Everything you need is in the tick marks — count them before you calculate anything. Equal marks mean equal lengths, and that is the only "measurement" either picture gives you. Nothing here has a length in centimetres, so both answers come out as pure fractions with no units.
Fig. $\displaystyle 6.43$ — the triangleRead the marks: the left side is cut in half, but the long right side is cut into three equal pieces. Then get the shaded part by taking the two white corners away from the whole triangle.Name the corners: \(\displaystyle A\) at the bottom-left, \(\displaystyle B\) at the bottom-right, \(\displaystyle C\) at the top.
Now count what is drawn on each slanted side.
Side \(\displaystyle AC\) (the steep left side) has one dot on it, and the two pieces it makes each carry three strokes. Equal marks, equal lengths — so that dot, call it \(\displaystyle D\), is the midpoint of \(\displaystyle AC\):
\[AD = DC = \tfrac{1}{2}\,AC
\]
Side \(\displaystyle CB\) (the long slanted right side) has two dots on it, and the three pieces they make each carry two strokes. So \(\displaystyle CB\) is cut into three equal parts. Call the dots \(\displaystyle E\) (the one nearer \(\displaystyle C\)) and \(\displaystyle F\) (the one nearer \(\displaystyle B\)):
\[CE = EF = FB = \tfrac{1}{3}\,CB
\]
Watch out — this is the step people get wrong here. It is tempting to glance at the picture, see a dot on each slanted side, and write "midpoint, midpoint". Count again: the left side has
one dot, the right side has
two. Two dots make
three pieces, not two, so there is no midpoint on \(\displaystyle CB\) at all. And the different stroke counts (three on the left, two on the right) are not saying "the left pieces equal the right pieces" — a tick mark only says
the pieces wearing that same mark are equal to each other.
The shaded piece is the quadrilateral \(\displaystyle ADEF\): from \(\displaystyle A\) up to \(\displaystyle D\), across to \(\displaystyle E\), down to \(\displaystyle F\), then straight back to \(\displaystyle A\). Look again at the picture and notice two things: the base \(\displaystyle AB\) is
not an edge of the shaded piece, and the corner \(\displaystyle B\) is
not inside it. The bottom-right of the picture is one big white triangle.
So the whole triangle is made of exactly three pieces:
\[\triangle ABC \;=\; \underbrace{ADEF}_{\text{shaded}} \;+\; \underbrace{\triangle CDE}_{\text{white, top}} \;+\; \underbrace{\triangle AFB}_{\text{white, bottom-right}}
\]
I will work out the two white pieces and subtract them.
The one tool I need — the "same height" rule.Formula used — Area of a triangle: \(\displaystyle \text{Area} = \tfrac{1}{2} \times b \times h\), where \(\displaystyle b\) is any one side you choose as the base, and \(\displaystyle h\) is the perpendicular (straight-down) distance from the opposite corner to that base.
What follows from it:
if two triangles have the same height, their areas are in the same ratio as their bases. The \(\displaystyle \tfrac{1}{2}\) is the same in both and the \(\displaystyle h\) is the same in both, so only \(\displaystyle b\) can change the answer.
Step $\displaystyle 1$ — the white triangle \(\displaystyle CDE\) at the top.Get there in two hops.
Hop 1. Compare \(\displaystyle \triangle CDE\) with \(\displaystyle \triangle CAE\). Their bases \(\displaystyle CD\) and \(\displaystyle CA\) both lie along the same straight line \(\displaystyle AC\), and both triangles have their third corner at the same point \(\displaystyle E\). Same corner and same line means the perpendicular distance from \(\displaystyle E\) to line \(\displaystyle AC\) is one single height serving both. So
\[\frac{\text{area}(CDE)}{\text{area}(CAE)}=\frac{CD}{CA}=\frac{1}{2}
\]
Hop 2. Compare \(\displaystyle \triangle CAE\) with the whole \(\displaystyle \triangle CAB\). Their bases \(\displaystyle CE\) and \(\displaystyle CB\) both lie along the same straight line \(\displaystyle CB\), and both have their third corner at the same point \(\displaystyle A\). Same height again. So
\[\frac{\text{area}(CAE)}{\text{area}(CAB)}=\frac{CE}{CB}=\frac{1}{3}
\]
Multiply the two hops:
\[\text{area}(CDE)=\frac{1}{2}\times\frac{1}{3}\times\text{area}(ABC)=\frac{1}{6}\,\text{area}(ABC)
\]
Step $\displaystyle 2$ — the white triangle \(\displaystyle AFB\) at the bottom right.Compare \(\displaystyle \triangle AFB\) with the whole \(\displaystyle \triangle ACB\). Their bases \(\displaystyle FB\) and \(\displaystyle CB\) lie along the same straight line \(\displaystyle CB\), and both have their third corner at the same point \(\displaystyle A\). Same height, so
\[\frac{\text{area}(AFB)}{\text{area}(ACB)}=\frac{FB}{CB}=\frac{1}{3}
\qquad\Longrightarrow\qquad
\text{area}(AFB)=\frac{1}{3}\,\text{area}(ABC)
\]
Step $\displaystyle 3$ — subtract the two white pieces.\[\text{area}(ADEF)=\text{area}(ABC)-\text{area}(CDE)-\text{area}(AFB)
\]
\[=\text{area}(ABC)\left(1-\frac{1}{6}-\frac{1}{3}\right)
\]
Put the fractions over a common denominator of \(\displaystyle 6\), using \(\displaystyle 1=\dfrac{6}{6}\) and \(\displaystyle \dfrac{1}{3}=\dfrac{2}{6}\):
\[1-\frac{1}{6}-\frac{1}{3}=\frac{6}{6}-\frac{1}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}
\]
So the shaded quadrilateral is \(\displaystyle \dfrac{1}{2}\) of the triangle.
