Areas of Parallelograms and Triangles CBSE Class 9 Mathematics Worksheets Set 07

Exercise I

Question 1. ABCD is a trapezium with AB ∥ DC, AB = 7 cm, AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find DC and the area of the trapezium.
Answer:
Draw perpendiculars from A and B to DC. Each right triangle has hypotenuse 5 cm and height 4 cm, so its base is √(25 − 16) = 3 cm.
DC = 3 + 7 + 3 = 13 cm.
Area = ½ × (7 + 13) × 4 = 40 cm².
In simple words: The slant sides cut off two 3-4-5 triangles, one at each end.

Teacher's Note:
1. Use the parallel sides' sum for the trapezium area.
2. The 3-4-5 triplet appears again and again in such problems.

Question 2. The diagonals of a parallelogram ABCD meet at O. A line through O meets AB at X and CD at Y. Show that ar(AXYD) = ½ ar(ABCD).
Answer:
In △OXB and △OYD: OB = OD, ∠XOB = ∠YOD (vertical) and ∠OBX = ∠ODY (alternate), so △OXB ≅ △OYD and ar(△OXB) = ar(△OYD).
ar(AXYD) = ar(AXOD) + ar(△ODY) = ar(AXOD) + ar(△OXB) = ar(△ABD).
A diagonal halves the parallelogram, so ar(AXYD) = ½ ar(ABCD).
In simple words: Swapping a triangle for an equal one turns the quadrilateral into half of the parallelogram.

Teacher's Note:
1. Cut and replace equal pieces without changing the area.
2. The same method shows that any line through the centre bisects the area.

Question 3. P is a point inside a parallelogram ABCD. Show that (i) ar(△APB) + ar(△PCD) = ½ ar(ABCD) (ii) ar(△APD) + ar(△PBC) = ar(△APB) + ar(△PCD).
Answer:
(i) Draw EF through P parallel to AB, with E on AD and F on BC. △APB and parallelogram ABFE are on the same base AB and between the same parallels, so ar(△APB) = ½ ar(ABFE). Likewise ar(△PCD) = ½ ar(EFCD).
Adding, ar(△APB) + ar(△PCD) = ½ ar(ABCD).
(ii) The four triangles make up the parallelogram, so ar(△APD) + ar(△PBC) = ar(ABCD) − ½ ar(ABCD) = ½ ar(ABCD), which equals ar(△APB) + ar(△PCD).
In simple words: Opposite pairs of triangles each take half the area.

Teacher's Note:
1. The helper line EF through P is the key step.
2. The result does not depend on where P is.

Question 4. A point O inside a rectangle ABCD is joined to the vertices. Prove that the sum of the areas of one pair of opposite triangles equals the sum for the other pair.
Answer:
A rectangle is a parallelogram, so by the previous result ar(△OAB) + ar(△OCD) = ½ ar(ABCD).
Also ar(△OAD) + ar(△OBC) = ½ ar(ABCD). The two sums are equal.
In simple words: Each opposite pair adds up to half the rectangle.

Teacher's Note:
1. Use the parallelogram result for the special case of a rectangle.
2. Check with O at the centre: all four triangles are equal.

Question 5. PQRS and ABRS are parallelograms on the same base SR (A and B lie on the line PQ), and X is any point on BR. Show that (i) ar(PQRS) = ar(ABRS) (ii) ar(△AXS) = ½ ar(PQRS).
Answer:
(i) The parallelograms PQRS and ABRS are on the same base SR and between the same parallels PQ and SR, so their areas are equal.
(ii) △AXS and parallelogram ABRS are on the same base AS and between the parallels AS and BR, so ar(△AXS) = ½ ar(ABRS) = ½ ar(PQRS).
In simple words: A triangle on the same base and between the same parallels as a parallelogram has half its area.

Teacher's Note:
1. Name the common base and the pair of parallels each time.
2. X can slide along BR without changing the area.

Question 6. The medians of △ABC meet at G. Show that ar(△AGB) = ar(△AGC) = ar(△BGC) = ⅓ ar(△ABC).
Answer:
A median bisects the area of a triangle. In △ABC, median AD gives ar(△ABD) = ar(△ACD), and in △GBC, GD is a median, so ar(△GBD) = ar(△GCD). Subtracting, ar(△AGB) = ar(△AGC).
In the same way, using median BE, ar(△AGB) = ar(△BGC).
The three equal areas add up to ar(△ABC), so each is ⅓ ar(△ABC).
In simple words: The centroid joins to the vertices and splits the triangle into three equal parts.

