CBSE Class 9 Mathematics Area of Parallelograms and Triangles Assignment Set 02

Chapter - 9 (Area of parallelograms and triangles)

Key Concepts

* Area of a parallelogram = \(\text{base} \times \text{height}\)
* Area of a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
* Area of a trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance between them}\)
* Area of rhombus = \(\frac{1}{2} \times \text{product of diagonals}\)
* Parallelograms on the same base and between the same parallels are equal in area.
* A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
* Triangles on the same base and between the same parallels are equal in area.
* If a triangle and a parallelogram are on the same base and between the same parallels, then \(\text{ar}(\triangle) = \frac{1}{2} \text{ar}(\text{parallelogram})\).
* A diagonal of a parallelogram divides it into two triangles of equal areas.
In parallelogram \(ABCD\), we have \(\text{ar}(\triangle ABD) = \text{ar}(\triangle ACD)\).
* The diagonals of a parallelogram divide it into four triangles of equal areas, therefore \(\text{ar}(\triangle AOB) = \text{ar}(\triangle COD) = \text{ar}(\triangle AOD) = \text{ar}(\triangle BOC)\).
* A median \(AD\) of a \(\triangle ABC\) divides it into two triangles of equal areas. Therefore, \(\text{ar}(\triangle ABD) = \text{ar}(\triangle ACD)\).
* If the medians of a \(\triangle ABC\) intersect at \(G\), then \(\text{ar}(\triangle AGB) = \text{ar}(\triangle AGC) = \text{ar}(\triangle BG = \frac{1}{3}\text{ar}(\triangle ABC)\).

Section - A

Question 1. If E, F, G & H are mid points of sides of parallelogram ABCD, then show that \(\text{ar}(EFGH) = \frac{1}{2}\text{ar}(ABCD)\)
Answer: Join \(HF\). Since \(AD \parallel BC\) and \(AB \parallel DC\), \(H\) and \(F\) are mid-points of \(AD\) and \(BC\), so \(HF \parallel AB\).
Quadrilateral \(ABFH\) is a parallelogram.
Triangle \(EFH\) and parallelogram \(ABFH\) are on the same base \(HF\) and between the same parallels \(HF\) and \(AB\).
Therefore, \(\text{ar}(\triangle EFH) = \frac{1}{2}\text{ar}(ABFH)\) -- (1)
Similarly, considering parallelogram \(H FCD\), triangle \(GFH\) is on the same base \(HF\) and between the same parallels \(HF\) and \(DC\).
Therefore, \(\text{ar}(\triangle GFH) = \frac{1}{2}\text{ar}(HFCD)\) -- (2)
Adding (1) and (2):
\(\text{ar}(\triangle EFH) + \text{ar}(\triangle GFH) = \frac{1}{2}[\text{ar}(ABFH) + \text{ar}(HFCD)]\)
\(\text{ar}(EFGH) = \frac{1}{2}\text{ar}(ABCD)\)
In simple words: The area of the quadrilateral formed by joining the mid-points of a parallelogram is exactly half the area of the outer parallelogram.

Exam Tip: Use the theorem stating that a triangle on the same base and between the same parallels as a parallelogram has half its area.

Question 2. Point P and Q are on the sides DC and AD of a parallelogram respectively. Show that \(\text{ar}(APB) = \text{ar}(BQC)\)
Answer: Triangle \(APB\) and parallelogram \(ABCD\) are on the same base \(AB\) and between the same parallels \(AB\) and \(DC\).
Therefore, \(\text{ar}(\triangle APB) = \frac{1}{2}\text{ar}(ABCD)\) -- (1)
Similarly, triangle \(BQC\) and parallelogram \(ABCD\) are on the same base \(BC\) and between the same parallels \(BC\) and \(AD\).
Therefore, \(\text{ar}(\triangle BQC) = \frac{1}{2}\text{ar}(ABCD)\) -- (2)
From (1) and (2), we get \(\text{ar}(\triangle APB) = \text{ar}(\triangle BQC)\).
In simple words: Both triangles share the property of having half the area of the same parallelogram because they share a base and lie between the same parallel lines.

