RS Aggarwal Class 9 Mathematics Solutions Chapter 10 Area

Exercise 10A

Question 1. Find the area of the parallelogram ABCD.
Answer: The area of triangle ABD is calculated using the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Substituting base = 5 cm and height = 7 cm, we get \( \text{Area of } \triangle ABD = \frac{1}{2} \times 5 \times 7 = \frac{35}{2} \text{ cm}^2 \). Similarly, the area of triangle CED is \( \frac{1}{2} \times 5 \times 7 = \frac{35}{2} \text{ cm}^2 \). Since the diagonal BD divides the parallelogram ABCD into two triangles of equal area, the total area equals \( \frac{35}{2} + \frac{35}{2} = 35 \text{ cm}^2 \).
In simple words: A parallelogram can be split into two equal triangles by drawing a diagonal. Find each triangle's area, then add them to get the parallelogram's area.

Exam Tip: When a diagonal divides a parallelogram, it creates two congruent triangles with equal areas - this property is key to solving many area problems.

Question 2. Find the area of parallelogram ABCD if AB = 10 cm, DL is perpendicular to AB, and DL = 6 cm. Also, find AD if BM is perpendicular to AD and BM = 8 cm.
Answer: Using the formula for the area of a parallelogram with base and height, the area equals \( AB \times DL = 10 \times 6 = 60 \text{ cm}^2 \). To find the length of side AD, we use the same area with a different base-height pair. Since the area of a parallelogram remains constant regardless of which base we choose, we have \( \text{Area} = AD \times BM \). Therefore, \( 60 = AD \times 8 \), which gives us \( AD = \frac{60}{8} = 7.5 \text{ cm} \).
In simple words: The area of a shape stays the same no matter which base and height you use. If you know the area and one height, you can find the matching base by dividing.

Exam Tip: Remember that a parallelogram has two different base-height pairs - using this flexibility helps you find missing sides when area is known.

Question 3. Find the area of rhombus ABCD where the diagonals AC = 24 cm and BD = 16 cm intersect at point O.
Answer: The diagonals of a rhombus are perpendicular to each other. Therefore, the area of triangle ACD is \( \frac{1}{2} \times AC \times OD = \frac{1}{2} \times 24 \times 8 = 96 \text{ cm}^2 \), where OD is half of diagonal BD. Similarly, the area of triangle ABC is \( \frac{1}{2} \times AC \times OB = \frac{1}{2} \times 24 \times 8 = 96 \text{ cm}^2 \). The total area of the rhombus is found by adding these two triangle areas: \( 96 + 96 = 192 \text{ cm}^2 \).
In simple words: A rhombus's diagonals cut it into four right triangles. The area formula is half the product of both diagonals: \( \frac{1}{2} \times d_1 \times d_2 \).

Exam Tip: The key property is that diagonals of a rhombus meet at right angles - always use this to set up the perpendicular height for triangle calculations.

Question 4. Find the area of trapezium ABCD where AB || CD, AB = 9 cm, CD = 6 cm, and CE is a perpendicular drawn to AB through C with CE = 8 cm.
Answer: The area of a trapezium is calculated using the formula \( \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance between them} \). Substituting the given values: \( \text{Area} = \frac{1}{2} \times (9 + 6) \times 8 = \frac{1}{2} \times 15 \times 8 = 60 \text{ cm}^2 \).
In simple words: To find a trapezium's area, add its two parallel sides, multiply by the perpendicular distance between them, then divide by 2.

Exam Tip: The perpendicular distance between parallel sides is crucial - make sure it's truly perpendicular, not a slant side.

Question 5. (i) Find the area of quadrilateral ABCD where DC = 17 cm, CB = 8 cm, and AB = 9 cm with the right angle at B.
Answer: In right-angled triangle DBC, using the Pythagorean theorem: \( DB^2 = DC^2 - CB^2 = 17^2 - 8^2 = 289 - 64 = 225 \text{ cm}^2 \), so \( DB = 15 \text{ cm} \). The area of triangle DBC is \( \frac{1}{2} \times 15 \times 8 = 60 \text{ cm}^2 \). In right-angled triangle DAB: \( AB^2 = DB^2 - AD^2 \), which gives \( 9^2 = 15^2 - AD^2 \), so \( AD^2 = 225 - 81 = 144 \text{ cm}^2 \), meaning \( AD = 12 \text{ cm} \). The area of triangle DAB is \( \frac{1}{2} \times 12 \times 9 = 54 \text{ cm}^2 \). Total area of quadrilateral ABCD is \( 60 + 54 = 114 \text{ cm}^2 \).
In simple words: Break the quadrilateral into two right triangles using a diagonal. Find each triangle's area separately, then add them together.

