Chapter 6: Measuring Space: Perimeter and AreaPage 142–143All 11 Questions on This Page

Class 9 Maths Chapter 6 Exercise Set 6.2 Solutions

Step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 6 Exercise Set 6.2 (Questions 1–11, Page 142–143). Covers area of triangles using base and height, area of trapeziums using the Pythagorean theorem, triangle area using Heron's formula, area of a rhombus using diagonals, equal-area triangles between parallel lines, median area division, midpoint parallelograms in 4-gons, interior point partitions of a square, and parallel line area transfers.

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Key Concepts for Exercise Set 6.2

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Area of a Triangle (Base & Height)

When any side of a triangle is chosen as the base and its perpendicular height from the opposite vertex is known, the area is half the product of base and height.

Area = 1/2 × base × height

Equal Heights Between Parallel Lines

If a triangle's base lies on one of two parallel lines and its opposite vertex lies on the other parallel line, the perpendicular height is simply the constant distance between those parallel lines.

Height = Perpendicular distance between parallel lines

Area of a Trapezium

A trapezium has one pair of parallel sides. Its area is half the product of the sum of the parallel sides and the perpendicular height between them.

Area = 1/2 × (sum of parallel sides) × height

Isosceles Trapezium & Pythagorean Theorem

In an isosceles trapezium (equal non-parallel sides), perpendiculars dropped from the shorter base split the base difference symmetrically: (a − b)/2 on each side. The height is found using the Pythagorean theorem.

h² + [(a − b)/2]² = c²

Semi-perimeter of a Triangle

The semi-perimeter s is half of the triangle's total perimeter. It is the starting value for calculating area via Heron's formula.

s = (a + b + c) / 2 = Perimeter / 2

Heron's Formula

Heron's formula computes the area of any triangle directly from the lengths of its three sides (a, b, c) and its semi-perimeter (s), without needing an altitude.

Area = √[s(s − a)(s − b)(s − c)]

Ratio of Sides of a Triangle

When the sides of a triangle are in a ratio p : q : r and the total perimeter P is known, represent the sides as px, qx, and rx. Solve px + qx + rx = P to determine the value of x.

px + qx + rx = Perimeter

Area of a Rhombus (from Diagonals)

A rhombus has perpendicular diagonals that bisect each other. Its area is half the product of the lengths of its two diagonals.

Area = 1/2 × d₁ × d₂

Equal-Area Triangles Between Parallel Lines

Triangles having the same (or equal) base and lying between the same two parallel lines have identical perpendicular heights, and therefore always have equal areas.

Area(△₁) = Area(△₂) ⟹ Ratio = 1 : 1

Median Divides a Triangle into Equal Areas

A median of a triangle connects any vertex to the midpoint of the opposite side. It bisects the base while sharing the same altitude, thereby dividing the triangle into two triangles of equal area.

area(△₁) = area(△₂) = 1/2 × area(△)

Midpoint Theorem

The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.

EF ∥ AC and EF = 1/2 × AC

Midpoints of a 4-gon & Varignon's Parallelogram

Joining the midpoints of the sides of any 4-gon (quadrilateral) in order forms a parallelogram. By decomposing the 4-gon into four corner triangles and the inner parallelogram, the corner triangles total half the area, leaving the inner parallelogram with exactly half the area of the 4-gon.

Area of parallelogram EFGH = 1/2 × Area of the given 4-gon

Triangles with Equal Bases and Heights

Triangles on equal bases (such as segments formed by a midpoint or median) and having identical altitudes to those bases have equal areas.

Base₁ = Base₂ and Height₁ = Height₂ ⟹ area(△₁) = area(△₂)

Area Partition of a Square from an Interior Point

Joining any interior point P of a square to all four vertices divides the square into two pairs of opposite triangles (red and green). Each pair sums to exactly half the area of the square, giving an area ratio of 1 : 1.

area(△PAB) + area(△PCD) = area(△PBC) + area(△PDA) = 1/2 a²

Question 1

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Find the area of triangle ADE in Fig. 6.31.

Fig. 6.31: Triangle ADE in Rectangle ABCDFig. 6.31

Fig. 6.31: Triangle ADE inside rectangle ABCD with AD = 8 cm and DC = 10 cm

Fig. 6.31: Finding the area of triangle ADE inside rectangle ABCD.

Solution

Understanding the geometry from Fig. 6.31:

•ABCD is a rectangle with vertical side AD and horizontal side DC = 10 cm.

•The opposite vertical side is BC = 8 cm. In a rectangle, opposite sides are equal, so AD = BC = 8 cm.

•Point E lies on the opposite vertical side BC.