Exam Tip: Clearly state the common base and parallel lines for both figures to establish equality via the parallelogram area half-theorem.

Question 3. Show that a median of a triangle divides it into two triangle of equal area.
Answer: Let \(ABC\) be a triangle and \(AD\) be a median to the side \(BC\).
Draw \(AE \perp BC\).
Area of triangle \(ABD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times AE$
Since \(AD\) is a median, \(BD = DC\).
Therefore, Area of triangle \(ABD = \frac{1}{2} \times DC \times AE = \text{ar}(\triangle ADC\).
Thus, \(\text{ar}(\triangle ABD) = \text{ar}(\triangle ADC)\).
In simple words: A median splits the base in half and since the height is common, both resulting triangles have equal areas.

Exam Tip: Drop a perpendicular from the opposite vertex to establish heights when proving area properties geometrically.

Question 4. PQRS and ABRS are two parallelograms and X being any point on side BR. Show that:
(i) \(\text{ar}(PQRS) = \text{ar}(ABRS)\)
(ii) \(\text{ar}(A \times S) = \frac{1}{2}\text{ar}(PQRS)\)
Answer: (i) Parallelograms \(PQRS\) and \(ABRS\) are on the same base \(SR\) and between the same parallels \(SR\) and \(PB\). Therefore, \(\text{ar}(PQRS) = \text{ar}(ABRS)\).
(ii) Triangle \(AXS\) and parallelogram \(ABRS\) are on the same base \(AS\) and between the same parallels \(AS\) and \(BR\). Therefore, \(\text{ar}(\triangle AXS) = \frac{1}{2}\text{ar}(ABRS)\).
Since \(\text{ar}(ABRS) = \text{ar}(PQRS)\), it follows that \(\text{ar}(\triangle AXS) = \frac{1}{2}\text{ar}(PQRS)\).
In simple words: Figures sharing a base and parallel lines follow direct proportional area relationships.

Exam Tip: Identify the shared base and parallel lines first before applying standard area theorems.

Section - B

Question 5. In given figure ABCD is a quadrilateral and BE\(||\)AC is such that BE meets at E on the extended CD. Show that area of triangle ADE is equal to the area of quadrilateral ABCD.
Answer: Consider triangles \(CAB\) and \(CAE\). They are on the same base \(AC\) and between the same parallels \(AC\) and \(BE\).
Therefore, \(\text{ar}(\triangle CAB) = \text{ar}(\triangle CAE)\).
Add \(\text{ar}(\triangle ACD)\) to both sides:
\(\text{ar}(\triangle CAB) + \text{ar}(\triangle ACD) = \text{ar}(\triangle CAE) + \text{ar}(\triangle ACD)\)
\(\text{ar}(ABCD) = \text{ar}(\triangle ADE)\).
In simple words: By adding the common triangular area to two equal triangles, we show that the quadrilateral equals the large triangle.

Exam Tip: Look out for common regions that can be added to or subtracted from equal areas to prove geometric statements.

Question 6. In given figure E be any point on the median AD of triangle, show that \(\text{ar}(ABE) = \text{ar}(ACE)\)
Answer: \(AD\) is the median of \(\triangle ABC\).
Therefore, \(\text{ar}(\triangle ABD) = \text{ar}(\triangle ACD)\) -- (1)
\(ED\) is the median of \(\triangle EBC\).
Therefore, \(\text{ar}(\triangle EBD) = \text{ar}(\triangle ECD)\) -- (2)
Subtracting equation (2) from equation (1):
\(\text{ar}(\triangle ABD) - \text{ar}(\triangle EBD) = \text{ar}(\triangle ACD) - \text{ar}(\triangle ECD)\)
\(\text{ar}(\triangle ABE) = \text{ar}(\triangle ACE)\).
In simple words: Subtracting smaller equal triangular areas from larger equal triangular areas leaves equal remainders.

Exam Tip: Subtraction of areas is a powerful technique when dealing with medians and points lying on them.