Exam Tip: Always identify which angles are right angles - these let you use the Pythagorean theorem to find missing sides.

Question 6. (ii) Find the area of trapezium PQRS where PQ || SR, SR = 16 cm, PQ = 8 cm, and diagonal QR = 17 cm with perpendicular RT drawn to PQ.
Answer: In right-angled triangle QRT, applying the Pythagorean theorem: \( RT^2 = QR^2 - QT^2 = 17^2 - 8^2 = 289 - 64 = 225 \text{ cm}^2 \), so \( RT = 15 \text{ cm} \). The area of the trapezium is calculated as \( \frac{1}{2} \times (\text{PQ} + \text{SR}) \times RT = \frac{1}{2} \times (8 + 16) \times 15 = \frac{1}{2} \times 24 \times 15 = 180 \text{ cm}^2 \).
In simple words: First find the perpendicular height between the parallel sides using the Pythagorean theorem. Then apply the trapezium area formula.

Exam Tip: When a diagonal is given in a trapezium, it often helps you find the height needed for the area formula.

Question 7. Given quadrilateral ABCD where BD is a diagonal and AL ⊥ BD and CM ⊥ BD, prove that the area of quadrilateral ABCD = \( \frac{1}{2} \times BD \times (AL + CM) \).
Answer: Consider the quadrilateral ABCD with diagonal BD. Since AL is perpendicular to BD, triangle BAD has area \( \frac{1}{2} \times BD \times AL \). Similarly, since CM is perpendicular to BD, triangle BCD has area \( \frac{1}{2} \times BD \times CM \). The total area of the quadrilateral is the sum of these two triangular areas: \( \text{Area of quadrilateral ABCD} = \frac{1}{2} \times BD \times AL + \frac{1}{2} \times BD \times CM = \frac{1}{2} \times BD \times (AL + CM) \).
In simple words: A diagonal splits a quadrilateral into two triangles. If both perpendicular distances to this diagonal are known, add them and multiply by half the diagonal.

Exam Tip: This proof shows how perpendiculars from opposite vertices to a diagonal create two separate triangles whose areas combine simply.

Question 8. In the diagram, D and L are points inside the parallelogram ABCD such that the diagonal BD divides it into triangles. If Area of triangle ABD = \( \frac{1}{2} \times 14 \times 8 = 56 \text{ cm}^2 \) and Area of triangle CBD = \( \frac{1}{2} \times 14 \times 6 = 42 \text{ cm}^2 \), find the total area of quadrilateral ABCD.
Answer: The area of triangle ABD is given as \( \frac{1}{2} \times 14 \times 8 = 56 \text{ cm}^2 \). The area of triangle CBD is given as \( \frac{1}{2} \times 14 \times 6 = 42 \text{ cm}^2 \). The total area of quadrilateral ABCD is the sum of both triangular areas: \( 56 + 42 = 98 \text{ cm}^2 \).
In simple words: When a quadrilateral is divided by a diagonal into two triangles, just add the two triangle areas to get the total.

Exam Tip: Always check that the diagonal actually divides the quadrilateral into non-overlapping triangles before adding their areas.

Question 9. Given triangles ADC and DCB with the same base CD and lying between parallel lines DC and AB, prove that the area of triangle ADC equals the area of triangle DCB.
Answer: When two triangles share the same base and both lie between the same pair of parallel lines, they have equal areas. Here, triangles ADC and DCB both have CD as their base. Since both triangles lie between the parallel lines DC and AB, they have the same perpendicular distance (height) from the base CD to the opposite parallel line. Therefore, using the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), both triangles have equal areas: \( \text{ar}(\triangle CDA) = \text{ar}(\triangle CDB) \). From this equality, subtracting the area of the common triangle OCD from both sides shows that \( \text{ar}(\triangle AOD) = \text{ar}(\triangle BOC) \).
In simple words: Triangles with the same base and between the same two parallel lines have equal area because they share the same height.

Exam Tip: This is a fundamental property - look for triangles with common bases and parallel line conditions to identify equal areas quickly.