Teacher's Note:
1. Equal bases with a common height give equal areas.
2. Each of the six small triangles around G has area ⅙ ar(△ABC).

Question 7. D is the mid-point of AB in △ABC and P is any point on BC. CQ ∥ PD meets AB at Q. Prove that ar(△BPQ) = ½ ar(△ABC).
Answer:
Join CD. △PDC and △PDQ are on the same base PD and between the parallels PD and CQ, so ar(△PDC) = ar(△PDQ).
ar(△BPQ) = ar(△BPD) + ar(△PDQ) = ar(△BPD) + ar(△PDC) = ar(△BDC).
CD is a median of △ABC, so ar(△BDC) = ½ ar(△ABC). Hence ar(△BPQ) = ½ ar(△ABC).
In simple words: Replacing one triangle by an equal one turns △BPQ into the half triangle BDC.

Teacher's Note:
1. Use the parallel line CQ ∥ PD to swap triangles.
2. A median always halves the area.

Question 8. E is the mid-point of the median AD of △ABC. Show that ar(△BED) = ¼ ar(△ABC).
Answer:
AD is a median of △ABC, so ar(△ABD) = ½ ar(△ABC).
BE is a median of △ABD, so ar(△BED) = ½ ar(△ABD) = ¼ ar(△ABC).
In simple words: Halve the area twice.

Teacher's Note:
1. Two medians, two halvings.
2. This is a quick check for the 1/4 pattern.

Question 9. In a parallelogram ABCD, AB = 10 cm and the altitudes corresponding to AB and AD are 7 cm and 8 cm. Find AD.
Answer:
Area = AB × altitude = 10 × 7 = 70 cm².
Area is also AD × 8, so AD = 70 ÷ 8 = 8.75 cm.
In simple words: The same area can be found from either base.

Teacher's Note:
1. The longer side has the shorter altitude.
2. Check: 8.75 × 8 = 70.

Question 10. E, F, G, H are the mid-points of the sides AB, BC, CD, DA of a parallelogram ABCD. Show that ar(EFGH) = ½ ar(ABCD).
Answer:
Join H and F. AH = BF and AH ∥ BF (since AD = BC and AD ∥ BC), so ABFH is a parallelogram and HF ∥ AB. Similarly HFCD is a parallelogram.
△EHF and parallelogram ABFH are on the same base HF and between the parallels HF and AB, so ar(△EHF) = ½ ar(ABFH). In the same way ar(△GHF) = ½ ar(HFCD).
Adding, ar(EFGH) = ½ ar(ABCD).
In simple words: The inner parallelogram has half the area of the outer one.

Teacher's Note:
1. The helper line HF is parallel to AB and CD.
2. The same result holds for any quadrilateral.

Exercise II

Question 11. A quadrilateral ABCD is such that the diagonal BD divides its area into two equal parts. Prove that BD bisects AC.
Answer:
Let AC meet BD at O. Draw AM ⊥ BD and CN ⊥ BD. ar(△ABD) = ½ × BD × AM and ar(△CBD) = ½ × BD × CN. They are equal, so AM = CN.
In △AMO and △CNO: ∠AMO = ∠CNO = 90°, ∠AOM = ∠CON (vertical) and AM = CN, so they are congruent (AAS).
Hence AO = CO, that is, BD bisects AC.
In simple words: Equal areas on the same base force equal altitudes, and equal altitudes give the mid-point.

Teacher's Note:
1. Drop perpendiculars from A and C to the diagonal.
2. The converse (diagonal bisecting the other gives equal areas) also holds.

Question 12. PQ is parallel to the side BC of △ABC (P on AB, Q on AC). BX ∥ CA and CY ∥ BA meet the line PQ produced at X and Y. Show that ar(△ABX) = ar(△ACY).
Answer:
BX ∥ QC and BC ∥ XQ, so BXQC is a parallelogram. Similarly BC ∥ PY and CY ∥ BP, so BCYP is a parallelogram.
These parallelograms are on the same base BC and between the same parallels BC and XY, so their areas are equal.
△ABX and parallelogram BXQC are on the same base BX and between the parallels BX and AC, so ar(△ABX) = ½ ar(BXQC). Similarly ar(△ACY) = ½ ar(BCYP). Hence ar(△ABX) = ar(△ACY).
In simple words: Two equal parallelograms each hold a triangle of exactly half their area.

Teacher's Note:
1. Identify the two parallelograms first.
2. Use the triangle-and-parallelogram half rule twice.

