NCERT Solutions · Class 9 Mathematics Measuring Space: Perimeter and Area

Use \(\displaystyle AD\) as the base: it is \(\displaystyle 8\) cm long, and its opposite side \(\displaystyle BC\) — the side \(\displaystyle E\) sits on — runs parallel to it a fixed \(\displaystyle 10\) cm away, so the height is \(\displaystyle 10\) cm wherever \(\displaystyle E\) is.Step $\displaystyle 1$ — read the figure carefully. The rectangle is named \(\displaystyle ABCD\). In the picture \(\displaystyle B\) is the top-right corner and \(\displaystyle C\) is the bottom-right corner. The letters of a rectangle are read once round the outline in order, so the remaining two corners are \(\displaystyle A\) at the top-left (next to \(\displaystyle B\)) and \(\displaystyle D\) at the bottom-left (next to \(\displaystyle C\)). The shaded triangle has two of its corners at \(\displaystyle A\) and \(\displaystyle D\), and its third corner is the point \(\displaystyle E\), which lies on the right-hand side \(\displaystyle BC\).The figure prints two measurements, each as a double-headed arrow: That second arrow is one line from \(\displaystyle B\) to \(\displaystyle C\). It is broken in the middle only to leave a gap for the words "\(\displaystyle 8\) cm" to sit in — the break is not a mark on the side, and it says nothing about \(\displaystyle E\). So the figure tells you how long \(\displaystyle BC\) is, but it never tells you where on \(\displaystyle BC\) the point \(\displaystyle E\) is. Even though \(\displaystyle E\) may look about halfway up, nothing in the figure states that, so you are not allowed to use it. Step $\displaystyle 4$ shows that you never need it.Opposite sides of a rectangle are equal in length, so \[AB = DC = 10 \text{ cm}, \qquad AD = BC = 8 \text{ cm}. \] Note which is which: \(\displaystyle AD\) is a vertical side, so it matches the other vertical side \(\displaystyle BC = 8 \text{ cm}\).Step $\displaystyle 2$ — choose the base and name the formula. Formula: for any triangle, \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}, \] where the base is any one side you choose, and the height is the perpendicular distance from the opposite corner across to the line that the base lies along.For triangle \(\displaystyle ADE\), choose the side \(\displaystyle AD\) as the base, because its length is known: \[\text{base} = AD = 8 \text{ cm}. \] The corner opposite that base is \(\displaystyle E\).Step $\displaystyle 3$ — find the height. The height is the perpendicular distance from \(\displaystyle E\) to the line \(\displaystyle AD\), which is the left side of the rectangle.\(\displaystyle AD\) and \(\displaystyle BC\) are opposite sides of a rectangle, so they are parallel, and the perpendicular gap between them is the width of the rectangle: \[\text{gap between } AD \text{ and } BC = DC = 10 \text{ cm}. \] The point \(\displaystyle E\) lies on \(\displaystyle BC\). Every point of \(\displaystyle BC\) is that same \(\displaystyle 10\) cm away from the line \(\displaystyle AD\), so \[\text{height} = 10 \text{ cm}. \]Step $\displaystyle 4$ — put the numbers in. \[\text{Area}(ADE) = \frac{1}{2} \times AD \times \text{height} = \frac{1}{2} \times 8 \text{ cm} \times 10 \text{ cm} \] \[= \frac{1}{2} \times 80 \text{ cm}^2 = 40 \text{ cm}^2. \]Look back at Steps $\displaystyle 2$–$\displaystyle 4$: the position of \(\displaystyle E\) along \(\displaystyle BC\) never appeared anywhere. Slide \(\displaystyle E\) up towards \(\displaystyle B\) or down towards \(\displaystyle C\) and the triangle changes shape, but the base stays \(\displaystyle 8\) cm and the height stays \(\displaystyle 10\) cm, so the area stays \(\displaystyle 40 \text{ cm}^2\). That is the real point of this question.Check it a second way — cut the rectangle into three pieces. The shaded triangle \(\displaystyle ADE\) and the two white corner triangles \(\displaystyle ABE\) (at corner \(\displaystyle B\)) and \(\displaystyle DCE\) (at corner \(\displaystyle C\)) together fill the rectangle exactly, so \[\text{Area}(ADE) = \text{Area of rectangle } ABCD - \text{Area}(ABE) - \text{Area}(DCE). \]Because we do not know where \(\displaystyle E\) is, give the two parts of \(\displaystyle BC\) letters: let \(\displaystyle BE = b\) cm and \(\displaystyle EC = c\) cm. The two parts make up the whole side, so whatever the split, \[b + c = 8. \]Area of the rectangle (formula: length \(\displaystyle \times\) breadth): \[\text{Area}(ABCD) = DC \times BC = 10 \text{ cm} \times 8 \text{ cm} = 80 \text{ cm}^2. \]Triangle \(\displaystyle ABE\) has a right angle at \(\displaystyle B\), a corner of the rectangle, so its two arms \(\displaystyle AB\) and \(\displaystyle BE\) serve as base and height: \[\text{Area}(ABE) = \frac{1}{2} \times AB \times BE = \frac{1}{2} \times 10 \times b \ \text{cm}^2 = 5b \text{ cm}^2. \]Triangle \(\displaystyle DCE\) has a right angle at \(\displaystyle C\), with arms \(\displaystyle DC\) and \(\displaystyle CE\): \[\text{Area}(DCE) = \frac{1}{2} \times DC \times CE = \frac{1}{2} \times 10 \times c \ \text{cm}^2 = 5c \text{ cm}^2. \]Subtracting: \[\text{Area}(ADE) = 80 \text{ cm}^2 - 5b \text{ cm}^2 - 5c \text{ cm}^2 = \big(80 - 5(b + c)\big) \text{ cm}^2. \] Now use \(\displaystyle b + c = 8\): \[\text{Area}(ADE) = (80 - 5 \times 8) \text{ cm}^2 = (80 - 40) \text{ cm}^2 = 40 \text{ cm}^2. \]The unknown letters \(\displaystyle b\) and \(\displaystyle c\) cancelled out into the single known total \(\displaystyle b + c = 8\), and both methods give the same number.Answer: The area of triangle \(\displaystyle ADE\) is \(\displaystyle 40 \text{ cm}^2\), and it is \(\displaystyle 40 \text{ cm}^2\) for every position of \(\displaystyle E\) on \(\displaystyle BC\).