Question 10. In triangle ABC with points D and E on sides AB and AC respectively, if triangles DBE and DCE have the same base DE and lie between parallel lines BC and DE, prove that the area of triangle DBE equals the area of triangle DCE. Also prove that the area of triangle ABE equals the area of triangle ACD.
Answer: Since triangles DBE and DCE share the same base DE and both lie between parallel lines BC and DE, they have equal areas by the theorem on triangles with the same base and between the same parallels: \( \text{ar}(\triangle DBE) = \text{ar}(\triangle DCE) \). Adding the area of triangle ADE to both sides of this equation gives \( \text{ar}(\triangle DBE) + \text{ar}(\triangle ADE) = \text{ar}(\triangle DCE) + \text{ar}(\triangle ADE) \), which simplifies to \( \text{ar}(\triangle ABE) = \text{ar}(\triangle ACD) \).
In simple words: When triangles share a base and are bounded by parallel lines, equal areas follow. Adding the same triangle to both sides preserves equality.

Exam Tip: The parallel lines property is crucial - always verify that the triangles truly lie between the stated parallel lines before claiming equal areas.

Question 11. In triangle ABC, points D and E lie on AB and AC such that the area of triangle BCE equals the area of triangle BCD. Prove that DE is parallel to BC.
Answer: Given that triangles BCE and BCD have equal areas, both triangles share the same base BC. Since they have the same base and equal areas, the perpendicular distances from points E and D to line BC must be equal. This means that points D and E lie on a line parallel to BC. Therefore, DE is parallel to BC.
In simple words: If two triangles with the same base have equal areas, their opposite vertices must be at the same distance from the base, making them on a parallel line.

Exam Tip: Equal areas combined with a shared base often implies parallel lines - this is a powerful converse to use in proofs.

Question 13. In parallelogram ABCD with point O inside, prove that the area of triangle OAB plus the area of triangle OCD equals half the area of the parallelogram.
Answer: Draw lines PQ through O parallel to AB and RS through O parallel to AD. Triangles AOB and parallelogram ABPQ share the same base AB and lie between parallel lines AB and PQ, so \( \text{ar}(\triangle AOB) = \frac{1}{2} \text{ar}(\text{parallelogram ABPQ}) \). Similarly, triangles COD and parallelogram PQCD share the same base CD and lie between parallel lines CD and PQ, so \( \text{ar}(\triangle COD) = \frac{1}{2} \text{ar}(\text{parallelogram PQCD}) \). Adding both equations: \( \text{ar}(\triangle AOB) + \text{ar}(\triangle COD) = \frac{1}{2}[\text{ar}(\text{parallelogram ABPQ}) + \text{ar}(\text{parallelogram PQCD})] = \frac{1}{2} \text{ar}(\text{parallelogram ABCD}) \).
In simple words: By drawing lines through O parallel to two adjacent sides, you create intermediate parallelograms that help decompose the original area into manageable parts.

Exam Tip: Auxiliary construction (drawing extra lines) is often the key to proving area relationships - choose directions parallel to the sides of the main figure.

Question 14. In quadrilateral ABCD, a line through D is drawn parallel to AC, meeting BC produced at point P. Prove that the area of triangle ABP equals the area of quadrilateral ABCD.
Answer: Since triangles ACP and ACD share the same base AC and lie between parallel lines AC and DP (where DP is parallel to AC), they have equal areas: \( \text{ar}(\triangle ACP) = \text{ar}(\triangle ACD) \). Adding the area of triangle ABC to both sides gives \( \text{ar}(\triangle ACP) + \text{ar}(\triangle ABC) = \text{ar}(\triangle ACD) + \text{ar}(\triangle ABC) \), which simplifies to \( \text{ar}(\triangle ABP) = \text{ar}(\text{quadrilateral ABCD}) \).
In simple words: By extending the quadrilateral to a larger triangle through a parallel line construction, you create equal areas that relate the new and original figures.

Exam Tip: When a line is drawn parallel to a diagonal or side, it creates triangles with the original that often have related areas - this is a powerful proof technique.

Question 15. In triangle ABC where AD is the median and P is a point on AD, prove that the area of triangle BDP equals the area of triangle CDP.
Answer: In triangle BPC, since PD is the median from vertex P to side BC, the median divides the triangle into two triangles of equal area: \( \text{ar}(\triangle BPD) = \text{ar}(\triangle CPD) \). Alternatively, since BD = DC (as D is the midpoint of BC), triangles ABD and ACD have equal areas. Subtracting the common triangle APD from both yields \( \text{ar}(\triangle ABD) - \text{ar}(\triangle APD) = \text{ar}(\triangle ACD) - \text{ar}(\triangle APD) \), which simplifies to \( \text{ar}(\triangle BDP) = \text{ar}(\triangle CDP) \).
In simple words: A median divides a triangle into two equal-area parts. Any point on a median maintains this equal-area property with the base endpoints.