•Triangle ADE has AD as its base, and vertex E lies on line BC which is parallel to AD.

•The perpendicular height from vertex E to the line containing base AD is equal to the width of the rectangle: AB = DC = 10 cm.

Given:

Base AD = 8 cm

Perpendicular height from E to AD = 10 cm

We know,

Area of triangle = 12 × base × height

Taking AD as the base and the perpendicular distance from E to AD as the height:

Area of triangle ADE = 12 × 8 × 10

= 4 × 10

= 40 cm²

Therefore,

Area = 40 cm²

Note

Alternative Geometric Insight: • Triangle ADE and rectangle ABCD share the exact same base AD and lie between the same parallel lines AD and BC. • By the theorem of areas, the area of a triangle on the same base and between the same parallels as a rectangle is equal to half the area of the rectangle: Area(△ADE) = 1/2 × Area(Rectangle ABCD) = 1/2 × (8 × 10) = 40 cm².

Answer:Area = 40 cm²

Question 2

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The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

Solution

Given:

Parallel sides of trapezium:

a = 40 cm

b = 20 cm

Each non-parallel side:

c = 26 cm

Finding the height of the trapezium:

Since the non-parallel sides are equal, the trapezium is isosceles.

Dropping perpendiculars from the ends of the shorter parallel side (20 cm) to the longer parallel side (40 cm) divides the excess base equally on both sides:

Difference between parallel sides:

40 − 20 = 20 cm

Horizontal part on each side:

= 202

= 10 cm

Now consider one of the right-angled triangles formed by a non-parallel side (hypotenuse = 26 cm), the perpendicular height h, and the horizontal segment (10 cm).

Using Pythagoras' theorem:

h² + 10² = 26²

Therefore,

h² = 26² − 10²

= 676 − 100

= 576

Taking square root on both sides:

h = √576

= 24 cm

So, the height of the trapezium is:

h = 24 cm

We know,

Area of trapezium = 12 × (sum of parallel sides) × height

Substituting the values:

Area = 12 × (40 + 20) × 24

= 12 × 60 × 24

= 30 × 24

= 720 cm²

Therefore,

Area = 720 cm²

Answer:Area = 720 cm²

Question 3

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Find the area of a triangle, given that its sides are 8 cm and 11 cm, and its perimeter is 32 cm.

Solution

Given:

Two sides of the triangle are:

a = 8 cm

b = 11 cm

Perimeter = 32 cm

Finding the third side:

Let the third side be c.

Perimeter = a + b + c

8 + 11 + c = 32

19 + c = 32

c = 32 − 19

c = 13 cm

So, the three sides are:

a = 8 cm, b = 11 cm, c = 13 cm

Semi-perimeter of the triangle:

s = a + b + c2

= 322

= 16 cm

Using Heron's formula:

Area = √[s(s − a)(s − b)(s − c)]

Substituting the values:

Area = √[16(16 − 8)(16 − 11)(16 − 13)]

= √[16 × 8 × 5 × 3]

Simplifying under the square root:

= √[16 × (4 × 2) × 15]

= √[64 × 30]

= 8√30 cm²

Calculating the decimal approximation:

Using √30 ≈ 5.4772:

8 × 5.4772 ≈ 43.82 cm²

Therefore,

Area = 8√30 cm² ≈ 43.82 cm²

Answer:Area = 8√30 cm² ≈ 43.82 cm²

Question 4

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The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

Solution

Given:

Ratio of sides = 3 : 5 : 7

Perimeter = 300 m

Finding the side lengths:

Let the three sides be:

3x, 5x, and 7x

Since the perimeter is 300 m:

3x + 5x + 7x = 300

15x = 300

x = 30015

x = 20

Therefore, the three sides are:

a = 3x = 3 × 20 = 60 m

b = 5x = 5 × 20 = 100 m

c = 7x = 7 × 20 = 140 m

Semi-perimeter of the triangle:

s = a + b + c2

= 60 + 100 + 1402

= 3002

= 150 m

Using Heron's formula:

Area = √[s(s − a)(s − b)(s − c)]

Substituting the values:

Area = √[150(150 − 60)(150 − 100)(150 − 140)]

= √(150 × 90 × 50 × 10)

Factoring under the square root:

= √6,750,000

= √(2,250,000 × 3)

= 1500√3 m²

Calculating the decimal approximation:

Using √3 ≈ 1.73205:

1500 × 1.73205 ≈ 2598.08 m²

Therefore,

Area = 1500√3 m² ≈ 2598.08 m²

Answer:Area = 1500√3 m² ≈ 2598.08 m²

Question 5

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One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

