Areas of Parallelograms and Triangles CBSE Class 9 Mathematics Worksheets Set 04

Questions with Answers

Question 1. Show that the area of a triangle is half the product of any side and the corresponding altitude.
Answer:
In triangle ABC with altitude AD on BC, draw AE parallel to BC and CE parallel to BA to complete the parallelogram ABCE on base BC with height AD.
Area of the parallelogram = BC x AD, and its diagonal AC divides it into two congruent triangles.
So area(ABC) = \( \tfrac12 \) x BC x AD.
In simple words: A triangle is half of the parallelogram built on the same base and height.

Teacher's Note:
1. The diagonal of a parallelogram bisects its area.
2. The same holds for each of the three sides.

Question 2. Show that if a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half that of the parallelogram.
Answer:
Let triangle PAB and parallelogram ABCD be on base AB between the parallels AB and CD. Join the diagonal BD of the parallelogram.
Triangles PAB and DAB are on the same base AB and between the same parallels, so their areas are equal.
The diagonal BD divides the parallelogram into two equal triangles, so area(PAB) = area(DAB) = \( \tfrac12 \) area(ABCD).
In simple words: Replace the triangle by the half of the parallelogram that has the same area.

Teacher's Note:
1. Use equal areas on the same base between the same parallels.
2. A diagonal cuts a parallelogram into two equal parts.

Question 3. Show that the area of a trapezium is half the product of its height and the sum of the parallel sides.
Answer:
Let ABCD be a trapezium with AB parallel to DC, AB = a, DC = b and height h. Join the diagonal AC.
Triangle ABC has base AB = a and height h, so its area is \( \tfrac12 ah \). Triangle ACD has base DC = b and the same height h, so its area is \( \tfrac12 bh \).
Area(ABCD) = \( \tfrac12 h(a + b) \).
In simple words: Split the trapezium into two triangles that share the same height.

Teacher's Note:
1. The distance between the parallel sides is the height of both triangles.
2. Add the two triangle areas.

Question 4. If G is the centroid of triangle ABC, prove that area(GAB) = \( \tfrac13 \) area(ABC).
Answer:
Let the median CF meet AB at F (the midpoint of AB). G lies on CF with GF = \( \tfrac13 \) CF.
Triangles GAB and CAB have the same base AB, and their heights are the distances of G and C from AB, in the ratio GF : CF = 1 : 3.
So area(GAB) = \( \tfrac13 \) area(ABC).
In simple words: G is one third of the way up from AB, so the triangle on AB has one third of the area.

Teacher's Note:
1. The centroid divides the median in the ratio 2 : 1.
2. The same holds for GBC and GCA.

Question 5. Show that the lines joining the midpoints of the sides of a triangle divide it into four congruent triangles, and hence of equal area.
Answer:
Let D, E, F be the midpoints of BC, CA, AB. By the midpoint theorem EF = \( \tfrac12 \) BC = BD = DC, DE = \( \tfrac12 \) AB = AF = FB and DF = \( \tfrac12 \) AC = AE = EC.
Triangles AFE, FBD, EDC and DEF have corresponding sides equal in pairs, so they are congruent by SSS.
Hence each has one quarter of the area of triangle ABC.
In simple words: Each small triangle has sides equal to half of the sides of the big triangle.

Teacher's Note:
1. Use the midpoint theorem and SSS.
2. The medial triangle is similar to the original with ratio 1 : 2.

Question 6. The diagonals of a parallelogram ABCD intersect at O. A line through O meets AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
Answer:
In triangles POD and QOB: OD = OB (diagonals bisect each other), angle POD = angle QOB (vertically opposite) and angle PDO = angle QBO (alternate angles). They are congruent by ASA.
Area(ABD) = \( \tfrac12 \) area(ABCD). Area(ABQP) = area(ABO) + area(AOP) + area(BOQ) = area(ABD) - area(POD) + area(QOB) = area(ABD).
So area(ABQP) = \( \tfrac12 \) area(ABCD) and the other part is also half.
In simple words: The line through the centre swaps equal pieces, so the halves are equal.

Teacher's Note:
1. A parallelogram is symmetrical about the intersection point of its diagonals.
2. Congruent triangles have equal areas.

Question 7. In triangle ABC, O is any point on the median AD. Show that area(ABO) = area(ACO).
Answer:
Since D is the midpoint of BC, the median AD divides triangle ABC into triangles ABD and ACD of equal area. Likewise OD is a median of triangle OBC, so area(OBD) = area(OCD).
Subtracting, area(ABD) - area(OBD) = area(ACD) - area(OCD), that is area(ABO) = area(ACO).
In simple words: Remove equal areas from equal areas.

Teacher's Note:
1. A median halves the area of the triangle it divides.
2. Triangles on equal bases BD and DC with the same height have equal areas.