Question 13. O is any point on the diagonal BD of a parallelogram ABCD. Prove that ar(△OAB) = ar(△OBC).
Answer:
Let the diagonals meet at M, the mid-point of AC. In △BAC, BM is a median, so ar(△BAM) = ar(△BCM). In △OAC, OM is a median, so ar(△OAM) = ar(△OCM).
Subtracting (or adding, if O lies on the other side of M), ar(△OAB) = ar(△OBC).
In simple words: Medians halve areas, and the difference of two equal pairs is equal.

Teacher's Note:
1. The same method proves ar(△AOB) = ar(△AOD) when O is on diagonal AC.
2. Draw the figure with O on each side of M to check both cases.

Question 14. D, E, F are the mid-points of BC, CA, AB of △ABC. Prove that BDEF is a parallelogram whose area is half that of △ABC, and that ar(△DEF) = ¼ ar(△ABC).
Answer:
By the mid-point theorem EF ∥ BC with EF = ½BC = BD, and DE ∥ AB with DE = ½AB = BF. So BDEF is a parallelogram.
The mid-points divide △ABC into four triangles AFE, FBD, EDC, DEF, which are congruent by SSS (each side is half a side of △ABC).
So ar(△DEF) = ¼ ar(△ABC) and ar(BDEF) = ar(△FBD) + ar(△DEF) = ½ ar(△ABC).
In simple words: The mid-point triangle splits the big triangle into four equal copies.

Teacher's Note:
1. Congruent triangles have equal areas.
2. BDEF is made of two of the four small triangles.

Question 15. The medians BE and CF of △ABC meet at G. Prove that the area of △GBC equals the area of quadrilateral AFGE.
Answer:
BE is a median, so ar(△ABE) = ar(△CBE), that is ar(AFGE) + ar(△BFG) = ar(△BGC) + ar(△CGE).
CF is a median, so ar(△BFC) = ar(△AFC), that is ar(△BFG) + ar(△BGC) = ar(AFGE) + ar(△CGE).
Subtracting the second equation from the first: ar(AFGE) − ar(△BGC) = ar(△BGC) − ar(AFGE), so 2 ar(AFGE) = 2 ar(△BGC).
Hence ar(△GBC) = ar(AFGE).
In simple words: Two median equations subtract to give equal areas.

Teacher's Note:
1. Write each median as an equation between two parts.
2. Each of the two equal areas is ⅓ ar(△ABC).

Question 16. Prove that the area of an equilateral triangle of side a is (√3/4)a².
Answer:
Draw the altitude AD to BC. It bisects BC, so BD = a/2.
AD = √(a² − a²/4) = (√3/2)a.
Area = ½ × a × (√3/2)a = (√3/4)a².
In simple words: Base a times height (√3/2)a, halved.

Teacher's Note:
1. The altitude of an equilateral triangle is also its median.
2. For a = 2 the area is √3.

Question 17. The base BC of △ABC is divided at D so that BD = ½DC. Prove that ar(△ABD) = ⅓ ar(△ABC).
Answer:
BD = ½DC gives BD : DC = 1 : 2, so BD = ⅓BC.
△ABD and △ABC have the same altitude from A. Their areas are in the ratio of their bases, so ar(△ABD) = ⅓ ar(△ABC).
In simple words: Same height, so area follows the base.

Teacher's Note:
1. Triangles on the same line with the same apex have areas in the ratio of the bases.
2. Then ar(△ADC) = ⅔ ar(△ABC).

Question 18. D is the mid-point of side BC of △ABC, E is the mid-point of BD and O is the mid-point of AE. Prove that ar(△BOE) = ⅛ ar(△ABC).
Answer:
AD is a median, so ar(△ABD) = ½ ar(△ABC). E is the mid-point of BD, so AE is a median of △ABD and ar(△ABE) = ½ ar(△ABD) = ¼ ar(△ABC).
In △ABE, BO is a median (O is the mid-point of AE), so ar(△BOE) = ½ ar(△ABE) = ⅛ ar(△ABC).
In simple words: Three halvings in a row.

Teacher's Note:
1. Count the halvings: 1/2, 1/4, 1/8.
2. Each median of a triangle halves its area.

Question 19. ABCD is a parallelogram and O is any point on the diagonal AC. Show that ar(△AOB) = ar(△AOD).
Answer:
Let the diagonals meet at M, so BM = DM. In △ABD, AM is a median, so ar(△ABM) = ar(△ADM). In △OBD, OM is a median, so ar(△OBM) = ar(△ODM).
Subtracting (or adding, if O lies beyond M), ar(△AOB) = ar(△AOD).
In simple words: Equal bases BM and DM with the same apex give equal areas.

Teacher's Note:
1. This is the same idea as the previous result for the other diagonal.
2. The point O may lie on either side of M.