Exam Tip: Medians are powerful tools - they always split triangles into equal areas, and this property carries through to other constructions involving them.

Question 16. In quadrilateral ABCD where the diagonals AC and BD intersect at O with BO = OD, prove that the area of triangle ABC equals the area of triangle ADC.
Answer: Since triangles ABC and ADC have the same base AC and both have equal heights from vertices B and D respectively to the line AC (because B and D are equidistant from the line AC due to BO = OD), the areas are equal. Specifically, triangles AOB and AOD share the same base AO. Since BO = OD, these triangles have equal heights from B and D to line AC, making \( \text{ar}(\triangle AOB) = \text{ar}(\triangle AOD) \). Similarly, \( \text{ar}(\triangle COB) = \text{ar}(\triangle COD) \). Adding both pairs: \( \text{ar}(\triangle AOB) + \text{ar}(\triangle COB) = \text{ar}(\triangle AOD) + \text{ar}(\triangle COD) \), which gives \( \text{ar}(\triangle ABC) = \text{ar}(\triangle ADC) \).
In simple words: When one diagonal is bisected, the triangles formed on either side of the other diagonal have equal areas.

Exam Tip: Diagonal bisection creates symmetry in area relationships - always check if a diagonal is bisected when comparing triangle areas.

Question 17. In triangle ABC where AD is a median and E is the midpoint of AD, prove that the area of triangle BED equals one-fourth the area of triangle ABC.
Answer: Since AD is the median of triangle ABC, it divides the triangle into two equal areas: \( \text{ar}(\triangle ABD) = \text{ar}(\triangle ACD) = \frac{1}{2} \text{ar}(\triangle ABC) \). Now, since BE is the median of triangle ABD (with E being the midpoint of AD), it divides triangle ABD into two equal areas: \( \text{ar}(\triangle ABE) = \text{ar}(\triangle BED) = \frac{1}{2} \text{ar}(\triangle ABD) \). Substituting the previous result: \( \text{ar}(\triangle BED) = \frac{1}{2} \times \frac{1}{2} \text{ar}(\triangle ABC) = \frac{1}{4} \text{ar}(\triangle ABC) \).
In simple words: Each median divides a triangle into two equal areas. Applying this twice - first with AD, then with BE - reduces the area by half each time.

Exam Tip: When multiple medians are involved, apply the median property successively to track how areas shrink by factors of one-half.

Question 18. In quadrilateral ABCD where the diagonals AC and BD intersect at O with BO = OD, prove that the area of triangle ABC equals the area of triangle ADC.
Answer: Since BO = OD, point O is the midpoint of diagonal BD. Triangles AOB and AOD share the same base AO and have equal perpendicular distances from B and D to the line AO (because O is the midpoint of BD). Therefore, \( \text{ar}(\triangle AOB) = \text{ar}(\triangle AOD) \). Similarly, triangles COB and COD share the same base CO and have the same perpendicular distance, so \( \text{ar}(\triangle COB) = \text{ar}(\triangle COD) \). Adding the corresponding equations: \( \text{ar}(\triangle AOB) + \text{ar}(\triangle COB) = \text{ar}(\triangle AOD) + \text{ar}(\triangle COD) \), which gives \( \text{ar}(\triangle ABC) = \text{ar}(\triangle ADC) \).
In simple words: When one diagonal bisects the other, the quadrilateral splits into two triangles of equal area on either side of the bisected diagonal.

Exam Tip: Always check whether a diagonal bisects another diagonal - this creates immediate area equalities that simplify proofs.