Step $\displaystyle 4$ — check it a completely different way.A fraction of a shape does not change if you stretch or shrink the whole picture, so I am allowed to pick convenient corners and just count. Take
\[A(0,0),\qquad B(6,0),\qquad C(0,6)
\]
Then \(\displaystyle D\), the midpoint of \(\displaystyle AC\), is \(\displaystyle (0,3)\). Going from \(\displaystyle C(0,6)\) towards \(\displaystyle B(6,0)\), each third step adds \(\displaystyle 2\) to \(\displaystyle x\) and takes \(\displaystyle 2\) off \(\displaystyle y\), so \(\displaystyle E=(2,4)\) and \(\displaystyle F=(4,2)\).
Formula used — Area of a right-angled triangle: \(\displaystyle \tfrac{1}{2}\times\text{base}\times\text{height}\). Here the base \(\displaystyle AB=6\) and the height \(\displaystyle AC=6\), so
\[\text{area}(ABC)=\tfrac{1}{2}\times 6\times 6=18 \text{ square units}
\]
Cut the shaded quadrilateral \(\displaystyle A(0,0),D(0,3),E(2,4),F(4,2)\) into two triangles along the line \(\displaystyle AE\):
\[\triangle ADE:\ \text{base } AD = 3 \text{ (along the } y\text{-axis)},\ \text{height} = 2 \text{ (the } x\text{-distance of } E)
\]
\[\text{area}(ADE)=\tfrac{1}{2}\times 3\times 2 = 3 \text{ square units}
\]
\[\triangle AEF:\ \text{area}=\tfrac{1}{2}\bigl|x_E\,y_F-x_F\,y_E\bigr|=\tfrac{1}{2}\bigl|2\times 2-4\times 4\bigr|=\tfrac{1}{2}\times 12=6 \text{ square units}
\]
\[\text{area}(ADEF)=3+6=9 \text{ square units}
\]
\[\frac{9}{18}=\frac{1}{2}\quad\checkmark
\]
Sense-check on the picture: the shaded quadrilateral looks like about half, and the two white corners together look like the other half. If you had wrongly treated \(\displaystyle CB\) as merely bisected, you would get \(\displaystyle \tfrac{3}{4}\) — and one look at how much white there is shows three-quarters is far too big.
Fig. $\displaystyle 6.44$ — the squarePut the square on a grid, write down the equation of each of the four slanted lines, solve them in pairs to find the four corners of the tilted shaded square, then compare its area with the big square's.First read the marks. Every side has one dot with equal ticks (two strokes) on both sides of it, so
each dot is the midpoint of its side.
Next read where the lines go. Each line starts at a
corner and ends at the
midpoint of a side that corner does not touch, and all four go round the same way.
Since the question asks only for a
fraction, the real size does not matter — enlarging the picture multiplies every area by the same number and leaves the ratio alone. I will pick a side of \(\displaystyle 10\) units to keep the numbers whole.
\[A(0,0)\ \text{bottom-left},\quad B(10,0)\ \text{bottom-right},\quad C(10,10)\ \text{top-right},\quad D(0,10)\ \text{top-left}
\]
Midpoints:
\[P(5,0)\ \text{on } AB,\quad Q(10,5)\ \text{on } BC,\quad R(5,10)\ \text{on } CD,\quad S(0,5)\ \text{on } DA
\]
The four drawn lines are \(\displaystyle A\to Q\), \(\displaystyle B\to R\), \(\displaystyle C\to S\) and \(\displaystyle D\to P\).