Solution

Let the shorter diagonal be:

d cm

Then the longer diagonal is:

2d cm

We know,

Area of rhombus = 12 × d₁ × d₂

Therefore,

128 = 12 × d × 2d

128 = d²

Therefore,

d = √128

= √(64 × 2)

= 8√2 cm

Calculating the decimal approximation:

Using √2 ≈ 1.414:

8 × 1.414 ≈ 11.31 cm

Therefore,

Shorter diagonal = 8√2 cm ≈ 11.31 cm

Answer:Shorter diagonal = 8√2 cm ≈ 11.31 cm

Question 6

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ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio: area(△PCD) : area(△QCD)

Parallelogram ABCD with Points P and Q on ABFig. 6.32 (Explanatory)

Parallelogram ABCD showing points P and Q on side AB, triangles PCD and QCD with common base CD and equal perpendicular height h

△PCD and △QCD share the common base CD and have identical perpendicular height h between the parallel lines AB and CD.

Solution

Since ABCD is a parallelogram:

AB ∥ CD

Points P and Q lie on AB.

Therefore, the perpendicular distances of P and Q from the parallel line CD are equal.

So, △PCD and △QCD have:

•the same base CD

•the same height, say h

We know,

Area of triangle = 12 × base × height

Therefore,

Area(△PCD) = 12 × CD × h

and

Area(△QCD) = 12 × CD × h

Taking the ratio:

Area(△PCD) / Area(△QCD)

= [1/2 × CD × h] / [1/2 × CD × h]

= 1

Therefore,

Area(△PCD) : Area(△QCD) = 1 : 1

Hence,

Area(△PCD) : Area(△QCD) = 1 : 1

Thus, the two triangles always have equal areas, regardless of where P and Q are chosen on AB.

Note

Geometric Theorem: • Triangles on the same base (or equal bases) and between the same parallel lines are equal in area. • Since both △PCD and △QCD share the common base CD and lie between parallel lines AB and CD, their areas must be equal, giving an area ratio of 1 : 1.

Answer:Area(△PCD) : Area(△QCD) = 1 : 1

Question 7

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O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Parallelogram PQRS with Point O on Diagonal PRFig. 6.33 (Explanatory)

Parallelogram PQRS (labeled clockwise with PQ at the top) showing diagonal PR, point O on PR, triangles PSO and PQO sharing base PO, and equal perpendicular heights h from S and Q to line PR

Parallelogram PQRS labeled clockwise with PQ at the top. Triangles PSO and PQO share common base PO and have equal perpendicular heights h from opposite vertices S and Q to line PR.

Solution

Given:

•PQRS is a parallelogram.

•O is any point on diagonal PR.

To prove:

area(△PSO) = area(△PQO)

Proof:

Since PQRS is a parallelogram, its diagonal PR divides it into two congruent triangles:

△PQR ≅ △PSR

Therefore, their altitudes to the common base PR are equal.

Let this common height be h.

Since O lies on PR, the segment PO lies on the same line as PR. Therefore, the perpendicular distances of S and Q from PO are also both h.

Now, △PSO and △PQO have the same base PO and the same height h.

We know,

Area of triangle = 12 × base × height

Therefore,

area(△PSO) = 12 × PO × h

and

area(△PQO) = 12 × PO × h

Hence,

area(△PSO) = area(△PQO)

Hence proved.

Answer:area(△PSO) = area(△PQO)

Question 8

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If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Midpoints of 4-gon ABCD Forming Parallelogram EFGHFig. 6.34 (Explanatory)

A general 4-gon ABCD with midpoints E, F, G, H of sides AB, BC, CD, DA joined to form inner parallelogram EFGH with dashed diagonals AC and BD

Midpoints E, F, G, H of 4-gon ABCD joined in order form parallelogram EFGH. The four corner triangles total half the area of ABCD, leaving parallelogram EFGH with area equal to half of ABCD.

Solution

Given:

•ABCD is a 4-gon (quadrilateral).

•Points E, F, G, and H are the midpoints of sides AB, BC, CD, and DA respectively.

•Joining E, F, G, and H in order forms a quadrilateral EFGH.

To prove:

1. EFGH is a parallelogram.

2. Area of parallelogram EFGH = 1/2 × Area of the given 4-gon ABCD.

Construction:

Join diagonals AC and BD of the 4-gon ABCD.

Part 1: Proving that EFGH is a Parallelogram

Consider △ABC:

E is the midpoint of AB and F is the midpoint of BC.

By the Midpoint Theorem:

EF ∥ AC and EF = 1/2 × AC --- (1)

Consider △ADC:

H is the midpoint of AD and G is the midpoint of CD.

