Mathbench

Chapter 5 — Geometry

Geometry is 17 to 20% of the test — eight questions — and on the ACT it is larger and more varied than on the SAT. Circles, solids, polygons and coordinate geometry all appear, and there is no formula sheet to lean on.

That last point governs this whole chapter. Every formula used here is in Chapter 8, with the reason it is true. Learn them there; use them here.

One habit worth building from the start: draw the picture. Almost every ACT geometry question describes a shape in words, and a sketch with the numbers written on it turns a confusing paragraph into an obvious calculation. It takes fifteen seconds and it is the single most reliable way to gain marks in this category.

Topics covered: angles in triangles and on lines · isosceles and equilateral triangles · polygons and quadrilaterals · similar figures · circles, arcs and sectors · the equation of a circle · solids and volume · coordinate geometry


Section 1 — Angles, Triangles and Polygons

The facts this section needs, all of them worth knowing by heart:

That last formula covers quadrilaterals (360°),(360°)\mathclose{,} pentagons (540°),(540°)\mathclose{,} hexagons (720°)(720°) and everything beyond, so it is one fact rather than five.


Q1Basic

Topic: The third angle of a triangle

A surveyor measures two of the three interior angles of a triangular plot of land. The plot is shown, with the two measured angles marked.

A triangle47°68°x°

What is the value of x?x\mathclose{?}

A) 6565

B) 112112

C) 115115

D) 245245

Show the worked solution

Answer: A

Explanation

The three angles of a triangle add to 180°:180°\mathclose{:}

47+68+x=18047 + 68 + x = 180 115+x=180115 + x = 180 x=65°x = 65°

Check: 47+68+65=180.47 + 68 + 65 = 180\mathclose{.} Correct.

Why each wrong option is wrong:

  • B, 112112 — subtracted only the 68°68° from 180 and forgot the 47°.47°\mathclose{.}
  • C, 115115 — added the two given angles and stopped. 115 is a step on the way, not the answer.
  • D, 245245 — subtracted from 360°,360°\mathclose{,} the total for a quadrilateral.

Always add all three of your angles at the end; it takes two seconds and catches every one of these.

Takeaway: the angles of a triangle add to 180°.180°\mathclose{.} Subtract the sum of the two you know, then add all three back to check.


Q2Medium

Topic: The base angles of an isosceles triangle

In triangle ABC,ABC\mathclose{,} AB=ACAB = AC and the measure of angle AA is 40°.40°\mathclose{.} What is the measure, in degrees, of angle B?B\mathclose{?}

F) 140140

G) 100100

H) 7070

J) 4040

Show the worked solution

Answer: H

Explanation

AB=AC,AB = AC\mathclose{,} so the angles opposite those sides — angles CC and BB — are equal. Call each of them x.x\mathclose{.}

40+x+x=18040 + x + x = 180 2x=1402x = 140 x=70°x = 70°

Check: 40+70+70=180.40 + 70 + 70 = 180\mathclose{.} Correct.

Which angles are equal? The ones opposite the equal sides. ABAB and ACAC both start at A,A\mathclose{,} so the equal angles are the other two, BB and C.C\mathclose{.} Getting this backwards is the usual error, and drawing the triangle makes it obvious.

Why each wrong option is wrong:

  • F, 140140 — found the 140°140° left over and forgot it is shared between two angles.
  • G, 100100 — treated angle CC as 40°40° as well.
  • J, 4040 — assumed the triangle is equilateral.

Takeaway: equal sides face equal angles. Subtract the odd angle from 180°,180°\mathclose{,} then halve what is left.


Q3Medium

Topic: An exterior angle of a triangle

In a triangle, the two interior angles not adjacent to a particular exterior angle measure 52°52° and 61°.61°\mathclose{.} What is the measure, in degrees, of that exterior angle?

A) 6767

B) 113113

C) 119119

D) 128128

Show the worked solution

Answer: B

Explanation

The exterior angle rule:

exterior angle=52+61=113°\text{exterior angle} = 52 + 61 = 113°

Why it works: the third interior angle is 180−52−61=67°,180 - 52 - 61 = 67°\mathclose{,} and the exterior angle sits on a straight line with it, so it is 180−67=113°.180 - 67 = 113°\mathclose{.} The rule is just those two steps combined — which is why it saves time.