Question 19. In triangle ABC where AD is the median and E is the midpoint of AD (so E is the midpoint of segment AD on the line from A to D), prove that the area of triangle BEC equals one-half the area of triangle ABC.
Answer: Since AD is the median of triangle ABC, it divides the triangle into two equal-area parts: \( \text{ar}(\triangle ABD) = \text{ar}(\triangle ACD) = \frac{1}{2} \text{ar}(\triangle ABC) \). Now, BE is the median of triangle ABD from vertex B to the midpoint E of side AD, which divides triangle ABD into two equal areas: \( \text{ar}(\triangle ABE) = \text{ar}(\triangle BED) = \frac{1}{2} \text{ar}(\triangle ABD) \). CE is the median of triangle ACD from vertex C to the midpoint E of side AD, so \( \text{ar}(\triangle ACE) = \text{ar}(\triangle CED) = \frac{1}{2} \text{ar}(\triangle ACD) \). Adding the two triangle areas with E as a vertex: \( \text{ar}(\triangle BED) + \text{ar}(\triangle CED) = \frac{1}{2} \text{ar}(\triangle ABD) + \frac{1}{2} \text{ar}(\triangle ACD) = \frac{1}{2} \times \frac{1}{2} \text{ar}(\triangle ABC) + \frac{1}{2} \times \frac{1}{2} \text{ar}(\triangle ABC) = \frac{1}{2} \text{ar}(\triangle ABC) \), so \( \text{ar}(\triangle BEC) = \frac{1}{2} \text{ar}(\triangle ABC) \).
In simple words: Midpoints of medians split the triangle in half again. By tracking areas through successive median applications, you find that a triangle formed by a midpoint covers exactly half the original.

Exam Tip: When points lie on medians, use median properties repeatedly to establish area ratios - each step typically halves or quarters the area.

Question 21. In parallelogram ABCD where O is any point on diagonal AC, prove that the area of triangle AOB equals the area of triangle AOD.
Answer: Construction: Join BD, intersecting AC at point P. Since the diagonals of a parallelogram bisect each other, P is the midpoint of both AC and BD. Triangles ODP and OBP share the point O and both have P on the line BD. Consider triangles ABP and ADP - they share base AP and since ABCD is a parallelogram, these triangles have equal areas (each being half of triangle ABD due to P bisecting BD). Since O lies on AC and triangles AOB and AOD share vertex O with base BD at P, and the perpendicular distance from O to line BD remains constant, the areas satisfy \( \text{ar}(\triangle AOB) = \text{ar}(\triangle AOD) \).
In simple words: In a parallelogram, when you pick any point on a diagonal and connect it to opposite vertices, the resulting triangles have equal area due to the diagonal bisection property.

Exam Tip: Parallelogram properties - especially that diagonals bisect each other - are key to establishing equal areas for triangles formed by interior points.

Question 22. Prove that if ABCD is a parallelogram with P, Q, R, S as the midpoints of sides AB, BC, CD, and DA respectively, then PQRS is a parallelogram with area equal to half the area of ABCD.
Answer: By the midpoint theorem, since P and Q are midpoints of AB and BC respectively, PQ is parallel to AC and PQ = \( \frac{1}{2} \)AC. Similarly, since S and R are midpoints of DA and CD, SR is parallel to AC and SR = \( \frac{1}{2} \)AC. Therefore, PQ || SR and PQ = SR. By the same logic, PS || BD and PS = \( \frac{1}{2} \)BD, while QR || BD and QR = \( \frac{1}{2} \)BD. Thus, PQRS is a parallelogram. For the area, triangles SPQ and DAS share base SP and midpoint properties ensure \( \text{ar}(\triangle SPQ) = \frac{1}{2} \text{ar}(\triangle DAS) \). Similarly for the other triangles. Since the four corner triangles (one at each corner of ABCD) together equal half the area of ABCD, the area of parallelogram PQRS equals \( \frac{1}{2} \text{ar}(\text{parallelogram ABCD}) \).
In simple words: Connecting the midpoints of a parallelogram's sides creates a smaller parallelogram inside with exactly half the area.

Exam Tip: The midpoint quadrilateral is always a parallelogram - always has half the area of the original, making it a powerful tool for area calculations.

Question 23. In pentagon ABCDE where EG is drawn parallel to DA, BA extended meets CF (drawn parallel to DB) at point G, and CF is drawn parallel to DB with BF meeting AB produced at F, prove that the area of pentagon ABCDE equals the area of triangle DGF.
Answer: Since EG is parallel to DA and triangles DGA and DGE share base DG with DA || EG, they have related areas by the parallel line property. Triangles DGA and DED have the same base D and parallel line properties establish area relationships. Through successive application of the property that triangles with same base and between parallels have equal area, and by adding component triangles correctly, we can show that the area enclosed by pentagon ABCDE equals the combined area built up through the triangle constructions to form triangle DGF. The parallel construction ensures areas are conserved and rearranged into the larger triangle DGF, whose area equals the original pentagon area.
In simple words: By drawing lines parallel to sides of the pentagon and extending the figure, you can transform the pentagon's area into a single triangle of equal area.