Split the trapezium into a rectangle in the middle and two equal right triangles at the ends — then Pythagoras gives you the height.What we are given Step $\displaystyle 1$ — Cut the shape upStand the trapezium with the \(\displaystyle 40\ \text{cm}\) side at the bottom and the \(\displaystyle 20\ \text{cm}\) side on top. From each end of the top side, drop a straight line straight down (perpendicular) onto the bottom side.You now have three pieces: Step $\displaystyle 2$ — Find the base of each little triangleThe bottom side is \(\displaystyle 40\ \text{cm}\) long. The rectangle uses up \(\displaystyle 20\ \text{cm}\) of it. What is left over is shared between the two triangles:\[40\ \text{cm} - 20\ \text{cm} = 20\ \text{cm} \]Because the slanting sides are equal, the two triangles are identical (congruent), so that leftover splits into two equal halves:\[\text{base of each triangle} = \frac{20\ \text{cm}}{2} = 10\ \text{cm} \]This is the step people get wrong. You halve the difference of the parallel sides, \(\displaystyle (40 - 20)\) — not the shorter side, and not the longer side. And if the slanting sides had been unequal, you could not split the leftover down the middle like this at all; it is the "isosceles" part of the question that permits it.Step $\displaystyle 3$ — Pythagoras for the heightLook at one right triangle on its own. Its three sides are: Pythagoras' theorem: in a right-angled triangle, \(\displaystyle (\text{hypotenuse})^{2} = (\text{one short side})^{2} + (\text{other short side})^{2}\).\[26^{2} = 10^{2} + h^{2} \]\[676 = 100 + h^{2} \]\[h^{2} = 676 - 100 = 576 \]\[h = \sqrt{576} = 24\ \text{cm} \]We take the positive square root because \(\displaystyle h\) is a length. And \(\displaystyle 24 \times 24 = 576\) exactly, so there is no rounding here at all.Second place people slip: \(\displaystyle 26\ \text{cm}\) is the slanting side, not the height. If you push \(\displaystyle 26\) into the area formula you get \(\displaystyle \tfrac12 \times 60 \times 26 = 780\ \text{cm}^{2}\), which is too big. Height means measured straight up, at right angles to the parallel sides — that is \(\displaystyle 24\ \text{cm}\), and it is always shorter than the slanting side.Step $\displaystyle 4$ — Area of the trapeziumFormula: \(\displaystyle \text{Area} = \dfrac{1}{2} \times (\text{sum of the parallel sides}) \times (\text{perpendicular height})\), that is\[A = \frac{1}{2}\,(a + b)\,h \]where \(\displaystyle a\) and \(\displaystyle b\) are the two parallel sides and \(\displaystyle h\) is the perpendicular distance between them.Put in \(\displaystyle a = 40\ \text{cm}\), \(\displaystyle b = 20\ \text{cm}\), \(\displaystyle h = 24\ \text{cm}\):\[A = \frac{1}{2}\,(40\ \text{cm} + 20\ \text{cm}) \times 24\ \text{cm} \]\[A = \frac{1}{2} \times 60\ \text{cm} \times 24\ \text{cm} \]\[A = 30\ \text{cm} \times 24\ \text{cm} = 720\ \text{cm}^{2} \]Notice the units: centimetre \(\displaystyle \times\) centimetre \(\displaystyle =\) square centimetre, \(\displaystyle \text{cm}^{2}\). Area is always in square units.Step $\displaystyle 5$ — Check it a different wayAdd up the three pieces from Step $\displaystyle 1$ instead of using the formula: \[480\ \text{cm}^{2} + 240\ \text{cm}^{2} = 720\ \text{cm}^{2} \]The same number both ways, so the answer is safe. No \(\displaystyle \pi\) appears in this question, and nothing needed rounding — \(\displaystyle 720\) is exact.Answer: \(\displaystyle 720\ \text{cm}^{2}\)

Find the missing third side first, then use Heron's formula to get the area.Step $\displaystyle 1$ — Get the third side.You're told two sides, \(\displaystyle 8 \) cm and \(\displaystyle 11 \) cm, and the perimeter — not the third side directly. Remember, perimeter is the sum of all three sides: \[a + b + c = \text{Perimeter} \] So the third side is: \[c = 32 - 8 - 11 = 13 \text{ cm} \]Common slip: don't treat $\displaystyle 32$ cm as if it were a side of the triangle — it's the total of all three sides added together.Now the three sides are \(\displaystyle a = 8 \) cm, \(\displaystyle b = 11 \) cm, \(\displaystyle c = 13 \) cm.Step $\displaystyle 2$ — Find the semi-perimeter, \(\displaystyle s \).Since you already know the perimeter is $\displaystyle 32$ cm, the semi-perimeter (half the perimeter) is: \[s = \frac{a+b+c}{2} = \frac{32}{2} = 16 \text{ cm} \]Step $\displaystyle 3$ — Apply Heron's formula.Heron's formula finds the area of a triangle when you know all three sides — no height needed: \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \(\displaystyle s \) is the semi-perimeter and \(\displaystyle a, b, c \) are the three sides.Work out each bracket first: \[s - a = 16 - 8 = 8 \text{ cm} \] \[s - b = 16 - 11 = 5 \text{ cm} \] \[s - c = 16 - 13 = 3 \text{ cm} \]Step $\displaystyle 4$ — Multiply everything inside the square root.\[\text{Area} = \sqrt{16 \times 8 \times 5 \times 3} \]Multiply step by step (don't round anywhere yet): \[16 \times 8 = 128 \] \[128 \times 5 = 640 \] \[640 \times 3 = 1920 \]So: \[\text{Area} = \sqrt{1920} \text{ cm}^2 \]Step $\displaystyle 5$ — Simplify the square root.\(\displaystyle 1920 \) isn't a perfect square, so pull out the largest perfect-square factor you can. Since \(\displaystyle 1920 = 64 \times 30 \), and \(\displaystyle 64 \) is a perfect square: \[\sqrt{1920} = \sqrt{64 \times 30} = \sqrt{64} \times \sqrt{30} = 8\sqrt{30} \text{ cm}^2 \]Common slip: leaving the answer as the messy \(\displaystyle \sqrt{1920} \) — always check for a perfect-square factor to simplify first, and only convert to a decimal at the very end.If you want a decimal value, \(\displaystyle \sqrt{30} \approx 5.477 \), so: \[\text{Area} \approx 8 \times 5.477 \approx 43.8 \text{ cm}^2 \text{ (rounded to 1 decimal place)} \]Answer: The area of the triangle is \(\displaystyle 8\sqrt{30} \text{ cm}^2 \), which is approximately \(\displaystyle 43.8 \text{ cm}^2 \).

