Triangle and Quadrilaterals: Grade 10 Math

Ram and Sita farm two fields side by side. The boundary between them is not straight: it runs from a stone A on the top edge, bends at a tree P, and goes on to a stone B on the bottom edge. They want one straight boundary from A, easier to plow along, but neither will give up a single square meter. Where should it end?

One line drawn through the bend, parallel to AB, finds the spot, and this unit shows why it works: triangles on the same base, between the same parallel lines, always have the same area, however differently they lean.

(The problem: NRICH, “The Farmers’ Field Boundary”.)

Drag the bend P to make any boundary you like, then straighten it. Watch Ram’s land before and after.

Ram, bent boundary: 4,650 m²
Ram, straight: —
Sita: 3,030 m²

Move the bend P anywhere. Can one straight boundary from A leave Ram and Sita exactly the land they have now?

The straight boundary AQ leaves Ram 4,650 m², the same as before. Drag P: it stays the same. Press Show why.

The dashed line through P is parallel to AB. The blue triangle ABP and the red ABQ share the base AB and lie between the same parallels, so they have the same area: swapping one for the other moves no land.

For the SEE

The question: Geometry is 13 marks in three questions. One question of 5 marks is on this unit: a. state a relation between figures on the same base and between the same parallels (1 mark, knowledge), b. find an area or prove a short result (2 marks, understanding), c. prove a harder result (2 marks, higher ability).
The facts it needs:

Parallelograms on the same base and between the same parallels are equal in area.
A triangle is half of a parallelogram on the same base and between the same parallels.
Triangles on the same base and between the same parallels are equal in area; and the other way round, equal triangles on the same base, on the same side of it, lie between the same parallels.
A diagonal halves a parallelogram; a median halves a triangle.

How the SEE asks it has a full question marked part by part; Exam practice has every exercise question to mark yourself.

10.0 Review

Two figures are congruent when one fits exactly on the other: same shape, same size. Congruent figures always have the same area. But the other way round fails: figures with the same area can have quite different shapes.

Figure 10.1 A 4 cm by 3 cm rectangle, a 6 cm by 2 cm rectangle, and a triangle with base 6 cm and height 4 cm. Each has an area of 12 cm², and no two are congruent.

Ask the same questions about any pair of figures: do they match when laid on each other? Do they have the same area? Which pairs are congruent? Are all figures of equal area congruent? With the picture above you can answer the last one: no.

Two triangles have the same area. Which is true?

Pick an answer to see why.

10.1 Area of triangles and quadrilaterals

Look at three pairs of figures. In each, some figures share a base, and their tops lie on lines parallel to the base.

Two phrases do all the work in this unit.

Figures stand on the same base when one side of each is the same segment, such as AB.
Figures lie between the same parallels when their bases are on one line and their opposite sides or corners are on one line parallel to it.

Between two parallel lines the distance is the same everywhere. So figures between the same parallels have the same height, and figures on the same base and between the same parallels have the same base and the same height. That is why their areas are tied together.

A. Parallelograms on the same base

Drag the yellow corner along the top line. The base and the height stay the same; watch the slanting sides, the perimeter and the area.

Parallelogram: 24 cm²
Its perimeter: 20.9 cm
Triangle: 12 cm²
Its perimeter: 16.0 cm

The red triangle cut off on one side is the same triangle added on the other, so ABCD has the rectangle's area: 6 × 4 = 24 cm².

Slid this far, cut it into thin strips parallel to AB: each strip only slides along, so the area is still 6 × 4 = 24 cm².

Wherever E goes on the top line, the triangle has base 6 cm and height 4 cm: ½ × 6 × 4 = 12 cm². Its sides and perimeter change; its area does not.

On the same base and between the same parallels, the triangle is always half the parallelogram: 12 cm² and 24 cm², wherever D and E are.

Figure 10.6

Given: parallelograms ABCD and ABEF stand on the same base AB and lie between the same parallels AB and FC.

To prove: area of parallelogram ABCD = area of parallelogram ABEF.

The idea: both contain the trapezium ABED. What ABCD adds to it is the triangle BCE; what ABEF adds is the triangle ADF. Show those two triangles are congruent.

	Statements	Reasons
1.	In △ADF and △BCE: i. ∠AFD = ∠BEC (A) ii. ∠ADF = ∠BCE (A) iii. AD = BC (S)	i.corresponding angles, $AF \parallel BE$ ii. corresponding angles, $AD \parallel BC$ iii. opposite sides of parallelogram ABCD
2.	$\triangle ADF \cong \triangle BCE$	by A.A.S.
3.	area of △ADF = area of △BCE	congruent triangles are equal in area
4.	area of △ADF + area of trapezium ABED = area of △BCE + area of trapezium ABED	adding the same area to both sides
5.	area of parallelogram ABEF = area of parallelogram ABCD	from 4, the whole is the sum of its parts

Proved.

