GSEB Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4

Question 1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Answer:
Given: We have a parallelogram named ABCD and a rectangle named ABEF. Both share the same base, AB, and they cover the same amount of space (equal areas).
To prove: Our goal is to show that the border length (perimeter) of parallelogram ABCD is longer than the border length of rectangle ABEF.
Proof: The parallelogram ABCD and the rectangle ABEF are both built on the same base line AB. They are also located between two parallel lines, AB and FC.
The perimeter of parallelogram ABCD is calculated as \( 2(AB + AD) \). Similarly, the perimeter of rectangle ABEF is found by \( 2(AB + AF) \).
Consider the triangle ADF. Here, angle AFD is \( 90^\circ \). This means angle ADF is an acute angle, less than \( 90^\circ \). By the angle sum property of a triangle, if angle AFD is right, then angle AFD must be greater than angle ADF. This implies that the side AD is longer than the side AF, because the side opposite to a larger angle in a triangle is always longer.
This leads to \( AB + AD \) being greater than \( AB + AF \). Therefore, if we multiply both sides by 2, we find that \( 2(AB + AD) \) is greater than \( 2(AB + AF) \). As a result, the perimeter of parallelogram ABCD is greater than the perimeter of rectangle ABEF.
In simple words: In a triangle where one angle is 90 degrees, the longest side is always opposite that 90-degree angle. Because a parallelogram's slanted side is longer than the rectangle's straight height (when sharing a base and parallel lines), the parallelogram will have a larger total outside length.

Exam Tip: Remember that for figures sharing the same base and between the same parallels, parallelograms are "stretched" compared to rectangles, making their non-parallel sides longer than the perpendicular height.

Question 2. In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔΑΒD) = ar(ΔADE) = ar(ΔΑΕC). Can you now answer the question that you have left in the 'Introduction, of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
Answer:
Remark: Note that by taking \( BD = DE = EC \), the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide \( \triangle ABC \) into n triangles of equal areas.
Solution:
Let's draw a line AM that is perpendicular to BC. This line AM serves as the height for all three triangles: ABD, ADE, and AEC. Since they share the same base line and the same vertex A, their heights are identical.
\( ar(\triangle ABD) = \frac{1}{2} \times BD \times AM \) (Equation 1)
\( ar(\triangle ADE) = \frac{1}{2} \times DE \times AM \) (Equation 2)
\( ar(\triangle AEC) = \frac{1}{2} \times EC \times AM \) (Equation 3)
Since we know that BD, DE, and EC are all equal, and AM (the height) is common to all three, comparing equations (1), (2), and (3) shows that the areas of triangle ABD, triangle ADE, and triangle AEC are all equal.
Therefore, yes, Budhia's field has indeed been divided into three sections of the same size.
In simple words: If you split the base of a triangle into equal parts and draw lines from the top corner to each dividing point, you create smaller triangles. All these smaller triangles will have the same height (from the top corner to the base) and the same length of base (because you divided it equally). So, their areas will be equal.

Exam Tip: A key property to remember is that triangles sharing the same vertex and having bases on the same line (where the bases are equal in length) will have equal areas.

Question 3. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(AADE) = ar(ABCF)
Answer:
Solution:
Since ABCD is a parallelogram, its opposite sides are parallel. So, \( AD \parallel BC \) (Equation 1).
Similarly, DCFE is also a parallelogram, meaning its opposite sides are parallel. Thus, \( DE \parallel CF \) (Equation 2).
Furthermore, ABFE is a parallelogram. This tells us that its opposite sides are equal in length, so \( AE = BF \) (Equation 3).
Using the information from equations (1), (2), and (3), we can see that triangle ADE is congruent to triangle BCF (by SSS congruency).
Therefore, \( ar(\triangle ADE) = ar(\triangle BCF) \).
Two congruent triangles are equal in areas.
In simple words: When you have multiple parallelograms sharing sides and heights, you can use their properties (like opposite sides being equal and parallel) to show that certain triangles or quadrilaterals within them are identical in shape and size, which means they also have the same area.