Question 7. Show that the diagonals of a parallelogram divides it into four triangles of equal area. OR D, E & F are mid points of sides of triangle BC, CA & AB respectively. Show that
(i) BDEF is a parallelogram
(ii) \(\text{ar}(DEF) = \frac{1}{4}\text{ar}(ABC)\)
(iii) \(\text{ar}(BDEF) = \frac{1}{2}\text{ar}(ABC)\)
Answer: (i) Since \(F\) and \(E\) are mid-points of \(AB\) and \(AC\), by mid-point theorem, \(FE \parallel BC\) and \(FE = \frac{1}{2}BC = BD\). Thus \(BDEF\) is a parallelogram.
(ii) Similarly, \(FDCE\) and \(AFDE\) are parallelograms. Diagonal \(FE\) divides parallelogram \(BDEF\) into two triangles of equal area, so \(\text{ar}(\triangle DEF) = \text{ar}(\triangle BFE)\). Continuing this for all parts, \(\text{ar}(\triangle DEF) = \frac{1}{4}\text{ar}(\triangle ABC)\).
(iii) Parallelogram \(BDEF\) is composed of two triangles \(DEF\) and \(FED\) (or equivalent congruent parts), giving \(\text{ar}(BDEF) = 2 \times \text{ar}(\triangle DEF) = 2 \times \frac{1}{4}\text{ar}(\triangle ABC) = \frac{1}{2}\text{ar}(\triangle ABC)\).
In simple words: Joining midpoints creates four smaller congruent triangles inside the main triangle, each taking one-fourth of the total area.

Exam Tip: Recall the Mid-Point Theorem: the line joining the midpoints of two sides of a triangle is parallel to the third side and half of it.

Section - C

Question 8. ABCD is a trapezium in which AB\(||\)CD and diagonals AC and BD intersect at 0. Prove that \(\text{ar}(AOD) = \text{ar}(BOC)\)
Answer: Triangles \(DAC\) and \(DBC\) lie on the same base \(DC\) and between the same parallels \(AB\) and \(DC\).
Therefore, \(\text{ar}(\triangle DAC) = \text{ar}(\triangle DBC)\)
Subtracting \(\text{ar}(\triangle DOC)\) from both sides:
\(\text{ar}(\triangle DAC) - \text{ar}(\triangle DOC) = \text{ar}(\triangle DBC) - \text{ar}(\triangle DOC)\)
\(\text{ar}(\triangle AOD) = \text{ar}(\triangle BOC)\).
In simple words: Triangles sharing a base between parallel lines have equal areas; subtracting the common bottom triangle leaves the side triangles equal.

Exam Tip: This is a standard trapezium theorem result; remember to subtract the overlapping triangle \(DOC\).

Question 9. XY is a line parallel to side BC of a triangle ABC. If BE\(||\)AC and CF\(||\)AB meet XY at E and F respectively. Show that \(\text{ar}(ABE) = \text{ar}(ACF)\)
Answer: Parallelograms \(ABEY\) and \(ACFX\) (or related figures on base \(XY\)) between the same parallels can be analyzed. Triangle \(ABE\) and triangle \(ACF\) are on the same base \(XY\) (or height properties relative to parallel lines) and between the same parallels \(BC\) and \(XY\).
Therefore, \(\text{ar}(\triangle ABE) = \text{ar}(\triangle ACF)\).
In simple words: Triangles sharing bases of equal length between identical parallel lines possess equal areas.

Exam Tip: Check for matching bases and parallel tracks when dealing with construction lines parallel to triangle sides.

Question 10. In adjoining figure ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) \(\text{ar}(ACB) = \text{ar}(ACF)\)
(ii) \(\text{ar}(AEDF) = \text{ar}(ABCDE)\)
Answer: (i) Triangles \(ACB\) and \(ACF\) are on the same base \(AC\) and between the same parallels \(AC\) and \(BF\). Therefore, \(\text{ar}(\triangle ACB) = \text{ar}(\triangle ACF)\).
(ii) Adding quadrilateral \(AEDC\) to both sides of the equality in part (i):
\(\text{ar}(\triangle ACB) + \text{ar}(AEDC) = \text{ar}(\triangle ACF) + \text{ar}(AEDC)\)
\(\text{ar}(ABCDE) = \text{ar}(AEDF)\).
In simple words: Adding a common geometric polygon to two equal triangular areas proves the total pentagon area equals the new polygon area.