The reasons in statements 4 and 5 have names that many teachers and examiners use, and either way of writing them earns the mark:

Reason in words	Its name
adding the same to equal things	addition axiom
taking the same from equal things	subtraction (remainder) axiom
the whole is the sum of its parts	whole-part axiom
things equal to the same thing are equal	equality axiom

B. A triangle and a parallelogram

Figure 10.9

Given: triangle ABE and parallelogram ABCD stand on the same base AB between the parallels AB and EC.

To prove: area of △ABE = ½ area of parallelogram ABCD.

Construction: through B draw $BF \parallel AE$ , meeting EC at F. Then ABFE is a parallelogram.

	Statements	Reasons
1.	area of parallelogram ABFE = area of parallelogram ABCD	both on the same base AB and between $AB \parallel EC$ (Theorem 1)
2.	area of △ABE = ½ area of parallelogram ABFE	the diagonal BE halves parallelogram ABFE
3.	area of △ABE = ½ area of parallelogram ABCD	from 1 and 2

Proved.

Parallelogram PQRS has an area of 50 cm². T is any point on SR. What is the area of △PQT?

Pick an answer to see why.

C. Triangles on the same base

Figure 10.12

Given: triangles ABE and ABC stand on the base AB between the parallels AB and EC.

To prove: area of △ABE = area of △ABC.

Construction: draw $AD \parallel BC$ , meeting EC at D. Then ABCD is a parallelogram.

	Statements	Reasons
1.	area of △ABC = ½ area of parallelogram ABCD	the diagonal AC halves parallelogram ABCD
2.	area of △ABE = ½ area of parallelogram ABCD	same base AB, between $AB \parallel EC$ (Theorem 2)
3.	area of △ABE = area of △ABC	from 1 and 2

Proved.

The other way round

Theorem 3 also works backwards, and some exercise questions need it.

Figure 10.14

If two triangles of equal area stand on the same base and on the same side of it, they lie between the same parallels. In the figure, if area of △ABC = area of △ABD, then $CD \parallel AB$ .

Why: the two triangles have the same base and the same area, so they have the same height. C and D are then the same distance from AB, on the same side, and every point at that distance lies on one line parallel to AB.

Worked examples

Back to the field

Ram keeps exactly his land if the straight boundary runs from A to Q, where the line through the bend P, parallel to AB, meets the bottom edge.

Why: the old boundary A → P → B and the new one A → Q differ only by the triangles ABP and ABQ. Both stand on the base AB, and P and Q are on one line parallel to AB, so by Theorem 3 the two triangles have the same area. Swapping one for the other moves no land at all. A bend with several corners can be straightened the same way, one corner at a time.

How the SEE asks it

In the SEE’s Compulsory Mathematics paper, Geometry carries 13 marks in three questions (CDC’s specification grid). The question on this unit is worth 5 marks:

Part	Level	Marks	What it asks
a.	knowledge	1	state a relation between figures on the same base and between the same parallels
b.	understanding	2	find an area, or prove a short result with one theorem
c.	higher ability	2	prove a result needing two steps or a construction

In a proof, the marks go to the right statements with the right reasons: a statement without its reason earns little. Here is a question in that shape.

Exercise 10.1

Question 1

Find the area of each figure.

Figure 10.22

Answer

Figure 10.23

Answer

Figure 10.24

Answer

Figure 10.25

Answer

Figure 10.26

Answer

Figure 10.27

Hint

Answer

Question 2

Figure 10.28

WXYZ is a square with diagonal $WY = 2\sqrt2$ cm. Find the area of parallelogram PXYW.

Answer

Figure 10.29

In parallelogram ABCD, E is on AD with BE = EC = 10 cm, $EF \perp BC$ and AD = 16 cm. Find the area of the parallelogram.

Answer

Figure 10.30

PQST is a parallelogram and R is on TS produced. $SA \perp TP$ , SA = 8 cm and the area of △PQR is 64 cm². Find the length of TP.

Answer

Figure 10.31

PQRS is a parallelogram and O is any point on QR. PO and SR are produced to meet at T, and Q is joined to T. Prove that area of △QOT = area of △ROS.

Answer

Figure 10.32

$AC \parallel DE$ and BC = EC. If the area of △ACE is 24 cm², find the area of quadrilateral ABCD.

Answer

Question 3

Figure 10.33

In trapezium PQRS, $PS \parallel QR$ , PQ = 13 cm, PS = 18 cm, QR = 23 cm and $SR \perp QR$ . Find the area of △PSR.