Exam Tip: For problems involving multiple overlapping or connected parallelograms, carefully list out all equal and parallel sides. This helps in identifying congruent triangles or shapes with equal areas.

Question 4. In the figure, ABCD is a parallelogram, and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(ΔΒΡC) = ar(ADPQ).
Answer:
Solution:
Join AC.
Consider triangles AQAC and AQDC. They share the same base, QC, and are located between the same parallel lines, AD and QC.
Because two triangles on the same base (or equal bases) and between the same parallel lines have equal areas, the area of triangle AQAC is equal to the area of triangle AQDC. (Equation 1)
Since \( ar(\triangle AQAC) = ar(\triangle AQDC) \), and \( \triangle QPC \) is a common region, subtracting \( ar(\triangle QPC) \) from both triangles gives \( ar(\triangle PAC) = ar(\triangle QDP) \). (Equation 2)
Now, consider triangles PAC and PBC. They share the same base PC and are located between the parallel lines AB and DC.
Because triangles on the same base and between parallel lines have equal areas, we know that the area of triangle APAC is equal to the area of triangle PBC. (Equation 3)
Combining our findings from equations (2) and (3), we conclude that the area of triangle BPC (which is the same as PBC) is equal to the area of triangle ADPQ (which is the same as QDP). Thus, \( ar(\triangle BPC) = ar(ADPQ) \).
In simple words: When lines inside a parallelogram cross each other, or when sides are extended, you can find triangles that share the same base and are between parallel lines. Such triangles always have the same area. By subtracting common areas, you can show that other specific triangles have equal areas too.

Exam Tip: When comparing areas of triangles in complex figures, look for opportunities to apply the property that triangles on the same base and between the same parallels have equal areas. Subtracting a common area can then lead to new area equalities.