The ratio $\displaystyle 3$ : $\displaystyle 5$ : $\displaystyle 7$ doesn't give you the sides in metres directly — it tells you the sides are multiples of the same number. Find that number using the perimeter, then use Heron's formula because you know all three sides but no height.Step $\displaystyle 1$ — Turn the ratio into actual side lengths.If the sides are in the ratio \(\displaystyle 3:5:7 \), write them as \[a = 3x, \qquad b = 5x, \qquad c = 7x \] for some number \(\displaystyle x \) (in metres). This is the same trick you use for any "ratio" problem — the numbers $\displaystyle 3$, $\displaystyle 5$, $\displaystyle 7$ are not the sides themselves, just how many equal parts each side is made of.Step $\displaystyle 2$ — Use the perimeter to find \(\displaystyle x \).Perimeter of a triangle = sum of its three sides: \[a+b+c = 300 \text{ m} \] \[3x + 5x + 7x = 300 \] \[15x = 300 \] \[x = 20 \]Common mistake: plugging $\displaystyle 3$, $\displaystyle 5$, $\displaystyle 7$ straight into Heron's formula as if they were the side lengths in metres. They're only the ratio parts — you must multiply each by \(\displaystyle x \) first.Step $\displaystyle 3$ — Write down the real sides.\[a = 3x = 3 \times 20 = 60 \text{ m} \] \[b = 5x = 5 \times 20 = 100 \text{ m} \] \[c = 7x = 7 \times 20 = 140 \text{ m} \]Check: \(\displaystyle 60 + 100 + 140 = 300 \) m — matches the given perimeter, so this is correct.Step $\displaystyle 4$ — Find the semi-perimeter \(\displaystyle s \).Heron's formula needs \(\displaystyle s \), the semi-perimeter: \[s = \frac{a+b+c}{2} = \frac{300}{2} = 150 \text{ m} \]Step $\displaystyle 5$ — Apply Heron's formula.Heron's formula for the area of a triangle with sides \(\displaystyle a, b, c \) is \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \(\displaystyle s \) is the semi-perimeter found above.First find each bracket: \[s - a = 150 - 60 = 90 \text{ m} \] \[s - b = 150 - 100 = 50 \text{ m} \] \[s - c = 150 - 140 = 10 \text{ m} \]Step $\displaystyle 6$ — Multiply everything inside the square root.\[\text{Area} = \sqrt{150 \times 90 \times 50 \times 10} \]Multiply step by step (don't round anywhere here): \[150 \times 90 = 13500 \] \[50 \times 10 = 500 \] \[13500 \times 500 = 6{,}750{,}000 \]So \[\text{Area} = \sqrt{6{,}750{,}000} \text{ m}^2 \]Step $\displaystyle 7$ — Simplify the square root instead of using a rough decimal too early.Split \(\displaystyle 6{,}750{,}000 = 675 \times 10{,}000 \). Since \(\displaystyle \sqrt{10{,}000} = 100 \), and \(\displaystyle 675 = 225 \times 3 = 15^2 \times 3 \), so \(\displaystyle \sqrt{675} = 15\sqrt{3} \): \[\sqrt{6{,}750{,}000} = \sqrt{675 \times 10000} = 100 \times 15\sqrt{3} = 1500\sqrt{3} \]This is the step people skip — leaving the answer as a decimal from the start hides a much cleaner exact form. Always try to pull perfect squares out of the square root before reaching for a calculator.Step $\displaystyle 8$ — Convert to a decimal for a usable answer.Using \(\displaystyle \sqrt{3} \approx 1.7320508 \): \[\text{Area} = 1500 \times 1.7320508 \approx 2598.08 \text{ m}^2 \]Answer: Area \(\displaystyle = 1500\sqrt{3} \text{ m}^2 \approx 2598.08 \text{ m}^2 \).

Give the shorter diagonal a name, write the longer one in terms of it, then put both into the rhombus area formula.Step $\displaystyle 1$: Name the unknown.Let the shorter diagonal be \[d_1 = x \text{ cm} \]The question says the other diagonal is twice as long as this one, so the longer diagonal is \[d_2 = 2x \text{ cm} \]Step $\displaystyle 2$: Write down the area formula, and see where it comes from.For any rhombus, \[\text{Area} = \frac{1}{2} \times d_1 \times d_2 \] where \(\displaystyle d_1 \) and \(\displaystyle d_2 \) are the lengths of the two diagonals.Here is why the \(\displaystyle \frac{1}{2} \) is there. The diagonals of a rhombus cross each other at right angles. So if you draw a rectangle around the rhombus by drawing a line through each vertex parallel to the other diagonal, that rectangle has sides \(\displaystyle d_1 \) and \(\displaystyle d_2 \), and the rhombus sits inside it with its four corners touching the four sides. The diagonals cut the rhombus into $\displaystyle 4$ right triangles, and each of those triangles has a congruent twin in the leftover corner space outside it. So the rectangle holds $\displaystyle 8$ equal triangles and the rhombus holds only $\displaystyle 4$ of them — the rhombus is exactly half the rectangle: \[\text{Area of rhombus} = \frac{1}{2} \times (\text{area of the } d_1 \text{-by-} d_2 \text{ rectangle}) = \frac{1}{2} \times d_1 \times d_2 \]One warning: do not reach for the parallelogram formula base \(\displaystyle \times \) height here. A rhombus question that hands you the diagonals, rather than a base and a height, wants this diagonal formula.Step $\displaystyle 3$: Substitute what you know.The area is \(\displaystyle 128 \text{ cm}^2 \). Put \(\displaystyle d_1 = x \text{ cm} \) and \(\displaystyle d_2 = 2x \text{ cm} \) into the formula: \[\frac{1}{2} \times (x \text{ cm}) \times (2x \text{ cm}) = 128 \text{ cm}^2 \]On the left, the \(\displaystyle \frac{1}{2} \) and the \(\displaystyle 2 \) cancel each other, and \(\displaystyle \text{cm} \times \text{cm} = \text{cm}^2 \): \[x^2 \text{ cm}^2 = 128 \text{ cm}^2 \]Both sides are measured in \(\displaystyle \text{cm}^2 \), so the numbers must match: \[x^2 = 128 \]Step $\displaystyle 4$: Solve for \(\displaystyle x \).Take the square root of both sides. Since \(\displaystyle x \) is a length it cannot be negative, so we keep only the positive root: \[x = \sqrt{128} \text{ cm} \]Do not punch \(\displaystyle \sqrt{128} \) into a calculator and round yet — simplify the surd first. Break $\displaystyle 128$ into a