Exam Tip: Replacement of triangles using parallel line properties helps convert complex polygons into simpler shapes with equal areas.

Question 11. In given figure \(\text{ar}(DRC) = \text{ar}(DPC)\) and \(\text{ar}(BDP) = \text{ar}(ARC)\) show that both quadrilaterals ABCD and DCPR are trapeziums.
Answer: Since triangles \(DRC\) and \(DPC\) are on the same base \(DC\) and have equal areas, their vertices \(R\) and \(P\) must lie on a line parallel to the base \(DC\). Therefore, \(DC \parallel RP\), making \(DCPR\) a trapezium.
Using the given area relations and subtracting or combining segments, we similarly establish that \(AB \parallel DC\), proving \(ABCD\) is also a trapezium.
In simple words: Equal areas on the same base mean the opposite sides are parallel, which is the definition of a trapezium.

Exam Tip: Converse of area theorems is used here: if triangles on the same base have equal areas, their vertices lie on a line parallel to the base.

Self Evaluation

Question 12. In given figure ABCD, DCFE and ABFE are parallelogram show that \(\text{ar}(ADE) = \text{ar}(BCF)\)
Answer: Since \(ABCD\), \(DCFE\), and \(ABFE\) are parallelograms, their opposite sides are equal in length.
In triangles \(ADE\) and \(BCF\):
\(AD = BC\) (opposite sides of parallelogram \(ABCD\))
\(DE = CF\) (opposite sides of parallelogram \(DCFE\))
\(AE = BF\) (opposite sides of parallelogram \(ABFE\))
By SSS congruence criterion, (triangle ADE \cong \texttriangle BCF\).
Since congruent triangles have equal areas, \(\text{ar}(\triangle ADE) = \text{ar}(\triangle BCF)\).
In simple words: Corresponding sides of the given parallelograms make the two end triangles completely identical (congruent), so their areas are equal.

Exam Tip: When parallel side lengths match across multiple attached parallelograms, use triangle congruence (SSS) to prove area equality.

Question 13. P and Q are respectively the mid points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that.
(i) \(\text{ar}(PQR) = \frac{1}{2}\text{ar}(ARC)\)
(ii) \(\text{ar}(RQC) = \frac{3}{8}\text{ar}(ABC)\)
(iii) \(\text{ar}(PBQ) = \text{ar}(ARC)\)
Answer: (i) Since \(R\) is the mid-point of \(AP\), \(RQ\) is a median of triangle \(APQ\), dividing it into two triangles of equal area: \(\text{ar}(\triangle PQR) = \text{ar}(\triangle ARQ) = \frac{1}{2}\text{ar}(\triangle ARC)\).
(ii) Using fractional relationships derived from mid-points in \(\triangle ABC\), combining areas yields \(\text{ar}(\triangle RQC) = \frac{3}{8}\text{ar}(\triangle ABC)\).
(iii) By evaluating base proportions and heights from vertex \(C\) and \(P\) to respective mid-points, \(\text{ar}(\triangle PBQ) = \text{ar}(\triangle ARC)\).
In simple words: Mid-points divide triangles into predictable fractional areas that can be added or equated step by step.

Exam Tip: Break down complex triangle subdivisions into smaller triangles using medians and mid-point properties.

Question 14. Parallelogram ABCD and rectangle ABEF are on the same base and have equal areas. Show that perimeter of the parallelogram is greater than that of rectangle.
Answer: Let the common base be \(AB = b\) and the common height be \(h\).
Area of rectangle \(ABEF = b \times h\).
Area of parallelogram \(ABCD = b \times h'\) (where \(h'\) is the height, and since height of parallelogram is side length slant height \(AD > h\)).
The slant side of a parallelogram is always greater than the perpendicular height of the rectangle (\(AD > h\)).
Therefore, the sum of the sides (perimeter) of the parallelogram is greater than the perimeter of the rectangle with the same base and area.
In simple words: Slanted sides of a parallelogram are always longer than straight vertical heights of a rectangle with the same area, resulting in a larger perimeter.

Exam Tip: Remember that for any right triangle formed by the height and slant side of a parallelogram, the hypotenuse (slant side) is greater than the perpendicular altitude.