Why each wrong option is wrong:

  • A, 6767 — gave the third interior angle, the one the exterior angle sits next to.
  • D, 128128 and C, 119119 — used only one of the two given angles.

Takeaway: an exterior angle equals the sum of the two remote interior angles. If you forget the rule, find the third angle and subtract from 180°180° — same answer, one more step.


Q4Medium

Topic: The interior angles of a polygon

What is the measure, in degrees, of each interior angle of a regular hexagon?

F) 6060

G) 120120

H) 180180

J) 720720

Show the worked solution

Answer: G

Explanation

Total interior angles:

(6−2)×180=4×180=720°(6 - 2) \times 180 = 4 \times 180 = 720°

The hexagon is regular, so all six angles are equal:

7206=120°\frac{720}{6} = 120°

Why each wrong option is wrong:

  • F, 6060 — gave the exterior angle. The exterior angles of any polygon add to 360°,360°\mathclose{,} so each is 3606=60°\frac{360}{6} = 60° — and indeed 120+60=180,120 + 60 = 180\mathclose{,} as they sit on a straight line together.
  • H, 180180 — used 6×1806 \times 180 for the total instead of (6−2)×180.(6 - 2) \times 180\mathclose{.} An angle of 180°180° is a straight line, not a corner.
  • J, 720720 — gave the total for all six angles.

Takeaway: (n−2)×180(n-2) \times 180 is the total. Divide by nn for one angle of a regular polygon. The exterior angles always total 360°,360°\mathclose{,} whatever nn is.


Q5Hard

Topic: Similar triangles in a shadow problem

A flagpole casts a shadow 9 metres long at the same moment that a 2-metre post casts a shadow 1.5 metres long on level ground. How tall is the flagpole, in metres?

A) 1818

B) 1212

C) 9.59.5

D) 6.756.75

Show the worked solution

Answer: B

Explanation

Draw both triangles. Each has a vertical object and a horizontal shadow, and the sun makes the same angle in both — so they are similar, and matching sides are in the same ratio.

Set up the proportion with height on top in both:

flagpole heightflagpole shadow=post heightpost shadow\frac{\text{flagpole height}}{\text{flagpole shadow}} = \frac{\text{post height}}{\text{post shadow}}

h9=21.5\frac{h}{9} = \frac{2}{1.5}

h=9×21.5=9×43=12 mh = 9 \times \frac{2}{1.5} = 9 \times \frac{4}{3} = 12 \text{ m}

Sanity check: the post is taller than its shadow (2(2 against 1.5),1.5)\mathclose{,} so the flagpole must be taller than its 9-metre shadow. 12 is. Correct.

Why each wrong option is wrong:

  • A, 1818 — multiplied the 9 by the post's height and never divided by the post's shadow.
  • C, 9.59.5 — added: the post is half a metre taller than its shadow, so the flagpole was made half a metre taller than its own. Similar triangles scale by a ratio, never by a difference.
  • D, 6.756.75 — set the proportion up upside down, which makes the flagpole shorter than its shadow and fails the sanity check.

Takeaway: similar shapes give equal ratios. Put the same measurement on top of both fractions, then check the answer is on the right side of the number you started with.


Section 2 — Circles

Four facts carry almost every ACT circle question:

circumference=2πrarea=πr2\text{circumference} = 2\pi r \qquad \text{area} = \pi r^{2}

An arc is part of the circumference and a sector is a slice of the area. Both are the same fraction of the whole, and that fraction is central angle360:\dfrac{\text{central angle}}{360}\mathclose{:}

arc length=θ360×2πrsector area=θ360×πr2\text{arc length} = \frac{\theta}{360} \times 2\pi r \qquad \text{sector area} = \frac{\theta}{360} \times \pi r^{2}

And the equation of a circle with centre (h,k)(h, k) and radius r:r\mathclose{:}

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Note the minus signs in the brackets and that the right-hand side is r2,r^2\mathclose{,} not r.r\mathclose{.} Both are where the marks go.