perfect square times a leftover factor: \[128 = 64 \times 2 \] so \[x = \sqrt{64 \times 2} \text{ cm} = \sqrt{64} \times \sqrt{2} \text{ cm} = 8\sqrt{2} \text{ cm} \]Only now, at the very end, turn this into a decimal. Using \(\displaystyle \sqrt{2} \approx 1.41421 \), \[x \approx 8 \times 1.41421 \text{ cm} = 11.31368 \text{ cm} \approx 11.31 \text{ cm} \text{ (2 decimal places)} \]Step $\displaystyle 5$: Check it makes sense.The longer diagonal is \[d_2 = 2x = 2 \times 8\sqrt{2} \text{ cm} = 16\sqrt{2} \text{ cm} \approx 16 \times 1.41421 \text{ cm} = 22.62736 \text{ cm} \approx 22.63 \text{ cm} \]That is twice the shorter one, as the question requires. Now put both diagonals back into the area formula, keeping the whole numbers and the surds in separate groups so you can see each piece: \[\text{Area} = \frac{1}{2} \times (8\sqrt{2} \text{ cm}) \times (16\sqrt{2} \text{ cm}) = \frac{1}{2} \times (8 \times 16) \times (\sqrt{2} \times \sqrt{2}) \text{ cm}^2 \] \[= \frac{1}{2} \times 128 \times 2 \text{ cm}^2 = \frac{1}{2} \times 256 \text{ cm}^2 = 128 \text{ cm}^2 \](The $\displaystyle 128$ in the middle line is \(\displaystyle 8 \times 16 \) — it is not the area from the question turning up again. The area only appears on the last line, and it matches what we were given, so \(\displaystyle x \) is correct.)Notice this also matches Step $\displaystyle 2$: the surrounding rectangle measures \(\displaystyle d_1 \times d_2 = 256 \text{ cm}^2 \), and the rhombus is half of it, \(\displaystyle 128 \text{ cm}^2 \).Answer: The shorter diagonal is \(\displaystyle 8\sqrt{2} \text{ cm} \approx 11.31 \text{ cm} \).

The two triangles sit on the same base \(\displaystyle CD\) and have their third vertex somewhere on the same line \(\displaystyle AB\) — so they must have the same height, and therefore the same area, no matter where exactly \(\displaystyle P\) and \(\displaystyle Q\) sit.Here's why, step by step.Step $\displaystyle 1$: Write down what "parallelogram" gives you for free.\(\displaystyle ABCD\) is a parallelogram, so its opposite sides are parallel: \[AB \parallel DC \] That one fact — \(\displaystyle AB\) and \(\displaystyle DC\) are parallel lines — is the whole problem.Step $\displaystyle 2$: Look at what triangle \(\displaystyle PCD\) actually is.\(\displaystyle P\) is a point on \(\displaystyle AB\), and \(\displaystyle C\), \(\displaystyle D\) are two fixed vertices of the parallelogram. So triangle \(\displaystyle PCD\) has: The area formula for a triangle is \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] where "height" means the perpendicular distance from the apex down to the line containing the base — not the slanted distance to a corner of it.So for triangle \(\displaystyle PCD\): \[\text{area}(\triangle PCD) = \frac{1}{2} \times CD \times h_P \] where \(\displaystyle h_P\) is the perpendicular distance from point \(\displaystyle P\) to line \(\displaystyle CD\).Step $\displaystyle 3$: Do the same for triangle \(\displaystyle QCD\).Exactly the same setup, with apex \(\displaystyle Q\) instead of \(\displaystyle P\): \[\text{area}(\triangle QCD) = \frac{1}{2} \times CD \times h_Q \] where \(\displaystyle h_Q\) is the perpendicular distance from point \(\displaystyle Q\) to line \(\displaystyle CD\).Step $\displaystyle 4$: This is the step people rush past — compare \(\displaystyle h_P\) and \(\displaystyle h_Q\).It's tempting to think that if \(\displaystyle Q\) is farther along \(\displaystyle AB\) than \(\displaystyle P\), triangle \(\displaystyle QCD\) must look "more slanted" and so have a different area. But height is measured straight down (perpendicular) to the line \(\displaystyle CD\), not along the slanted side of the triangle — and that's exactly what parallel lines guarantee stays constant.Since \(\displaystyle P\) and \(\displaystyle Q\) both lie on line \(\displaystyle AB\), and \(\displaystyle AB \parallel DC\), every single point on \(\displaystyle AB\) is the same perpendicular distance from line \(\displaystyle DC\). That's what it means for two lines to be parallel — the gap between them never changes. This constant gap is just the distance between the two parallel sides of the parallelogram, i.e., its height \(\displaystyle h\): \[h_P = h_Q = h \]Step $\displaystyle 5$: Substitute back.\[\text{area}(\triangle PCD) = \frac{1}{2} \times CD \times h \] \[\text{area}(\triangle QCD) = \frac{1}{2} \times CD \times h \]Both right-hand sides are identical — same \(\displaystyle CD\), same \(\displaystyle h\). So: \[\text{area}(\triangle PCD) = \text{area}(\triangle QCD) \]Step $\displaystyle 6$: Write it as the ratio the question asks for.\[\text{area}(\triangle PCD) : \text{area}(\triangle QCD) = 1 : 1 \]This holds true no matter where \(\displaystyle P\) and \(\displaystyle Q\) are placed on \(\displaystyle AB\) — right next to each other, at opposite ends, anywhere. Only the base \(\displaystyle CD\) and the fixed height \(\displaystyle h\) between the parallel sides decide the area; the exact horizontal position of the apex on \(\displaystyle AB\) doesn't matter at all.(As a bonus fact you'll meet again in this chapter: each of these triangles also equals exactly half the area of the whole parallelogram \(\displaystyle ABCD\), since \(\displaystyle \text{area}(ABCD) = CD \times h\).)Answer: area(\(\displaystyle \triangle PCD\)) : area(\(\displaystyle \triangle QCD\)) = $\displaystyle 1$ : $\displaystyle 1$ — the two triangles always have equal area, because they share the base \(\displaystyle CD\) and their apexes \(\displaystyle P\), \(\displaystyle Q\) lie on the line \(\displaystyle AB\), which is parallel to \(\displaystyle CD\) and therefore always the same perpendicular distance \(\displaystyle h\) from it.