Exam Tip: Parallel line constructions for area proofs often involve extending and rearranging shapes - track which triangles are being added or subtracted carefully.

Question 24. In triangle ABC where AD is the median, prove that the area of triangle ABD equals the area of triangle ACD.
Answer: Since AD is the median of triangle ABC, point D is the midpoint of side BC, meaning BD = DC. Both triangles ABD and ACD share the same height from vertex A perpendicular to the base BC. Using the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), since BD = DC and both triangles have the same height from A, their areas are equal: \( \text{ar}(\triangle ABD) = \frac{1}{2} \times BD \times h = \frac{1}{2} \times DC \times h = \text{ar}(\triangle ACD) \), where h is the perpendicular distance from A to BC.
In simple words: A median from any vertex to the opposite side always divides the triangle into two equal areas because it splits the base into two equal parts.

Exam Tip: This is one of the most fundamental area properties - every median divides its triangle into two equal-area parts, no exceptions.

Question 25. In triangle ABC where D is a point on BC such that BD = \( \frac{1}{2} \)DC, prove that the area of triangle ABD equals one-third the area of triangle ABC.
Answer: Since BD = \( \frac{1}{2} \)DC, we have BD + DC = BC, so \( \frac{1}{2} \)DC + DC = BC, which gives \( \frac{3}{2} \)DC = BC, thus DC = \( \frac{2}{3} \)BC and BD = \( \frac{1}{3} \)BC. Both triangles ABD and ABC share the same height from vertex A perpendicular to base BC. Using the area formula: \( \text{ar}(\triangle ABD) = \frac{1}{2} \times BD \times h = \frac{1}{2} \times \frac{1}{3} BC \times h = \frac{1}{3} \left( \frac{1}{2} \times BC \times h \right) = \frac{1}{3} \text{ar}(\triangle ABC) \), where h is the perpendicular distance from A to BC.
In simple words: When a point divides a side into a 1:2 ratio, the triangle formed from the smaller piece has one-third the area of the whole.

Exam Tip: Area ratios depend on base ratios when heights are shared - if a point divides the base in ratio m:n, the triangle areas are in ratio m:(m + n).

Question 26. In parallelogram ABCD where BD is the diagonal, prove that the area of triangle ABD equals the area of triangle BCD.
Answer: In parallelogram ABCD, the diagonal BD divides it into two triangles: ABD and BCD. Since ABCD is a parallelogram, opposite sides are equal: AB = CD and AD = BC. Additionally, the diagonal BD is common to both triangles. By the side-side-side (SSS) congruence criterion, triangles ABD and CDB are congruent. Therefore, their corresponding areas are equal: \( \text{ar}(\triangle ABD) = \text{ar}(\triangle BCD) \). Alternatively, using base and height: both triangles share base BD, and since opposite sides of a parallelogram are parallel, the perpendicular distances from A and C to line BD are equal, confirming \( \text{ar}(\triangle ABD) = \text{ar}(\triangle BCD) \).
In simple words: Any diagonal of a parallelogram divides it into two congruent triangles with equal areas - this is a basic property of parallelograms.

Exam Tip: Diagonals split parallelograms and other quadrilaterals into triangles - always check if these triangles are congruent, which guarantees equal areas.

Question 27. In triangle ABC where D is a point on BC such that BD:DC = m:n, prove that the area of triangle ABD divided by the area of triangle ACD equals m:n.
Answer: Let BD = mx and DC = nx for some positive value x. Both triangles ABD and ACD share the same height from vertex A perpendicular to the base BC. Using the area formula: \( \text{ar}(\triangle ABD) = \frac{1}{2} \times BD \times h = \frac{1}{2} \times mx \times h \) and \( \text{ar}(\triangle ACD) = \frac{1}{2} \times DC \times h = \frac{1}{2} \times nx \times h \), where h is the perpendicular distance from A to BC. Taking the ratio: \( \frac{\text{ar}(\triangle ABD)}{\text{ar}(\triangle ACD)} = \frac{\frac{1}{2} \times mx \times h}{\frac{1}{2} \times nx \times h} = \frac{m}{n} \), so \( \text{ar}(\triangle ABD) : \text{ar}(\triangle ACD) = m : n \).
In simple words: When a point divides a triangle's base in a certain ratio, the two resulting triangles have areas in that same ratio.

Exam Tip: This ratio property is fundamental - whenever a cevian or base-point divides a side, use ratios of bases (with common height) to find area ratios instantly.