Q6Medium

Topic: The area of a sector

A circular flower bed is to be divided into wedge-shaped planting sections. The figure shows one section: a sector with centre O,O\mathclose{,} a radius of 6 metres and a central angle of 60°.60°\mathclose{.}

A circle with two radii60°6OAB

What is the area of that sector, in square metres?

F) 2π2 \pi

G) 6π6 \pi

H) 12π12 \pi

J) 36π36 \pi

Show the worked solution

Answer: G

Explanation

A 60°60° sector is 60360=16\frac{60}{360} = \frac{1}{6} of the circle.

The whole circle's area:

πr2=π×36=36π\pi r^2 = \pi \times 36 = 36\pi

One sixth of it:

16×36π=6π square metres\frac{1}{6} \times 36\pi = 6\pi \text{ square metres}

Why each wrong option is wrong:

  • F, 2π2 \pi — used the arc-length formula, which gives a length, not an area. Check the units: an area needs r2r^2 in it.
  • H, 12π12 \pi — used 2πr22\pi r^2 in place of πr2,\pi r^2\mathclose{,} carrying the 2 over from the circumference formula.
  • J, 36π36 \pi — gave the area of the whole circle without taking the fraction.

Takeaway: a sector is θ360\frac{\theta}{360} of the circle. Use πr2\pi r^2 for area and 2πr2\pi r for arc length, and let the units tell you which you need.


Q7Medium

Topic: Reading the centre and radius from a circle's equation

A radio transmitter's coverage area is described in the xyxy-plane by (x−4)2+(y+3)2=25.(x - 4)^2 + (y + 3)^2 = 25\mathclose{.} What are its centre and radius?

A) centre (−4, 3),\left( -4, \ 3\right)\mathclose{,} radius 2525

B) centre (−4, 3),\left( -4, \ 3\right)\mathclose{,} radius 55

C) centre (4, −3),\left( 4, \ -3\right)\mathclose{,} radius 2525

D) centre (4, −3),\left( 4, \ -3\right)\mathclose{,} radius 55

Show the worked solution

Answer: D

Explanation

The standard form is

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

with centre (h,k)(h, k) and radius r.r\mathclose{.}

Match it up carefully. (x−4)2(x - 4)^2 gives h=4.h = 4\mathclose{.} For the yy term, rewrite (y+3)2(y + 3)^2 as (y−(−3))2,(y - (-3))^2\mathclose{,} so k=−3.k = -3\mathclose{.}

The right-hand side is r2=25,r^2 = 25\mathclose{,} so r=5r = 5 — take the square root.

Centre (4,−3),(4, -3)\mathclose{,} radius 5.

Why each wrong option is wrong:

  • B, centre (−4, 3),\left( -4, \ 3\right)\mathclose{,} radius 55 — copied the signs as they appear in the brackets. The formula has a minus built in, so a +3+3 inside means −3-3 in the centre.
  • C, centre (4, −3),\left( 4, \ -3\right)\mathclose{,} radius 2525 — gave 25 as the radius. The equation gives r2.r^2\mathclose{.}
  • A, centre (−4, 3),\left( -4, \ 3\right)\mathclose{,} radius 2525 — both errors together.

Takeaway: the signs in the brackets flip, and the number on the right is r2.r^2\mathclose{.} Rewrite (y+3)(y+3) as (y−(−3))(y-(-3)) if the sign flip is not automatic yet.


Q8Hard

Topic: Finding the radius from a diameter's endpoints

In the standard (x,y)(x, y) coordinate plane, the endpoints of a diameter of a circle are (1,2)(1, 2) and (7,10).(7, 10)\mathclose{.} What is the radius of the circle?

F) 55

G) 1010

H) 1414

J) 5050

Show the worked solution

Answer: F

Explanation

Find the distance between the two points first:

d=(7−1)2+(10−2)2=62+82=36+64=100=10d = \sqrt{(7-1)^2 + (10-2)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

That distance is the diameter, so the radius is half of it:

r=102=5r = \frac{10}{2} = 5

The 6, 8, 10 triangle here is a scaled 3, 4, 5 — worth recognising, because it saves the arithmetic entirely.