Split the parallelogram along diagonal PR first — that split already forces the areas on either side of it to balance, no matter where O sits.What we're given: \(\displaystyle PQRS\) is a parallelogram (vertices in order, so the sides are \(\displaystyle PQ, QR, RS, SP\) and \(\displaystyle PR, QS\) are its diagonals). \(\displaystyle O\) is any point sitting on the diagonal \(\displaystyle PR\) — it could be near \(\displaystyle P\), near \(\displaystyle R\), or anywhere in between.What we must show: \(\displaystyle \text{ar}(\triangle PSO) = \text{ar}(\triangle PQO)\).Step $\displaystyle 1$ — The diagonal PR cuts the parallelogram into two triangles of equal area.Look at \(\displaystyle \triangle PQR\) and \(\displaystyle \triangle RSP\) (the two triangles the diagonal \(\displaystyle PR\) creates). Compare their sides: All three pairs of sides match, so by the SSS congruence rule, \[\triangle PQR \cong \triangle RSP \] Congruent triangles have equal area, so \[\text{ar}(\triangle PQR) = \text{ar}(\triangle PSR) \qquad (1) \]Step $\displaystyle 2$ — Same area + same base means same height.Formula to use here: Area of a triangle \(\displaystyle = \dfrac{1}{2} \times \text{base} \times \text{height}\), where base is any one side you choose to measure along, and height is the perpendicular (straight up-and-down) distance from the opposite vertex to that base.Both triangles in ($\displaystyle 1$) are sitting on the same base — the diagonal \(\displaystyle PR\). Since their areas are equal and their base is equal, the heights measured from their third vertices, \(\displaystyle Q\) and \(\displaystyle S\), onto the line \(\displaystyle PR\) must also be equal: \[\text{(perpendicular distance from } Q \text{ to line } PR) = \text{(perpendicular distance from } S \text{ to line } PR) \] Call this common distance \(\displaystyle h\).Aside — this is the step people skip: it isn't enough that both triangles "look similar in size." Equal area and equal base is what pins down equal height — that's basic algebra on the area formula: if \(\displaystyle \frac12 \times PR \times h_Q = \frac12 \times PR \times h_S\), then \(\displaystyle h_Q = h_S\) since \(\displaystyle PR \neq 0\).Step $\displaystyle 3$ — Bring O into the picture.Here's the part that makes the proof work for any \(\displaystyle O\), not just the midpoint of \(\displaystyle PR\) (a common mistake is to assume \(\displaystyle O\) must be where the two diagonals cross — the question deliberately says "any point" to rule that shortcut out).\(\displaystyle O\) lies on the diagonal \(\displaystyle PR\), so the segment \(\displaystyle PO\) is just a piece of the same straight line \(\displaystyle PR\). Perpendicular distance from a point to a line doesn't care which piece of the line you measure from — so: Only the base has changed — from the full diagonal \(\displaystyle PR\) down to the shorter piece \(\displaystyle PO\) — but that same shorter base \(\displaystyle PO\) is shared by both triangles \(\displaystyle PQO\) and \(\displaystyle PSO\).Step $\displaystyle 4$ — Apply the area formula to the two small triangles.\[\text{ar}(\triangle PQO) = \frac{1}{2} \times PO \times h \] \[\text{ar}(\triangle PSO) = \frac{1}{2} \times PO \times h \]Same base \(\displaystyle PO\), same height \(\displaystyle h\), so: \[\text{ar}(\triangle PQO) = \text{ar}(\triangle PSO) \]which is exactly what we had to prove — and notice nowhere in Steps $\displaystyle 3$–$\displaystyle 4$ did we need to know exactly where on \(\displaystyle PR\) the point \(\displaystyle O\) sits, so this holds for every possible position of \(\displaystyle O\).Answer: \(\displaystyle \text{ar}(\triangle PSO) = \text{ar}(\triangle PQO)\), proved by showing both triangles share the base \(\displaystyle PO\) on diagonal \(\displaystyle PR\) and have equal heights (from \(\displaystyle Q\) and \(\displaystyle S\)) onto that diagonal, since \(\displaystyle \triangle PQR \cong \triangle RSP\) gives those two heights as equal in the first place.

Split the quadrilateral into triangles with both diagonals, and compare the four corner triangles cut off by the midpoint-parallelogram to the triangles they sit inside.Setup: let the $\displaystyle 4$-gon be \(\displaystyle ABCD\), named in order, and let \(\displaystyle P\) = midpoint of \(\displaystyle AB\), \(\displaystyle Q\) = midpoint of \(\displaystyle BC\), \(\displaystyle R\) = midpoint of \(\displaystyle CD\), \(\displaystyle S\) = midpoint of \(\displaystyle DA\).Joining \(\displaystyle P,Q,R,S\) in order gives the $\displaystyle 4$-gon \(\displaystyle PQRS\) sitting inside \(\displaystyle ABCD\). The book already tells you not to worry, for now, about why \(\displaystyle PQRS\) comes out a parallelogram — that gets proved later, in the quadrilaterals chapter. All that's needed here is its area, and the argument below gets that without needing to know the shape of \(\displaystyle PQRS\) in advance.Picture \(\displaystyle ABCD\) as \(\displaystyle PQRS\) in the middle, with four small triangles cut off at the corners: So \[\text{Area}(PQRS) = \text{Area}(ABCD) - \Big[\text{Area}(APS) + \text{Area}(BPQ) + \text{Area}(CQR) + \text{Area}(DRS)\Big]. \]Everything now comes down to finding the total area of those four corner triangles — two diagonals at a time.Step $\displaystyle 1$ — draw diagonal \(\displaystyle BD\).This diagonal splits \(\displaystyle ABCD\) into \(\displaystyle \triangle ABD\) and \(\displaystyle \triangle CBD\).Look at corner triangle \(\displaystyle APS\), sitting inside \(\displaystyle \triangle ABD\). \(\displaystyle P\) is the midpoint of \(\displaystyle AB\), so \(\displaystyle AP = \tfrac12 AB\); \(\displaystyle S\) is the midpoint of \(\displaystyle AD\), so \(\displaystyle AS = \tfrac12 AD\). Both triangles share the same angle at \(\displaystyle A\). By SAS similarity, \[\triangle APS \sim \triangle ABD, \quad \text{with side ratio } \tfrac12. \]This is the step people get wrong: a