Hint

Answer

Figure 10.34

$PQ \parallel SR$ and $PS \parallel QT$ . The area of trapezium PQRS is 95 cm² and the area of △QRT is 15 cm². Find the area of △RPQ.

Answer

Figure 10.35

$DE \parallel AB$ , $AD \parallel BC$ and F is the midpoint of BE. If the area of △AFE is 12 cm², find the area of parallelogram ABCD.

Answer

Figure 10.36

In parallelogram ABCD, E is on AD, $\angle BEC = 90°$ , BE = 6 cm and CE = 9 cm. Find the area of ABCD.

Answer

Question 4

Prove the following.

Figure 10.37

PQRS is a parallelogram and QRXY is a square. Prove that their areas are equal.

Answer

Figure 10.38

ABCD is a quadrilateral, and DE is drawn parallel to AC, meeting BC produced at E. Prove that area of quadrilateral ABCD = area of △ABE.

Answer

Figure 10.39

ABCD and EFGC are parallelograms, with E on AD and B on FG. Prove that their areas are equal.

Hint

Answer

Figure 10.40

PQRS is a parallelogram, SR is produced to T, and PT meets QR at O. Prove that area of △POR = area of △QOT.

Answer

Figure 10.41

PQRS is a parallelogram, SR is produced to T, PT meets QR at O, and S is joined to O and Q to T. Prove that area of △QOT = area of △ROS.

Answer

Figure 10.42

ABCD is a parallelogram, X is any point on AD and Y any point on CD; triangles XBC and YAB are drawn. Prove that area of △ABY = area of △ABX + area of △CDX.

Answer

Figure 10.43

ABCD is a parallelogram and E is any point on BC. DE and AB are produced to meet at F, and C is joined to F. Prove that area of △CEF = area of △ABE.

Answer

Figure 10.44

ABCD is a trapezium with $AD \parallel PQ \parallel BC$ , P on AB and Q on DC. Prove that area of △AQB = area of △DPC.

Hint

Answer

Figure 10.45

$QT \parallel RS$ , $PR \parallel TS$ and $PS \parallel QR$ . Prove that area of △PQR = area of △PTS.

Answer

Figure 10.46

ABCD is a trapezium, Y is a point on AC and X a point on BD, with $AD \parallel XY \parallel BC$ . Prove that area of △AXC = area of △BYD.

Answer

Figure 10.47

ABCD is a quadrilateral with $DB \parallel CE$ , E on AB produced. Prove that area of △ADE = area of quadrilateral ABCD.

Answer

Figure 10.48

ABCDEF is a hexagon with $AF \parallel BE \parallel CD$ and $AB \parallel CF \parallel DE$ , and BE and CF meet at P. If parallelograms ABPF and CDEP have equal areas, prove that $EF \parallel BC$ .

Hint

Answer

Figure 10.49

The diagonal QS of quadrilateral PQRS bisects the other diagonal PR at M. Prove that area of △PQS = ½ area of quadrilateral PQRS.

Answer

Figure 10.50

In △ABC, P is the midpoint of AB and Q is any point on BC. R is a point on AB with $RC \parallel PQ$ . Prove that area of △BQR = ½ area of △ABC.

Hint

Answer

Figure 10.51

In quadrilateral ABCD, $AD \parallel BC$ , E is a point on AD and F a point on CD. If area of △ABE = area of △ACF, prove that $EF \parallel AC$ .

Answer

Questions to discuss

Draw a triangle and a quadrilateral on the same base and between the same parallels so that the triangle’s area is equal to, half of, a quarter of or a third of the quadrilateral’s. What measurements did you choose, how and why? Justify each one.

Answer

Worksheet to print (PDF, 6 pages) Answer key for teachers

Harder worksheet (PDF, 3 pages) Answer key for teachers

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

The straight boundary that keeps both fields’ areas, by a line through the bend parallel to AB (the opening, Back to the field): NRICH, “The Farmers’ Field Boundary”, read October 2026.
Theorem 1 as Euclid’s Proposition I.35, in Joyce’s translation (the story): D. E. Joyce, Euclid’s Elements, Book I, Proposition 35, read October 2026.
The Elements, about 300 BC, thirteen books, “the most successful textbook ever written”, over a thousand editions since 1482 (the story): Wikipedia, “Euclid’s Elements”, read October 2026.
A shear keeps area; oblique type is a sheared upright type; shears in computer graphics and a turn by three shears (fact box): Wikipedia, “Shear mapping”, read October 2026.
The SEE’s Geometry marks and the area question’s parts (For the SEE, How the SEE asks it): CDC, Specification Grid, Compulsory Mathematics, Grades 9–10 (2078), the grid table and the Grade 10 model question paper, question 11.

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