Question 5. In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersect BC at F, show that:
(i) \( ar(ABDE) = \frac {1}{4} ar(\triangle ABC) \)
(ii) \( ar(ABDE) = \frac {1}{2} ar(\triangle BAE) \)
(iii) \( ar(\triangle ABC) = 2ar(ABEC) \)
(iv) \( ar(ABFE) = ar(AAFD) \)
(v) \( ar(ABFE) = 2ar(AFED) \)
(vi) \( ar(AFED) = \frac {1}{8} ar(AAFC) \)
Answer:
Solution:
\( \triangle ABC \) is an equilateral triangle.
\( \angle ABC = \angle BCA = \angle CAB = 60^\circ \) (1)
\( \triangle BDE \) is an equilateral triangle.
\( \angle BDE = \angle DEB = \angle EBD = 60^\circ \) (2)
\( \implies \angle ABC = \angle BDE \)
Alternate interior angles. Similarly, \( \angle ACB = \angle EBC \)
\( AC \parallel BE \) (4) (each \( 60^\circ \))
Now, \( \triangle CBA \) and \( \triangle CEA \) are on the same base AC and between the same parallels.
\( ar(\triangle CBA) = ar(\triangle CEA) \)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
(i)
\( \implies ar(\triangle ABC) = ar(ACD) + ar(ACED) + ar(\triangle ADE) \) (5)
In \( \triangle ABC \), AD is a median.
\( ar(\triangle ABD) = ar(\triangle ACD) = \frac{1}{2} ar(\triangle ABC) \) (6)
A median of a triangle divides it into two triangles of equal area.
In \( \triangle EBC \), ED is a median.
\( ar(\triangle ECD) = ar(\triangle EBD) = \frac{1}{2} ar(\triangle EBC) \) (7)
A median of a triangle divides it into two triangles of equal area.
\( \triangle DEA \) and \( \triangle DB E \) are on the same base DE and between the same parallels AB and DE.
\( ar(\triangle DEA) = ar(\triangle DBE) \) (8)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Using (6), (7) and (8), in (5) gives
\( ar(\triangle ABC) = \frac{1}{2} ar(\triangle ABC) + ar(\triangle BDE) + ar(ABDE) \)
\( \implies \frac{1}{2} ar(\triangle ABC) = 2ar(ABDE) \)
\( \implies ar(ABDE) = \frac{1}{4} ar(\triangle ABC) \)
(ii)
\( \triangle BAE \) and \( ABCE \) are on the same base BE and between the same parallels BE and AC.
\( ar(\triangle BAE) = ar(ABCE) \)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
\( \implies \frac{1}{2} ar(\triangle BAE) = 2ar(ABDE) \) [From (7)]
\( \implies ar(ABDE) = \frac{1}{4} ar(\triangle BAE) \)
(iii)
\( 2ar(ABEC) = 2ar(ABDE) \) [From (7)]
\( = 4ar(ABDE) \)
\( = ar(\triangle ABC) \) [From (i)]
(iv)
\( \triangle EBD \) and \( \triangle EAD \) are on the same base ED and between the same parallels AB and \( DE \).
\( ar(\triangle EBD) = ar(\triangle EAD) \)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
\( \implies ar(\triangle EBD) - ar(\triangle EFD) = ar(\triangle EAD) - ar(\triangle EFD) \)
Subtracting the same areas from both sides.
\( \implies ar(ABFE) = ar(AAFD) \)
(v)
\( ar(ABDE) = \frac{1}{4} ar(\triangle ABC) \) [From (i)]
\( = \frac{1}{4} \cdot 2 ar(\triangle ABD) \)
\( = \frac{1}{4} ar(AABD) \)
Bases of ABDE and AABD are the same.
Altitude of ABDE \( = \frac{1}{2} \) Altitude of \( \triangle ABD \) (9)
\( ar(\triangle BEF) = ar(\triangle AFD) \) (10) [From (iv)]
But Altitude of ABDE = Altitude of ABEF (Considering common vertex)
And, altitude of \( \triangle ABD \) = Altitude of \( \triangle AFD \) (Considering common vertex A)
Altitude of ABEF \( = \frac{1}{2} \) Altitude of \( \triangle AFD \) (11)
From (10) and (11),
\( BF = 2FD \) (12)
In ABFE and AFED,
\( BF = 2FD \)
and alt. (ABFE) = alt. (AFED)
\( ar(ABFE) = 2ar(AFED) \).
(vi)
Let the altitude of \( \triangle ABD \) be h.
Then, altitude of ABED \( = \frac{h}{2} \) [From (9)]
Now, \( ar(AFED) = \frac{1}{2} \cdot FD \cdot \frac{h}{2} = \frac{FD \cdot h}{4} \) (13)
\( ar(AAFC) = \frac{1}{2} \cdot FC \cdot h \)
\( = \frac{1}{2} (FD + DC) h \cdot \frac{1}{2} \)
\( = \frac{1}{2} (FD + BD) h \cdot \frac{1}{2} \)
\( = \frac{1}{2} (FD + BF + FD) h \cdot \frac{1}{2} \)
\( = \frac{1}{2} (2FD + BF) h \cdot \frac{1}{2} \)
\( = \frac{1}{2} (2FD + 2FD) h \) [From (12)]
\( = 2 \cdot FD \cdot h \) (14)
From (13) and (14), we obtain
\( ar(AFED) = \frac{1}{8} ar(\triangle AFC) \)
In simple words: When you have equilateral triangles and medians, you can use the properties of area relationships. For instance, a median divides a triangle into two equal areas. By carefully combining these area relationships, you can show how the area of one shape is a specific fraction of another larger shape.

Exam Tip: Break down complex figures into simpler triangles and use properties like "median divides a triangle into two equal areas" or "triangles on same base between parallels have equal areas" to find relationships between areas.