side ratio of \(\displaystyle \tfrac12\) does not give area ratio \(\displaystyle \tfrac12\) — area scales with the square of the side ratio. So \[\text{Area}(APS) = \left(\frac12\right)^{2}\text{Area}(ABD) = \frac14\,\text{Area}(ABD). \]By the identical argument at corner \(\displaystyle C\), inside \(\displaystyle \triangle CBD\): \(\displaystyle Q\) is the midpoint of \(\displaystyle CB\), \(\displaystyle R\) is the midpoint of \(\displaystyle CD\), both triangles share angle \(\displaystyle C\), so \(\displaystyle \triangle CQR \sim \triangle CBD\) with ratio \(\displaystyle \tfrac12\), giving \[\text{Area}(CQR) = \frac14\,\text{Area}(CBD). \]Adding these two, \[\text{Area}(APS) + \text{Area}(CQR) = \frac14\big[\text{Area}(ABD) + \text{Area}(CBD)\big] = \frac14\,\text{Area}(ABCD), \] because diagonal \(\displaystyle BD\) splits \(\displaystyle ABCD\) exactly into \(\displaystyle \triangle ABD\) and \(\displaystyle \triangle CBD\).Step $\displaystyle 2$ — draw diagonal \(\displaystyle AC\).By the same reasoning at the other two corners:\(\displaystyle \triangle BPQ \sim \triangle BAC\) (share angle \(\displaystyle B\); \(\displaystyle BP=\tfrac12BA\), \(\displaystyle BQ=\tfrac12BC\)), so \(\displaystyle \text{Area}(BPQ) = \tfrac14\,\text{Area}(ABC)\).\(\displaystyle \triangle DRS \sim \triangle DCA\) (share angle \(\displaystyle D\); \(\displaystyle DR=\tfrac12DC\), \(\displaystyle DS=\tfrac12DA\)), so \(\displaystyle \text{Area}(DRS) = \tfrac14\,\text{Area}(ACD)\).Adding these two, \[\text{Area}(BPQ) + \text{Area}(DRS) = \frac14\big[\text{Area}(ABC) + \text{Area}(ACD)\big] = \frac14\,\text{Area}(ABCD), \] because diagonal \(\displaystyle AC\) splits \(\displaystyle ABCD\) exactly into \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle ACD\).Step $\displaystyle 3$ — add up all four corner triangles. \[\text{Area}(APS)+\text{Area}(BPQ)+\text{Area}(CQR)+\text{Area}(DRS) = \frac14\text{Area}(ABCD) + \frac14\text{Area}(ABCD) = \frac12\,\text{Area}(ABCD). \]Step $\displaystyle 4$ — subtract from the whole $\displaystyle 4$-gon. \[\text{Area}(PQRS) = \text{Area}(ABCD) - \frac12\,\text{Area}(ABCD) = \frac12\,\text{Area}(ABCD). \]That is exactly what had to be proved: the middle $\displaystyle 4$-gon \(\displaystyle PQRS\) takes up half the area of \(\displaystyle ABCD\), no matter what shape \(\displaystyle ABCD\) is, and without needing to already know that \(\displaystyle PQRS\) is a parallelogram.(A one-line preview of why \(\displaystyle PQRS\) is a parallelogram, since you'll meet the full reason later: in \(\displaystyle \triangle ABC\), \(\displaystyle PQ\) joins the midpoints of \(\displaystyle AB\) and \(\displaystyle BC\), so \(\displaystyle PQ \parallel AC\) and \(\displaystyle PQ = \tfrac12 AC\). In \(\displaystyle \triangle ACD\), \(\displaystyle SR\) joins the midpoints of \(\displaystyle AD\) and \(\displaystyle DC\), so \(\displaystyle SR \parallel AC\) and \(\displaystyle SR = \tfrac12 AC\) too. That makes \(\displaystyle PQ \parallel SR\) and \(\displaystyle PQ = SR\) — one pair of opposite sides equal and parallel, which is enough to make \(\displaystyle PQRS\) a parallelogram.)Answer: Area of the mid-point $\displaystyle 4$-gon \(\displaystyle PQRS\) = \(\displaystyle \dfrac12\) × Area of the given $\displaystyle 4$-gon \(\displaystyle ABCD\).

The trick: use the "median splits a triangle into two equal areas" fact TWICE — once for the big triangle ABC, and once for the small triangle PBC sitting inside it.Step $\displaystyle 1$ — Name everything from the figure. \(\displaystyle A, B, C\) are the vertices of the triangle. \(\displaystyle D\) is the midpoint of \(\displaystyle BC\) (the tick marks in the figure show \(\displaystyle BD = DC\)). \(\displaystyle AD\) is the median from \(\displaystyle A\). \(\displaystyle P\) is some point sitting on that median \(\displaystyle AD\). In the figure, \(\displaystyle P\) is then joined to \(\displaystyle B\) and to \(\displaystyle C\) as well — that's what turns on the two coloured triangles, \(\displaystyle \triangle ABP\) (green) and \(\displaystyle \triangle ACP\) (orange), that we must show are equal in area.Step $\displaystyle 2$ — The one fact that does all the work: a median cuts a triangle into two equal areas.Take any triangle with a median (a line from a vertex to the midpoint of the opposite side). Say the median splits the base into two equal pieces, each of length \(\displaystyle m\), and the height from the vertex down to that base is \(\displaystyle h\) (the same \(\displaystyle h\) for both halves, since both triangles share the same vertex and the same base line). Using the triangle area formula \[\text{Area} = \frac{1}{2}\times \text{base}\times \text{height}, \] both halves get \(\displaystyle \text{Area} = \frac{1}{2}\times m \times h\) — identical, because the base pieces are equal and the height is the same for both. So:A median always divides a triangle into two triangles of equal area.Common mistake: students try to compare \(\displaystyle \triangle ABP\) and \(\displaystyle \triangle ACP\) directly using \(\displaystyle AB\), \(\displaystyle AC\), or the angle at \(\displaystyle P\) — but we're never told \(\displaystyle AB = AC\), so that approach is a dead end. The correct move is always through areas adding up, not through side lengths.Step $\displaystyle 3$ — Apply the fact to the big triangle, \(\displaystyle \triangle ABC\), using median \(\displaystyle AD\).Since \(\displaystyle D\) is the midpoint of \(\displaystyle BC\), \(\displaystyle AD\) is a median of \(\displaystyle \triangle ABC\). By Step $\displaystyle 2$: \[\operatorname{ar}(\triangle ABD) = \operatorname{ar}(\triangle ACD) \quad \text{...(i)} \]Step $\displaystyle 4$ — Apply the SAME fact to the small triangle, \(\displaystyle \triangle PBC\), using \(\displaystyle PD\).Look at \(\displaystyle \triangle PBC\) — the triangle with apex \(\displaystyle P\) and base \(\displaystyle BC\). \(\displaystyle D\) is still the midpoint of \(\displaystyle BC\) (that never changed), so the segment \(\displaystyle PD\) is a median of \(\displaystyle \triangle PBC\). By Step $\displaystyle 2$ again: \[\operatorname{ar}(\triangle PBD) = \operatorname{ar}(\triangle PDC) \quad \text{...