Step $\displaystyle 1$ — write each line as \(\displaystyle y=mx+c\).Formula used — Gradient (steepness): \(\displaystyle m=\dfrac{y_2-y_1}{x_2-x_1}\), the "rise" divided by the "run" between two points on the line. Then \(\displaystyle c\) is the \(\displaystyle y\)-value where the line crosses the \(\displaystyle y\)-axis; find it by putting one known point into \(\displaystyle y=mx+c\).
Line \(\displaystyle AQ\), from \(\displaystyle A(0,0)\) to \(\displaystyle Q(10,5)\):
\[m=\frac{5-0}{10-0}=\frac{1}{2},\qquad \text{it passes through } (0,0) \text{ so } c=0
\qquad\Longrightarrow\qquad y=\tfrac{1}{2}x
\]
Line \(\displaystyle BR\), from \(\displaystyle B(10,0)\) to \(\displaystyle R(5,10)\):
\[m=\frac{10-0}{5-10}=\frac{10}{-5}=-2,\qquad 0=-2(10)+c \Rightarrow c=20
\qquad\Longrightarrow\qquad y=-2x+20
\]
Line \(\displaystyle CS\), from \(\displaystyle C(10,10)\) to \(\displaystyle S(0,5)\):
\[m=\frac{5-10}{0-10}=\frac{-5}{-10}=\frac{1}{2},\qquad \text{it passes through } (0,5) \text{ so } c=5
\qquad\Longrightarrow\qquad y=\tfrac{1}{2}x+5
\]
Line \(\displaystyle DP\), from \(\displaystyle D(0,10)\) to \(\displaystyle P(5,0)\):
\[m=\frac{0-10}{5-0}=-2,\qquad \text{it passes through } (0,10) \text{ so } c=10
\qquad\Longrightarrow\qquad y=-2x+10
\]
Step $\displaystyle 2$ — find the four corners of the shaded shape.Two lines cross where their \(\displaystyle y\)-values are equal, so set the right-hand sides equal and solve for \(\displaystyle x\).
Corner \(\displaystyle W\), where \(\displaystyle AQ\) meets \(\displaystyle DP\):
\[\tfrac{1}{2}x=-2x+10
\]
Multiply every term by \(\displaystyle 2\) to clear the fraction:
\[x=-4x+20 \;\Rightarrow\; 5x=20 \;\Rightarrow\; x=4,\qquad y=\tfrac{1}{2}(4)=2
\qquad\Longrightarrow\qquad W(4,2)
\]
Corner \(\displaystyle X\), where \(\displaystyle AQ\) meets \(\displaystyle BR\):
\[\tfrac{1}{2}x=-2x+20 \;\Rightarrow\; x=-4x+40 \;\Rightarrow\; 5x=40 \;\Rightarrow\; x=8,\qquad y=\tfrac{1}{2}(8)=4
\qquad\Longrightarrow\qquad X(8,4)
\]
Corner \(\displaystyle Y\), where \(\displaystyle BR\) meets \(\displaystyle CS\):
\[-2x+20=\tfrac{1}{2}x+5 \;\Rightarrow\; -4x+40=x+10 \;\Rightarrow\; 30=5x \;\Rightarrow\; x=6,\qquad y=-2(6)+20=8
\qquad\Longrightarrow\qquad Y(6,8)
\]
Corner \(\displaystyle Z\), where \(\displaystyle CS\) meets \(\displaystyle DP\):
\[\tfrac{1}{2}x+5=-2x+10 \;\Rightarrow\; x+10=-4x+20 \;\Rightarrow\; 5x=10 \;\Rightarrow\; x=2,\qquad y=\tfrac{1}{2}(2)+5=6
\qquad\Longrightarrow\qquad Z(2,6)
\]
Watch out: never read a crossing point off the drawing with a ruler and your eye. Solve the two equations together. The drawing is a guide; the algebra is the proof.
Step $\displaystyle 3$ — check the shape really is a square, and find its side.Formula used — Distance between two points: \(\displaystyle \text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). This is just Pythagoras' theorem, with the horizontal gap and the vertical gap as the two short sides of a right-angled triangle.
\[WX:\ \Delta x=8-4=4,\ \Delta y=4-2=2 \;\Rightarrow\; \sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}
\]
\[XY:\ \Delta x=6-8=-2,\ \Delta y=8-4=4 \;\Rightarrow\; \sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}
\]
\[YZ:\ \Delta x=2-6=-4,\ \Delta y=6-8=-2 \;\Rightarrow\; \sqrt{16+4}=\sqrt{20}
\]
\[ZW:\ \Delta x=4-2=2,\ \Delta y=2-6=-4 \;\Rightarrow\; \sqrt{4+16}=\sqrt{20}
\]
All four sides are the same length.