By the Midpoint Theorem:

HG ∥ AC and HG = 1/2 × AC --- (2)

From equations (1) and (2):

EF ∥ HG and EF = HG

Since a pair of opposite sides of quadrilateral EFGH is equal and parallel:

EFGH is a parallelogram.

Similarly, considering diagonal BD in △ABD and △CBD:

EH ∥ BD ∥ FG and EH = FG = 1/2 × BD.

Part 2: Proving Area(EFGH) = 1/2 × Area(ABCD)

The 4-gon ABCD is decomposed into the inner parallelogram EFGH and four corner triangles (△AEH, △BEF, △CFG, and △DGH):

Area(parallelogram EFGH) = Area(ABCD) − [Area(△AEH) + Area(△BEF) + Area(△CFG) + Area(△DGH)]

Finding the area of corner triangles along diagonal AC:

In △ABC, E and F are the midpoints of AB and BC.

Since △BEF is similar to △BAC with scale factor 1/2:

Area(△BEF) = 14 × Area(△ABC)

Similarly, in △ADC, G and H are the midpoints of CD and DA:

Area(△DGH) = 14 × Area(△ADC)

Adding these two opposite corner triangles:

Area(△BEF) + Area(△DGH) = 14 × Area(△ABC) + 14 × Area(△ADC)

= 14 × [Area(△ABC) + Area(△ADC)]

= 14 × Area(ABCD)--- (3)

Finding the area of corner triangles along diagonal BD:

In △ABD, E and H are the midpoints of AB and AD:

Area(△AEH) = 14 × Area(△ABD)

In △CBD, F and G are the midpoints of BC and CD:

Area(△CFG) = 14 × Area(△CBD)

Adding these two corner triangles:

Area(△AEH) + Area(△CFG) = 14 × Area(△ABD) + 14 × Area(△CBD)

= 14 × [Area(△ABD) + Area(△CBD)]

= 14 × Area(ABCD)--- (4)

Total area of all four corner triangles:

Adding equations (3) and (4):

[Area(△AEH) + Area(△BEF) + Area(△CFG) + Area(△DGH)]

= 1/4 × Area(ABCD) + 1/4 × Area(ABCD)

= 24 × Area(ABCD)

= 12 × Area(ABCD)

Calculating area of the inner parallelogram EFGH:

Area(parallelogram EFGH) = Area(ABCD) − 12 × Area(ABCD)

= 12 × Area(ABCD)

Therefore,

Area of parallelogram EFGH = 12 × Area of the given 4-gon

Hence proved.

Answer:Area of parallelogram EFGH = 12 × Area of the given 4-gon

Question 9

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In △ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area(△ABP) = area(△ACP).

Fig. 6.32: Median AD and Point P in Triangle ABCFig. 6.32

Fig. 6.32: Triangle ABC with midpoint D of BC, median AD, point P on AD, green triangle ABP and red triangle ACP

Fig. 6.32: Point P lies on median AD. Subtracting the equal areas of △PBD and △PCD from equal triangles △ABD and △ACD proves area(△ABP) = area(△ACP).

Solution

Given:

•D is the midpoint of BC.

•Therefore, BD = DC.

•Segment AD is a median of △ABC.

•P is any point on the median AD.

To prove:

area(△ABP) = area(△ACP)

Proof:

First, consider △PBD and △PCD:

•They have equal bases: BD = DC (since D is the midpoint of BC).

•Both triangles share vertex P, so they have the same perpendicular height from P to the line containing base BC.

Since triangles with equal bases and equal heights have equal areas:

area(△PBD) = area(△PCD)--- (1)

Now consider △ABD and △ACD:

•Segment AD is a median of △ABC.

•They have equal bases BD = DC and share the same perpendicular height from vertex A to line BC.

Therefore, a median divides a triangle into two triangles of equal area:

area(△ABD) = area(△ACD)--- (2)

Subtracting equation (1) from equation (2):

area(△ABD) − area(△PBD) = area(△ACD) − area(△PCD)

From the figure, since P lies on AD:

•area(△ABD) − area(△PBD) = area(△ABP)

•area(△ACD) − area(△PCD) = area(△ACP)

Therefore,

area(△ABP) = area(△ACP)

Hence proved.

Answer:area(△ABP) = area(△ACP)

Question 10

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Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (△PAB and △PCD) and the green region (△PBC and △PDA)?

Fig. 6.33: Square ABCD with Interior Point P and Colored RegionsFig. 6.33

Fig. 6.33: Square ABCD with interior point P joined to vertices A, B, C, D, showing red region (triangles PAB and PCD) and green region (triangles PBC and PDA)

Fig. 6.33: Point P inside square ABCD. Opposite triangles to parallel sides have heights summing to the side length a, showing both red and green regions equal half the square's area.