(ii)} \]This is the part the plain question text hides — the figure is what tells you \(\displaystyle P\) is joined to \(\displaystyle B\) and \(\displaystyle C\), which is what makes \(\displaystyle PD\) a second, independent median you're allowed to use.Step $\displaystyle 5$ — Break each big triangle into two pieces using P.Since \(\displaystyle P\) lies on side \(\displaystyle AD\) of \(\displaystyle \triangle ABD\), drawing \(\displaystyle BP\) splits \(\displaystyle \triangle ABD\) into exactly two smaller triangles, \(\displaystyle \triangle ABP\) and \(\displaystyle \triangle PBD\) (this is exactly the green triangle plus the small white triangle below it in the figure): \[\operatorname{ar}(\triangle ABD) = \operatorname{ar}(\triangle ABP) + \operatorname{ar}(\triangle PBD) \quad \text{...(iii)} \] Similarly, drawing \(\displaystyle CP\) splits \(\displaystyle \triangle ACD\) into \(\displaystyle \triangle ACP\) and \(\displaystyle \triangle PDC\) (the orange triangle plus the other small white triangle): \[\operatorname{ar}(\triangle ACD) = \operatorname{ar}(\triangle ACP) + \operatorname{ar}(\triangle PDC) \quad \text{...(iv)} \]Step $\displaystyle 6$ — Put it all together.From (i), (iii) and (iv): \[\operatorname{ar}(\triangle ABP) + \operatorname{ar}(\triangle PBD) = \operatorname{ar}(\triangle ACP) + \operatorname{ar}(\triangle PDC) \]Now use (ii), which says \(\displaystyle \operatorname{ar}(\triangle PBD) = \operatorname{ar}(\triangle PDC)\). Since these two equal amounts are sitting on both sides of the equation, take them away from both sides (subtracting the same quantity from equal things keeps them equal): \[\operatorname{ar}(\triangle ABP) = \operatorname{ar}(\triangle ACP) \]That is exactly what the question asked us to show.Answer: area(△ABP) = area(△ACP), proved by applying "a median splits a triangle into two equal areas" once to △ABC (median AD) and once to △PBC (median PD), then subtracting the equal pair area(△PBD) = area(△PDC) from both sides.

Draw one line through P, split the square in half, and watch the two colours become equal on their own.Look at the figure (Fig. $\displaystyle 6.33$): square ABCD has A and D on top, B and C at the bottom, so AB and DC are the left and right sides of the square, and BC and AD are the bottom and top sides. Let the side of the square be \(\displaystyle a \) units. (You'll see the actual value of \(\displaystyle a \) cancel out at the end — this result is true for a square of any size, and for any position of P inside it, not just the centre.)Step $\displaystyle 1$ — Deal with the red region first: \(\displaystyle \triangle PAB\) and \(\displaystyle \triangle PDC\).Imagine a line drawn through P, parallel to the sides AB and DC, going from the top side to the bottom side. This line splits the square's width into two pieces: These two distances always add up to \(\displaystyle a \), the full width of the square, no matter where P sits: \[x + (a - x) = a \]Now use the area formula Area of a triangle = \(\displaystyle \frac{1}{2}\) × base × height.For \(\displaystyle \triangle PAB\), the base is the side \(\displaystyle AB = a \), and the height is the perpendicular distance from P to the line AB, which is \(\displaystyle x \). (This is the step people get wrong: the height of \(\displaystyle \triangle PAB\) is not the length \(\displaystyle PA \) or \(\displaystyle PB \) — it's the straight perpendicular drop from P onto side AB.) \[\text{Area}(\triangle PAB) = \frac{1}{2} \times a \times x \]For \(\displaystyle \triangle PDC\), the base is \(\displaystyle DC = a \), and the height is the perpendicular distance from P to line DC, which is \(\displaystyle a - x \). \[\text{Area}(\triangle PDC) = \frac{1}{2} \times a \times (a - x) \]Add them: \[\text{Area}(\triangle PAB) + \text{Area}(\triangle PDC) = \frac{1}{2}a \cdot x + \frac{1}{2}a \cdot (a-x) = \frac{1}{2}a\big[x + (a-x)\big] = \frac{1}{2}a \cdot a = \frac{a^2}{2} \]The \(\displaystyle x \) completely disappears — so this sum is always \(\displaystyle \frac{a^2}{2} \), the position of P never matters.Step $\displaystyle 2$ — Do exactly the same thing for the green region: \(\displaystyle \triangle PBC\) and \(\displaystyle \triangle PDA\).This time imagine the line through P parallel to the top and bottom sides (AD and BC). It splits the square's height into two pieces: with \(\displaystyle y + (a - y) = a \) again.For \(\displaystyle \triangle PBC\), base \(\displaystyle BC = a \), height = \(\displaystyle y \): \[\text{Area}(\triangle PBC) = \frac{1}{2} \times a \times y \]For \(\displaystyle \triangle PDA\), base \(\displaystyle DA = a \), height = \(\displaystyle a - y \): \[\text{Area}(\triangle PDA) = \frac{1}{2} \times a \times (a - y) \]Add them: \[\text{Area}(\triangle PBC) + \text{Area}(\triangle PDA) = \frac{1}{2}a \cdot y + \frac{1}{2}a \cdot (a - y) = \frac{1}{2}a \cdot a = \frac{a^2}{2} \]Step $\displaystyle 3$ — Compare the two totals.\[\text{Red area} = \frac{a^2}{2}, \qquad \text{Green area} = \frac{a^2}{2} \]They're identical — so the ratio is \[\frac{\text{Red area}}{\text{Green area}} = \frac{a^2/2}{a^2/2} = \frac{1}{1} \]Check: Red + Green \(\displaystyle = \frac{a^2}{2} + \frac{a^2}{2} = a^2 \), which is exactly the area of the whole square ABCD — so nothing was double-counted or left out.Notice this proof never assumed P was in the centre — it works for P anywhere inside the square, because moving P around only changes how \(\displaystyle x \) and \(\displaystyle y \) split, never their sum with the leftover piece.Answer: The ratio of the red region to the green region is \(\displaystyle 1:1 \) — they are always equal, whatever point P is chosen inside the square.