Are the corners right angles? Two lines are perpendicular when their gradients multiply to \(\displaystyle -1\). Side \(\displaystyle WX\) lies along line \(\displaystyle AQ\), gradient \(\displaystyle \tfrac{1}{2}\); side \(\displaystyle XY\) lies along line \(\displaystyle BR\), gradient \(\displaystyle -2\).
\[\tfrac{1}{2}\times(-2)=-1
\]
So they meet at a right angle, and the same pair of gradients meets at every corner. Four equal sides and four right angles: it is a square.
Step $\displaystyle 4$ — the two areas.Formula used — Area of a square: \(\displaystyle \text{Area}=(\text{side})^2\).
\[\text{area(shaded square)}=\left(\sqrt{20}\right)^2=20 \text{ square units}
\]
(The square root and the squaring undo each other, so no decimals are needed anywhere.)
\[\text{area(big square)}=10\times 10=100 \text{ square units}
\]
Watch out: \(\displaystyle \sqrt{20}\) is the
side, not the area. And do not measure from \(\displaystyle W(4,2)\) straight across to \(\displaystyle Y(6,8)\) — that is a
diagonal, not a side. Squaring the diagonal would give \(\displaystyle 40\), which is double the true area.
Step $\displaystyle 5$ — the fraction.
\[\frac{\text{area(shaded)}}{\text{area(big square)}}=\frac{20}{100}=\frac{1}{5}
\]
because dividing the top and the bottom by \(\displaystyle 20\) gives \(\displaystyle \dfrac{1}{5}\).
Sense-check: \(\displaystyle \tfrac{1}{5}=0.2\), so the tilted square should look like about a fifth of the big one — and it does.
Both answers are fractions of an area, so they have no units.
Answer: In Fig. $\displaystyle 6.43$, \(\displaystyle \dfrac{1}{2}\) of the triangle is shaded; in Fig. $\displaystyle 6.44$, \(\displaystyle \dfrac{1}{5}\) of the square is shaded. 
Find the area of triangle ADE in Fig. 6.31.
In \(\displaystyle \triangle \mathrm{ABC}\), the midpoint of BC is D (Fig. $\displaystyle 6.32$). Median AD is drawn. P is any point on AD. Show that area \(\displaystyle (\triangle \mathrm{ABP})=\operatorname{area}(\triangle \mathrm{ACP})\).
Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. $\displaystyle 6.33$). What is the ratio of the areas of the red region ( \(\displaystyle \triangle \mathrm{PAB}\) and \(\displaystyle \triangle \mathrm{PCD}\) ) and the green region ( \(\displaystyle \triangle \mathrm{PBC}\) and \(\displaystyle \triangle \mathrm{PDA}\) )?
In \(\displaystyle \triangle \mathrm{ABC}, \mathrm{D}\) is the midpoint of \(\displaystyle \mathrm{AB} . \mathrm{P}\) is any point on BC , and Q is a point on AB such that CQ || PD. PQ is joined (Fig. $\displaystyle 6.34$). Prove that \(\displaystyle \operatorname{Area}(\triangle \mathrm{BPQ})=\frac{1}{2} \operatorname{Area}(\triangle \mathrm{ABC})\).
Fig. $\displaystyle 6.43$: What fraction of the triangle is shaded? Fig. $\displaystyle 6.44$: What fraction of the square is shaded?
Fig. $\displaystyle 6.45$: What fraction of the rectangle is covered by the circles? Fig. $\displaystyle 6.46$: What fraction of the rectangle is covered by the circles?
In Fig. $\displaystyle 6.50$, four semicircles have been drawn within the given square whose side is $\displaystyle 2$ units. The centres of these semicircles are the midpoints of the sides. They create a $\displaystyle 4$-petalled flower (shown in blue). Find the perimeter and the area of this flower.
In Fig. $\displaystyle 6.51$ we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is \(\displaystyle l\). Show that the area of the green region enclosed between the two circles is \(\displaystyle \frac{1}{4} \pi l^{2}\).
In Fig. $\displaystyle 6.52$, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
Fig. $\displaystyle 6.53$ shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius \(\displaystyle r\).
In Fig. $\displaystyle 6.54$, we see three triangles within a rectangle. The areas of the triangles are \(\displaystyle A, B, C\), as marked. Show that the area of the rectangle is \[\frac{2(A+C)(B+C)}{C} . \]