Why each wrong option is wrong:

  • G, 1010 — gave the diameter. Stopping one step early is the usual error in this question, which is exactly why it says "radius".
  • H, 1414 — added the horizontal and vertical differences without squaring. That is the distance you would walk along two sides, not the straight line.
  • J, 5050 — added the squares to reach 100 and halved that, never taking the square root.

Takeaway: the distance formula is Pythagoras in disguise. And read the last word — diameter or radius decides whether you halve.


Section 3 — Solids and Volume

The volumes worth knowing, all of which are in Chapter 8 with their reasons:

solid volume
box (rectangular prism) lwhlwh
cylinder πr2h\pi r^2 h
any prism (area of cross-section) ×\times length
sphere 43πr3\tfrac{4}{3}\pi r^3
cone 13πr2h\tfrac{1}{3}\pi r^2 h

The ACT also likes scaling: what happens to area and volume when a length changes. The rule is worth more than any single formula:

If every length is multiplied by k,k\mathclose{,} then every area is multiplied by k2k^2 and every volume by k3.k^3\mathclose{.}


Q9Medium

Topic: The volume of a cylinder

The cylindrical water tank shown has the radius and the height marked, in metres.

A cylinder of radius 3 and height 10310

What is its volume, in cubic metres?

A) 30π30 \pi

B) 60π60 \pi

C) 90π90 \pi

D) 900π900 \pi

Show the worked solution

Answer: C

Explanation

V=πr2h=π×32×10=π×9×10=90πV = \pi r^2 h = \pi \times 3^2 \times 10 = \pi \times 9 \times 10 = 90\pi

Square the radius first, then multiply by the height. Doing it in the other order is fine too, as long as only the radius gets squared.

Why each wrong option is wrong:

  • A, 30π30 \pi — forgot to square the radius.
  • B, 60π60 \pi — used 2πrh,2\pi r h\mathclose{,} which is the curved surface area, not a volume. A volume needs three lengths multiplied together, and this has only two.
  • D, 900π900 \pi — squared 3×103 \times 10 rather than just the 3.

Takeaway: πr2h\pi r^2 h — only the radius is squared. If your expression has fewer than three lengths multiplied together, it is not a volume.


Q10Hard

Topic: Scaling a solid

A spherical balloon is inflated until its radius is three times what it was before. By what factor does its volume increase?

F) 33

G) 99

H) 1212

J) 2727

Show the worked solution

Answer: J

Explanation

The volume of a sphere is 43πr3.\frac{4}{3}\pi r^3\mathclose{.} Replacing rr with 3r:3r\mathclose{:}

43π(3r)3=43π⋅27r3\frac{4}{3}\pi (3r)^3 = \frac{4}{3}\pi \cdot 27 r^3

so the volume is 27 times what it was. The bracket matters: (3r)3(3r)^3 cubes the 3 as well as the r.r\mathclose{.}

The general rule: multiply every length by kk and volume multiplies by k3.k^3\mathclose{.} Here k=3,k = 3\mathclose{,} so 33=27.3^3 = 27\mathclose{.}

Why each wrong option is wrong:

  • F, 33 — assumed volume grows in proportion to length.
  • G, 99 — used k2,k^2\mathclose{,} which is the factor for area, not volume.
  • H, 1212 — brought the 43\frac{4}{3} into the scaling. Constants in the formula do not affect how it scales; only the power of rr does.

Takeaway: lengths scale by k,k\mathclose{,} areas by k2,k^2\mathclose{,} volumes by k3.k^3\mathclose{.} Cube the bracket, not just the letter inside it.


Q11Medium

Topic: The surface area of a box

The closed cardboard box shown has the measurements marked, in units, and every face has to be covered.

A rectangular box 4 by 3 by 2423

What is its surface area, in square units?

