.

Result

37 units.

Given

Is the hypotenuse always the longest side of a right triangle? Justify.

To Find

Comparison of the hypotenuse with both legs

Reasoning

The legs a,b are positive and $c^2$ = $a^2$+$b^2$.

Solution

1. Since $b^2$ > 0, $c^2$ > $a^2$, so c > a.

2. Since $a^2$ > 0, $c^2$ > $b^2$, so c > b.

   Positive square roots preserve these inequalities.

Result

Yes: the hypotenuse is strictly longer than either leg.

Given

True or false: every Baudhāyana triple is primitive or a scaled version of a primitive triple.

To Find

Truth and justification of the primitive-reduction statement

Reasoning

Divide a positive integer triple by its greatest common divisor.

Solution

1. Let g = gcd(a,b,c).

   If g = 1, the triple is primitive.

2. If g > 1, write a = gA,b = gB,c = gC.

   The numbers A,B,C are positive integers with gcd 1.

3. Dividing $a^2$+$b^2$ = $c^2$ by $g^2$ gives $A^2$+$B^2$ = $C^2$.

   Thus (A,B,C) is primitive and the original triple is its scale-g version.

Result

True.

Given

Give five rectangles with integer side lengths and integer diagonals.

To Find

Five verified integer rectangles

Reasoning

A rectangle diagonal satisfies $d^2$ = $a^2$+$b^2$.

Solution

1. Sides 3,4 give d = 5: 9+16 = 25.

2. Sides 5,12 give d = 13: 25+144 = 169.

3. Sides 8,15 give d = 17: 64+225 = 289.

4. Sides 7,24 give d = 25: 49+576 = 625.

5. Sides 12,35 give d = 37: 144+1225 = 1369.

Result

Rectangles 3 × 4, 5 × 12, 8 × 15, 7 × 24 and 12 × 35 satisfy the requirement.

Given

Squares of sides 7 and 5 units.

To Find

A square of area $7^2$−$5^2$ and its exact side

Property

Construction: draw AB = 5. At B draw a perpendicular; intersect it at C with the radius-7 circle centred at A.

Solution

1. AC = 7 and ∠ABC = 90° by construction.

2. ${\mathrm{BC}}^2$ = ${\mathrm{AC}}^2$−${\mathrm{AB}}^2$ = 49−25 = 24.

3. Construct a square on BC.

   Its area is 24.

4. BC = $\sqrt{24}$ = 2$\sqrt{6}$.

Result

Required side = 2$\sqrt{6}$ units; area = 24 square units.

Given

An unlimited unit lattice; every vertex must have integer coordinates.

To Find

Possible areas, including 2,3,4,5

Property

For integers p,q not both zero, perpendicular equal side vectors are $\left(p,q\right)$ and $\left(-q,p\right)$.

Solution

1. Vertices $\left(0,0\right)$,$\left(p,q\right)$,(p−q,p+q),$\left(-q,p\right)$ form a square of area $p^2$+$q^2$.

2. Area 2: use $\left(p,q\right)$ = $\left(1,1\right)$.

   Area 4: $\left(2,0\right)$.

   Area 5: $\left(2,1\right)$.

3. For area 3, |p| and |q| can only be 0 or 1

   their squared sum is at most 2.

   Thus 3 is impossible.

4. Every lattice square has an integer side vector, so the condition is necessary as well as sufficient.

Result

Exactly the positive integers expressible as a sum of two integer squares occur.

Given

Equilateral △ABC with each side 6; altitude AD to BC.

To Find

Area of △ABC

Property

RHS congruence, the Baudhāyana–Pythagoras theorem and triangle area.

Solution

1. △ADB≅△ADC by RHS: equal hypotenuses AB = AC, common AD and right angles at D.

2. Thus BD = DC = 3.

3. ${\mathrm{AD}}^2$ = $6^2$−$3^2$ = 27, so AD = 3$\sqrt{3}$.

4. Area = $\frac{\left(6\times 3\sqrt{3}\right)}{2}$.

Result

Area = 9$\sqrt{3}$ square units.