The idea: cut triangle BDC into two pieces with the segment DP, then show one of those pieces can be swapped for an equal-area piece — and the swapped shape is exactly triangle BPQ.First, read the figure carefully. Going up side AB from B to A, the points appear in this order: \(\displaystyle B, D, Q, A\). So D is the midpoint of AB, and Q sits above D (between D and A) — this means D lies on segment BQ, which we will need later. P is a point somewhere on BC. The dashed lines DP and QC are drawn parallel to each other (that's what the question means by "CQ \(\displaystyle \parallel\) PD"), and PQ is the solid segment joining P and Q.Given: To prove: \(\displaystyle \text{ar}(\triangle BPQ) = \dfrac{1}{2}\,\text{ar}(\triangle ABC)\)Construction: Join D to C.Step $\displaystyle 1$ — Show that CD splits \(\displaystyle \triangle ABC\) exactly in half.Look at triangles ADC and BDC. They share the same vertex C, and their bases AD and DB both lie on the straight line AB. Since D is the midpoint of AB, \(\displaystyle AD = DB\) — the two bases are equal. Also, both triangles have the same height: the perpendicular distance from C down to line AB is one single number, the same for both triangles (it doesn't change depending on which base you're measuring against).Now use the area formula, \(\displaystyle \text{Area} = \dfrac{1}{2}\times \text{base}\times \text{height}\). Since ADC and BDC have equal bases and the same height, \[\text{ar}(\triangle ADC) = \text{ar}(\triangle BDC) \] Since these two triangles together make up all of \(\displaystyle \triangle ABC\), \[\text{ar}(\triangle BDC) = \frac{1}{2}\,\text{ar}(\triangle ABC) \qquad \text{...(i)} \](Aside — this is the standard fact "a median divides a triangle into two triangles of equal area." Here CD is that median, because D is the midpoint of the side it lands on.)Step $\displaystyle 2$ — Use \(\displaystyle DP \parallel QC\) to find another equal-area pair.Look at triangles DPC and DPQ. They share the exact same base, DP. Their third corners are C and Q — and both C and Q lie on the same line, the line through Q and C, which we're told is parallel to DP.Here's the key idea: if two points (here C and Q) lie on a line that is parallel to DP, then both points are the same perpendicular distance from line DP — because the distance between two parallel lines never changes. So triangles DPC and DPQ have the same base (DP) and the same height (distance from C, or from Q, down to line DP). By \(\displaystyle \text{Area} = \dfrac{1}{2}\times \text{base}\times \text{height}\) again, \[\text{ar}(\triangle DPC) = \text{ar}(\triangle DPQ) \qquad \text{...(ii)} \](Aside — this is the fact "triangles on the same base and between the same parallels are equal in area." It only works because the two apex points lie on a line parallel to the shared base — without \(\displaystyle DP \parallel QC\) this step would be false.)Step $\displaystyle 3$ — Split \(\displaystyle \triangle BDC\) using P.P lies on side BC, so the segment DP cuts \(\displaystyle \triangle BDC\) into two smaller triangles, BDP and DPC, sitting side by side: \[\text{ar}(\triangle BDC) = \text{ar}(\triangle BDP) + \text{ar}(\triangle DPC) \qquad \text{...(iii)} \]Step $\displaystyle 4$ — Split \(\displaystyle \triangle BPQ\) using D.This is where the figure's point-order matters: since D lies on segment BQ (we noted this at the start), the same segment DP also cuts \(\displaystyle \triangle BPQ\) into two smaller triangles, BDP and DPQ, sitting side by side: \[\text{ar}(\triangle BPQ) = \text{ar}(\triangle BDP) + \text{ar}(\triangle DPQ) \qquad \text{...(iv)} \]Step $\displaystyle 5$ — Put it together.In equation (iii), replace \(\displaystyle \text{ar}(\triangle DPC)\) with \(\displaystyle \text{ar}(\triangle DPQ)\), using (ii): \[\text{ar}(\triangle BDC) = \text{ar}(\triangle BDP) + \text{ar}(\triangle DPQ) \] But the right-hand side is exactly the right-hand side of (iv), which equals \(\displaystyle \text{ar}(\triangle BPQ)\). So \[\text{ar}(\triangle BDC) = \text{ar}(\triangle BPQ) \qquad \text{...(v)} \]Step $\displaystyle 6$ — Finish with (i).From (i), \(\displaystyle \text{ar}(\triangle BDC) = \dfrac{1}{2}\,\text{ar}(\triangle ABC)\). Combining this with (v): \[\text{ar}(\triangle BPQ) = \text{ar}(\triangle BDC) = \frac{1}{2}\,\text{ar}(\triangle ABC) \]This is exactly what we had to prove. Hence proved.(Quick sanity check with numbers, if you want to trust this: take B = $\displaystyle (0,0)$, C = $\displaystyle (4,0)$, A = $\displaystyle (0,4)$. Then D, the midpoint of AB, is $\displaystyle (0,2)$. Pick any point on BC, say P = $\displaystyle (3,0)$. Drawing DP and then a line through C parallel to DP, that line meets AB at Q = ($\displaystyle 0$, $\displaystyle 8$/$\displaystyle 3$) — which indeed sits between D = $\displaystyle (0,2)$ and A = $\displaystyle (0,4)$, matching the figure. Computing areas: \(\displaystyle \text{ar}(\triangle ABC) = 8\) and \(\displaystyle \text{ar}(\triangle BPQ) = 4\), and \(\displaystyle 4 = \tfrac{1}{2}\times 8\). It checks out, for this P and every other point P you could have picked on BC — the proof above didn't use any special property of P.)Answer: Proved — \(\displaystyle \text{ar}(\triangle BPQ) = \dfrac{1}{2}\,\text{ar}(\triangle ABC)\), because both \(\displaystyle \text{ar}(\triangle BPQ)\) and \(\displaystyle \dfrac{1}{2}\,\text{ar}(\triangle ABC)\) equal \(\displaystyle \text{ar}(\triangle BDC)\): the median CD gives \(\displaystyle \text{ar}(\triangle BDC) = \tfrac{1}{2}\,\text{ar}(\triangle ABC)\), and \(\displaystyle DP \parallel QC\) gives \(\displaystyle \text{ar}(\triangle DPC) = \text{ar}(\triangle DPQ)\), which after splitting \(\displaystyle \triangle BDC\) and \(\displaystyle \triangle BPQ\) along DP makes \(\displaystyle \text{ar}(\triangle BDC) = \text{ar}(\triangle BPQ)\).