A) 1818

B) 2424

C) 2626

D) 5252

Show the worked solution

Answer: D

Explanation

The six faces come in three pairs, one pair for each combination of two dimensions:

4×3=12(twice)4×2=8(twice)3×2=6(twice)4 \times 3 = 12 \quad (\text{twice}) \qquad 4 \times 2 = 8 \quad (\text{twice}) \qquad 3 \times 2 = 6 \quad (\text{twice})

surface area=2(12+8+6)=2×26=52\text{surface area} = 2(12 + 8 + 6) = 2 \times 26 = 52

Why each wrong option is wrong:

  • A, 1818 — added the dimensions rather than multiplying them in pairs.
  • B, 2424 — gave the volume, 4×3×2.4 \times 3 \times 2\mathclose{.} Check the units: a surface area is in square units, and multiplying three lengths gives cubic units.
  • C, 2626 — added one of each face and forgot the box has two of each.

Takeaway: a box has six faces in three matching pairs, so 2(lw+lh+wh).2(lw + lh + wh)\mathclose{.} If you find yourself multiplying all three dimensions together, you are calculating a volume.


Section 4 — Coordinate Geometry

Three formulas connect algebra to the picture, and all three come from Pythagoras or from counting:

distance=(x2−x1)2+(y2−y1)2\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

midpoint=(x1+x22,  y1+y22)\text{midpoint} = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)

slope=y2−y1x2−x1\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}

The midpoint is an average, so it adds; distance and slope both subtract. Mixing up which one adds is the commonest error here.


Q12Basic

Topic: The midpoint of a line segment

In the standard (x,y)(x, y) coordinate plane, a line segment has endpoints (4,9)(4, 9) and (8,15).(8, 15)\mathclose{.} What is the midpoint?

F) (−2, −3)\left( -2, \ -3\right)

G) (4, 6)\left( 4, \ 6\right)

H) (6, 12)\left( 6, \ 12\right)

J) (2, 3)\left( 2, \ 3\right)

Show the worked solution

Answer: H

Explanation

Average each coordinate separately:

x:  4+82=122=6x: \; \frac{4 + 8}{2} = \frac{12}{2} = 6

y:  9+152=242=12y: \; \frac{9 + 15}{2} = \frac{24}{2} = 12

Midpoint (6,12).(6, 12)\mathclose{.}

A good check: the midpoint must lie between the two endpoints in both coordinates. 66 is between 44 and 8;8\mathclose{;} 1212 is between 99 and 15.15\mathclose{.} Correct.

Why each wrong option is wrong:

  • F, (−2, −3)\left( -2, \ -3\right) and J, (2, 3)\left( 2, \ 3\right) — subtracted instead of adding. Both fail the "between" check: their yy-coordinates are nowhere near 9 to 15.
  • G, (4, 6)\left( 4, \ 6\right) — gave the differences without halving.

Takeaway: the midpoint adds and halves; distance and slope subtract. Then check your answer lies between the two endpoints.


Q13Hard

Topic: Counting points at a fixed distance

In the xyxy-plane, a point must lie exactly 5 units from the origin and also exactly 3 units from the horizontal axis. How many such points are there?

A) 11

B) 22

C) 33

D) 44

Show the worked solution

Answer: D

Explanation

"5 units from the origin" means

x2+y2=25x^2 + y^2 = 25

"3 units from the horizontal axis" means the distance from the xx-axis is 3, so

y=3ory=−3y = 3 \quad \text{or} \quad y = -3

Take y=3:y = 3\mathclose{:}

x2+9=25⟹x2=16⟹x=±4x^2 + 9 = 25 \quad\Longrightarrow\quad x^2 = 16 \quad\Longrightarrow\quad x = \pm 4

That gives (4,3)(4, 3) and (−4,3).(-4, 3)\mathclose{.} Now y=−3y = -3 gives the same two xx values, so (4,−3)(4, -3) and (−4,−3).(-4, -3)\mathclose{.}

Four points in total.

Why each wrong option is wrong:

  • A, 1,1\mathclose{,} B, 22 and C, 33 — each misses some of the sign combinations. There are two independent choices, ±\pm for xx and ±\pm for y,y\mathclose{,} so 2×2=4.2 \times 2 = 4\mathclose{.}

Takeaway: "a distance from an axis" allows both sides of it, and a squared equation gives both signs. Sketch the circle and the two lines — the four crossings are then impossible to miscount.


Q14Medium

Topic: The distance between two points

Two points are plotted in the standard (x,y)(x, y) coordinate plane, as shown.