Question 6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that \( ar(\triangle APB) \times ar(ACPD) = ar(\triangle APD) \times ar(\triangle BPC) \)
Answer:
Solution:
Construction: From vertex A, we draw a perpendicular line AE to the diagonal BD. Similarly, from vertex C, we draw a perpendicular line CF to the diagonal BD.
Proof:
The product of the areas of triangle APB and triangle CPD is \( ar(\triangle APB) \times ar(\triangle CPD) = \left( \frac{1}{2} \times PB \times AE \right) \times \left( \frac{1}{2} \times DP \times CF \right) \). This simplifies to \( \frac{1}{4} \times PB \times AE \times DP \times CF \). This is our first equation. (1)
Next, the product of the areas of triangle APD and triangle BPC is \( ar(\triangle APD) \times ar(\triangle BPC) = \left( \frac{1}{2} \times DP \times AE \right) \times \left( \frac{1}{2} \times PB \times CF \right) \). This also simplifies to \( \frac{1}{4} \times PB \times AE \times DP \times CF \). This is our second equation. (2)
Comparing equations (1) and (2), we can clearly see that the product of the areas of triangle APB and triangle CPD is equal to the product of the areas of triangle APD and triangle BPC. Thus, \( ar(\triangle APB) \times ar(\triangle CPD) = ar(\triangle APD) \times ar(\triangle BPC) \).
In simple words: When a quadrilateral's diagonals cross, they form four triangles. The rule is that if you multiply the areas of opposite triangles, these products will be equal. To prove this, draw lines straight down from the top corners to the diagonal, forming right-angled triangles to help with area calculations.

Exam Tip: For area product equalities involving intersecting diagonals, drawing altitudes to the diagonals is a standard construction. The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).

Question 7. 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:
1. \( ar(APRQ) = \frac {1}{2} ar(AARC) \)
2. \( ar(ARQC) = \frac {3}{8} ar(\triangle ABC) \)
3. \( ar(APBQ) = ar(AARC) \)
Answer:
Solution:
Construction: Join AQ and CP.
Proof:
1.
\( ar(APRQ) = ar(\triangle ARQ) \)
A median of a triangle divides it into two triangles of equal area.
\( = \frac{1}{2} ar(\triangle APQ) = \frac{1}{2} ar(\triangle BPQ) \)
\( = \frac{1}{2} ar(ACPQ) = \frac{1}{2} \cdot \frac{1}{2} ar(\triangle BPC) \)
\( = \frac{1}{4} ar(ABPC) = \frac{1}{4} \cdot \frac{1}{2} ar(\triangle ABC) \)
\( = \frac{1}{8} ar(\triangle ABC) \) (1)
\( \frac{1}{2} ar(\triangle ARC) = \frac{1}{2} \cdot \frac{1}{2} ar(\triangle APC) \)
\( = \frac{1}{4} ar(\triangle APC) = \frac{1}{4} \cdot \frac{1}{2} ar(\triangle ABC) \)
\( = \frac{1}{8} ar(\triangle ABC) \) (2)
From (1) and (2), we have
\( ar(APRQ) = \frac{1}{2} ar(\triangle ARC) \)
2.
\( ar(ARQC) = ar(ARBQ) \)
A median of a triangle divides it into triangles of equal areas.
\( = ar(APRQ) + ar(ABPQ) \)
\( = \frac{1}{8} ar(\triangle ABC) + \frac{1}{2} ar(APBC) \) [Using (1)]
\( = \frac{1}{8} ar(\triangle ABC) + \frac{1}{2} \cdot \frac{1}{2} ar(\triangle ABC) \)
\( = \frac{1}{8} ar(\triangle ABC) + \frac{1}{4} ar(\triangle ABC) \)
\( = \frac{3}{8} ar(\triangle ABC) \)
3.
\( ar(APBQ) = \frac{1}{4} ar(\triangle ABC) \) (3) [From (ii)]
\( ar(AARC) = \frac{1}{4} ar(\triangle ABC) \) (4) [From (i)]
From (3) and (4), we get \( ar(\triangle PBQ) = ar(\triangle ARC) \)
In simple words: When you have midpoints and medians in a triangle, these lines split the triangle into smaller parts with predictable area relationships. For example, a median divides a triangle into two triangles of equal area. By applying these rules repeatedly, you can show that one area is a specific fraction of another.