Question 8. Show that the area of a rhombus is half the product of the lengths of its diagonals.
Answer:
The diagonals of a rhombus bisect each other at right angles at O, forming four congruent right triangles with legs \( \tfrac{d_1}{2} \) and \( \tfrac{d_2}{2} \).
Each has area \( \tfrac12 \cdot \tfrac{d_1}{2} \cdot \tfrac{d_2}{2} \), so the four together have area \( 4 \cdot \tfrac{d_1 d_2}{8} \).
Area = \( \tfrac12 d_1 d_2 \).
In simple words: Four equal right triangles make up the rhombus.

Teacher's Note:
1. The diagonals of a rhombus are perpendicular bisectors of each other.
2. This also works for any quadrilateral with perpendicular diagonals.

Question 9. If each diagonal of a quadrilateral separates it into two triangles of equal area, show that the quadrilateral is a parallelogram.
Answer:
Let the diagonals of ABCD meet at O. Triangles ABC and ADC have equal area and the same base AC, so B and D are at equal distances from AC, on opposite sides.
Hence AC bisects BD, that is OB = OD. In the same way, triangles BAD and BCD give OA = OC.
The diagonals bisect each other, so ABCD is a parallelogram.
In simple words: Equal areas on the same base mean the opposite vertices are equally far from the diagonal.

Teacher's Note:
1. Show that each diagonal bisects the other.
2. A quadrilateral whose diagonals bisect each other is a parallelogram.

Question 10. Triangles ABC and DBC are on the same base BC, with A and D on opposite sides of BC, and area(ABC) = area(DBC). Show that BC bisects AD.
Answer:
Draw AL and DM perpendicular to BC. Equal areas on the same base give AL = DM.
Let AD meet BC at O. Triangles ALO and DMO have angle ALO = angle DMO = 90°, angle AOL = angle DOM (vertically opposite) and AL = DM, so they are congruent by AAS.
So AO = OD, that is BC bisects AD.
In simple words: Equal heights on both sides make the intersection point the midpoint.

Teacher's Note:
1. Use AAS congruence.
2. The point O may lie on BC produced.

Question 11. D and E are points on the sides AB and AC of triangle ABC such that area(BCE) = area(BCD). Prove that DE is parallel to BC.
Answer:
Triangles BCE and BCD have the same base BC and equal areas, so their altitudes from E and D to BC are equal.
Points D and E are on the same side of BC at equal distances from it, so DE is parallel to BC.
In simple words: Points at the same height above BC lie on a line parallel to BC.

Teacher's Note:
1. Equal areas on the same base mean equal heights.
2. The converse: if DE is parallel to BC the two triangles have equal areas.

Question 12. ABCD is a quadrilateral. A line through D parallel to AC meets BC produced at P. Prove that area(ABP) = area(ABCD).
Answer:
Triangles ACD and ACP are on the same base AC and between the same parallels AC and DP, so area(ACD) = area(ACP).
Adding area(ABC) to both sides: area(ABC) + area(ACD) = area(ABC) + area(ACP).
So area(ABCD) = area(ABP).
In simple words: Swap triangle ACD for the equal-area triangle ACP to get a triangle with the same area as the quadrilateral.

Teacher's Note:
1. This is how a quadrilateral is converted to an equal-area triangle.
2. Same base and same parallels give equal areas.

Question 13. Given two points A and B and a positive number x, find the locus of a point P such that the area of triangle PAB is x.
Answer:
Area(PAB) = \( \tfrac12 \) AB x (distance of P from AB) = x, so the distance of P from the line AB is \( \dfrac{2x}{AB} \).
The locus is the pair of lines parallel to AB, at distance 2x/AB on either side of it.
In simple words: P must stay at a fixed height from AB, giving two parallel lines.

Teacher's Note:
1. Fixed area on a fixed base means fixed height.
2. The two lines are on opposite sides of AB.

Question 14. Show that the areas of triangles having the same altitude are proportional to the lengths of their bases.
Answer:
Let triangles ABC and PQR have the same altitude h and bases BC and QR.
\( \dfrac{\text{area}(ABC)}{\text{area}(PQR)} = \dfrac{\tfrac12 \cdot BC \cdot h}{\tfrac12 \cdot QR \cdot h} = \dfrac{BC}{QR} \)
Hence the areas are proportional to the bases.
In simple words: With the same height the area depends only on the base.

Teacher's Note:
1. The factor h cancels.
2. This result is used to prove the basic proportionality theorem.

Question 15. Show that a straight line drawn parallel to one side of a triangle divides the other two sides proportionally.
Answer:
Let DE be parallel to BC in triangle ABC, with D on AB and E on AC. Join BE and CD.
Triangles ADE and BDE have the same altitude from E, so \( \dfrac{\text{area}(ADE)}{\text{area}(BDE)} = \dfrac{AD}{DB} \). Similarly \( \dfrac{\text{area}(ADE)}{\text{area}(CDE)} = \dfrac{AE}{EC} \).
Triangles BDE and CDE are on the same base DE and between the same parallels, so their areas are equal.
Therefore \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \).
In simple words: Compare areas on a common height to compare the lengths of the sides.

Teacher's Note:
1. This is the basic proportionality (Thales) theorem.
2. Same base and same parallels give equal areas of BDE and CDE.