A graph in the xy-planexy−4−3−2−1123456−112345678910

What is the distance between the two points?

F) 14\sqrt{14}

G) 1010

H) 1414

J) 100100

Show the worked solution

Answer: G

Explanation

The points are (−2,1)(-2, 1) and (4,9).(4, 9)\mathclose{.} Find the two gaps first:

across: 4−(−2)=6up: 9−1=8\text{across: } 4 - (-2) = 6 \qquad \text{up: } 9 - 1 = 8

Then Pythagoras:

62+82=36+64=100=10\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

Another 6, 8, 10 triangle. The 3-4-5 family turns up constantly on the ACT.

Why each wrong option is wrong:

  • F, 14\sqrt{14} — added before squaring.
  • H, 1414 — added the gaps, which is the distance walked along two sides of the right triangle rather than across it. It is always larger than the true distance.
  • J, 100100 — did everything right and forgot the square root.

Takeaway: find both gaps, square, add, then root. Watch the double negative in 4−(−2);4 - (-2)\mathclose{;} that is where the arithmetic usually goes wrong.


Q15Medium

Topic: Corresponding sides of similar triangles

The two triangles shown are similar, and their sides have the lengths marked, in units.

Two similar triangles: sides 8, 6 and 4, and a longest side of 2086420

What is the perimeter of the larger triangle?

A) 1818

B) 3030

C) 4545

D) 9090

Show the worked solution

Answer: C

Explanation

The scale factor comes from the pair of corresponding sides you know — the longest of each:

k=208=2.5k = \frac{20}{8} = 2.5

Perimeter scales by the same factor as any length:

(4+6+8)×2.5=18×2.5=45(4 + 6 + 8) \times 2.5 = 18 \times 2.5 = 45

Why each wrong option is wrong:

  • A, 1818 — gave the smaller triangle's perimeter.
  • B, 3030 — replaced the longest side and left the other two unscaled.
  • D, 9090 — matched the 20 with the shortest side of the smaller triangle, a factor of 5. The longest side corresponds to the longest.

Takeaway: find the scale factor from a matching pair, then apply it to every length. Perimeter scales by k;k\mathclose{;} area would scale by k2.k^2\mathclose{.}


Q16Medium

Topic: The area of a composite shape

The floor plan shown is made of two rectangles, and every corner is a right angle. The lengths marked are in metres.

An L-shaped floor: 14 along the bottom, 10 up the left, 8 along the top, 6 up the right146810

What is the area of the floor, in square metres?

F) 4848

G) 8484

H) 116116

J) 164164

Show the worked solution

Answer: H

Explanation

Two sides are not marked, and they are found from the ones that are. The top edge is 8 and the bottom is 14, so the step goes in by 14−8=6.14 - 8 = 6\mathclose{.} The left side is 10 and the right side is 6, so the step goes up by 10−6=4.10 - 6 = 4\mathclose{.}

Now cut the floor into two rectangles along the line of the step.

lower: 14×6=84upper: 8×4=32\text{lower: } 14 \times 6 = 84 \qquad \text{upper: } 8 \times 4 = 32

84+32=11684 + 32 = 116

A second route checks it: the whole 14 by 10 rectangle is 140, and the corner that is missing is 6 by 4, which is 24. 140−24=116140 - 24 = 116 ✓

Why each wrong option is wrong:

  • F, 4848 — the perimeter: 14+6+6+4+8+10.14 + 6 + 6 + 4 + 8 + 10\mathclose{.} That is the length of the walls, not the size of the floor.
  • G, 8484 — the lower rectangle only. The upper part was left out.
  • J, 164164 — 14×614 \times 6 plus 8×10.8 \times 10\mathclose{.} The second rectangle was taken the full 10 high, so the part where the two overlap was counted twice.

Takeaway: find every missing side first, then cut the shape into rectangles that do not overlap — or take the surrounding rectangle and subtract the corner that is missing. Do both and the two answers check each other.


Q17Hard

Topic: Areas of similar figures

Two similar rectangles have corresponding sides in the ratio 2:5.2: 5\mathclose{.} What is the ratio of their areas?