Exam Tip: Remember that a median always divides a triangle into two triangles of equal area. Use this fundamental property repeatedly to find relationships between areas of smaller triangles formed by medians and midpoints.

Question 8. In figure, ABC is a right triangle right angled at A. BCDE, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX \( \perp \) DE meets BC at Y. Show that:
(i) \( \triangle MBC - \triangle ABD \)
(ii) \( ar(BYXD) = 2ar(AMBC) \)
(iii) \( ar(BYXD) = ar(ABMN) \)
(iv) \( AFCB = \triangle ACE \)
(v) \( ar(CYXE) = 2ar(AFCB) \)
(vi) \( ar(CYXE) = ar(ACFG) \)
(vii) \( ar(BCED) = ar(ABMN) + ar(ACFG). \)
Answer:
Solution:
(i) In \( \triangle MBC \) and \( \triangle ABD \),
Side \( MB = AB \) (1) [Sides of a square]
Side \( BC = BD \) (2) [Sides of a square]
Angle \( MBA = \angle CBD \) [Each \( = 90^\circ \)]
\( \implies \angle MBA + \angle ABC = \angle CBD + \angle ABC \)
Adding the same \( \angle ABC \) to both sides.
\( \implies \angle MBC = \angle ABD \) (3)
Considering (1), (2) and (3), \( \triangle MBC \cong \triangle ABD \) [By congruence SAS rule]
(ii)
\( ar(BYXD) = 2 ar(\triangle ABD) \) (Since BD is parallel to AX and they share a common base BD)
\( = 2ar(\triangle MBC) \) [From part (i), \( ar(\triangle MBC) = ar(\triangle ABD) \)]
(iii)
\( ar(BYXD) = 2ar(\triangle ABD) \)
and \( ar(ABMN) = 2ar(\triangle MBC) \) (Since MB is parallel to NC, and common base is MB)
\( = 2ar(\triangle ABD) \) [From (i)]
\( \implies ar(BYXD) = ar(ABMN) \)
(iv)
In \( \triangle FCB \) and \( \triangle ACE \),
\( FC = AC \) [Sides of a square]
\( CB = CE \) [Sides of a square]
\( \angle FCA = \angle BCE \) [Each \( = 90^\circ \)]
\( \implies \angle FCA + \angle ACB = \angle BCE + \angle ACB \)
Adding the same on both sides.
\( \implies \angle FCB = \angle ACE \)
\( \implies \triangle FCB \cong \triangle ACE \) [SAS congruence rule]
(v)
\( ar(CYXE) = 2ar(\triangle ACE) \)
\( = 2ar(\triangle FCB) \) [From (iv)]
Since \( \triangle FCB \cong \triangle ACE \), their areas are equal. Therefore, \( ar(\triangle FCB) = ar(\triangle ACE) \).
(vi)
\( ar(CYXE) = 2ar(\triangle ACE) = 2ar(\triangle FCB) \)
And, \( ar(ACFG) = 2ar(\triangle FCB) \) (Since BG is parallel to CF and CF is the common base)
\( \implies ar(CYXE) = ar(ACFG) \)
(vii)
\( ar(BCED) = ar(CYXE) + ar(BYXD) \)
\( = ar(ACFG) + ar(ABMN) \) [From (iii) and (vi)]
\( = ar(ABMN) + ar(ACFG) \)
In simple words: This final part demonstrates Pythagoras' theorem using areas. The area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. You prove this by showing how the area of the large square can be split and then reassembled into the areas of the two smaller squares.

Exam Tip: This is a direct area-based proof of the Pythagorean theorem. Understand how the areas of the squares built on the sides of a right triangle sum up. The proof connects the large square's area to the sum of the two smaller squares' areas via intermediate quadrilaterals.