Solution

Let the side of the square ABCD be a.

From the figure, the two colored regions are:

•Red region = area(△PAB) + area(△PCD)

•Green region = area(△PBC) + area(△PDA)

Area of the Red Region:

Let the perpendicular distance of P from side AB be h.

Since ABCD is a square, the opposite sides AB and CD are parallel and separated by distance a.

Therefore, the perpendicular distance of P from CD is:

a − h

Using the triangle area formula (Area = 1/2 × base × height):

area(△PAB) = 12 × a × h

and

area(△PCD) = 12 × a × (a − h)

Adding these two areas to find the total red region:

Area of red region = 12·a·h + 12·a·(a − h)

= 12·a·[h + (a − h)]

= 12·a·[a]

= 12 a²

Area of the Green Region:

Let the perpendicular distance of P from side BC be k.

Since opposite sides BC and AD are parallel and separated by distance a, the perpendicular distance of P from AD is:

a − k

Using the triangle area formula:

area(△PBC) = 12 × a × k

and

area(△PDA) = 12 × a × (a − k)

Adding these two areas to find the total green region:

Area of green region = 12·a·k + 12·a·(a − k)

= 12·a·[k + (a − k)]

= 12·a·[a]

= 12 a²

Comparing the Two Regions:

Area of red region = 12 a²

Area of green region = 12 a²

Taking the ratio:

Red region : Green region = 1/2 a² : 1/2 a²

= 1 : 1

Thus, the red and green regions have equal areas, regardless of where P lies inside the square.

Answer:Red region : Green region = 1 : 1

Question 11

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In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined (Fig. 6.34). Prove that Area(△BPQ) = 1/2 Area(△ABC).

Fig. 6.34: Triangle ABC with Parallel Lines CQ and PDFig. 6.34

Fig. 6.34: Triangle ABC with midpoint D of AB, point P on BC, Q on AB such that CQ is parallel to PD, with shaded triangle BPQ

Fig. 6.34: Triangles PDQ and PDC share base PD and lie between parallel lines CQ ∥ PD, so area(△PDQ) = area(△PDC). Therefore, area(△BPQ) = area(△BCD) = 1/2 area(△ABC).

Solution

Given:

•In △ABC, point D is the midpoint of side AB.

•P is any point on BC.

•Q is a point on AB such that CQ ∥ PD.

•Line segment PQ is joined.

To prove:

area(△BPQ) = 12 area(△ABC)

Proof:

Join CD to form segment CD.

Since D is the midpoint of side AB, segment CD is a median of △ABC.

A median of a triangle divides it into two triangles of equal area:

area(△BCD) = 12 area(△ABC)--- (1)

Now consider △PDQ and △PDC:

•Both triangles have the same base PD.

•Vertices Q and C lie on the line QC, which is parallel to base PD (CQ ∥ PD).

•Therefore, both triangles have the same perpendicular height between the parallel lines PD and CQ.

Triangles on the same base and between the same parallel lines are equal in area:

area(△PDQ) = area(△PDC)--- (2)

Now, look at triangle BPQ in Fig. 6.34:

Triangle BPQ is composed of △BPD and △PDQ:

area(△BPQ) = area(△BPD) + area(△PDQ)

Substitute area(△PDC) in place of area(△PDQ) using equation (2):

area(△BPQ) = area(△BPD) + area(△PDC)

From the figure, the sum of △BPD and △PDC forms triangle BCD:

area(△BPD) + area(△PDC) = area(△BCD)

Therefore,

area(△BPQ) = area(△BCD)

Using equation (1), since area(△BCD) = 1/2 area(△ABC):

area(△BPQ) = 12 area(△ABC)

Hence proved.

Answer:area(△BPQ) = 12 area(△ABC)

Need to practice other sections in Chapter 6?

Return to Chapter 6 (Measuring Space: Perimeter and Area)→

💡Key Concepts for Exercise Set 6.2

Area of a Triangle (Base & Height)

Equal Heights Between Parallel Lines

Area of a Trapezium

Isosceles Trapezium & Pythagorean Theorem

Semi-perimeter of a Triangle

Heron's Formula

Ratio of Sides of a Triangle

Area of a Rhombus (from Diagonals)

Equal-Area Triangles Between Parallel Lines

Median Divides a Triangle into Equal Areas

Midpoint Theorem

Midpoints of a 4-gon & Varignon's Parallelogram

Triangles with Equal Bases and Heights

Area Partition of a Square from an Interior Point

Key Concepts for Exercise Set 6.2