A) 2 : 52 : 5

B) 4 : 54 : 5

C) 4 : 254 : 25

D) 8 : 1258 : 125

Show the worked solution

Answer: C

Explanation

Lengths are in the ratio 2 : 5,2 : 5\mathclose{,} so both dimensions scale by 52\frac{5}{2} going from the smaller to the larger. Area is length × width, so the factor applies twice:

area ratio=22 : 52=4 : 25\text{area ratio} = 2^2 : 5^2 = 4 : 25

Why each wrong option is wrong:

  • A, 2 : 52 : 5 — used the length ratio unchanged.
  • B, 4 : 54 : 5 — squared only the 2.
  • D, 8 : 1258 : 125 — cubed, which is the factor for volumes.

Takeaway: lengths scale by k,k\mathclose{,} areas by k2,k^2\mathclose{,} volumes by k3.k^3\mathclose{.} Square both parts of the ratio.


Q18Medium

Topic: Congruent triangles

Two triangles have three pairs of equal sides. Which of the following must be true of them?

F) Their corresponding angles are equal and their areas are equal

G) Their corresponding angles are equal but their areas may differ

H) Their areas are equal but their angles may differ

J) Neither the angles nor the areas need match

Show the worked solution

Answer: F

Explanation

Three pairs of equal sides is the SSS condition, and it makes the triangles congruent — identical in every respect, not merely the same shape.

So the corresponding angles are equal and the areas are equal. A triangle's three sides fix it completely: there is no way to build two different triangles from the same three lengths.

Why each wrong option is wrong:

  • G, Their corresponding angles are equal but their areas may differ — describes similar triangles, which share angles but may differ in size. Congruent is stronger.
  • H, Their areas are equal but their angles may differ — equal areas do not on their own force equal angles, but here the angles are equal too.
  • J, Neither the angles nor the areas need match — both do match.

Takeaway: similar means same shape, possibly different size. Congruent means identical. Three equal sides give congruence, and congruence gives you everything.


Q19Medium

Topic: An exterior angle of a regular polygon

What is the measure, in degrees, of each exterior angle of a regular octagon?

A) 4545

B) 135135

C) 360360

D) 10801080

Show the worked solution

Answer: A

Explanation

Walk once round any polygon and you turn through one full circle, so the exterior angles always total 360°,360°\mathclose{,} however many sides there are. A regular octagon has eight equal ones:

3608=45°\frac{360}{8} = 45°

Check it against the interior angle. The interior angles total (8−2)×180=1080°,(8 - 2) \times 180 = 1080°\mathclose{,} so each is 135°,135°\mathclose{,} and 135+45=180135 + 45 = 180 — an interior angle and its exterior angle sit on a straight line together. Correct.

Why each wrong option is wrong:

  • B, 135135 — the interior angle.
  • C, 360360 — the total of all eight exterior angles, not one of them.
  • D, 10801080 — the total of the interior angles.

Takeaway: exterior angles total 360°360° for every polygon, so each one in a regular polygon is 360n.\frac{360}{n}\mathclose{.} It is often the quickest route to the interior angle as well.


Q20Medium

Topic: The area of a parallelogram

A parallelogram-shaped panel has a base of 15 centimetres, slanted sides of 10 centimetres, and a perpendicular height of 8 centimetres. What is the area of the panel, in square centimetres?

F) 5050

G) 6060

H) 120120

J) 150150

Show the worked solution

Answer: H

Explanation

The area of a parallelogram is base times perpendicular height:

15×8=120 square centimetres15 \times 8 = 120 \text{ square centimetres}

The 10 cm slanted side is there to be ignored. Height always means the perpendicular distance between the two parallel sides, never the sloping edge — and on the ACT that extra length is offered precisely because it is tempting.

Why each wrong option is wrong:

  • F, 5050 — the perimeter, 2(15+10).2(15 + 10)\mathclose{.} That is a distance round the edge, in centimetres, not an area.
  • G, 6060 — halved it, which is the triangle formula. A parallelogram is two triangles, so it does not carry the half.
  • J, 150150 — used the slanted side, which overstates the area. The perpendicular height is always the shorter of the two.

Takeaway: a parallelogram is base × perpendicular height, with no half. A slanted side given alongside the height is there to be left alone.