NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration (Old NCERT Questions)

Part- 1

Try These

Question: (a) Match the following figures with their respective areas in the box.

(b) Write the perimeter of each shape.

Solution:

(b) (i) The given figure is a rectangle in which

Length = 14 cm

Breadth = 7 cm

∵ Perimeter of a rectangle = 2 * [Length + Breadth]

∴ Perimeter of the given figure = 2 * [14 cm + 7 cm]

= 2 * 21 cm = 42 cm

(ii) The figure is a square housing its side as 7 cm.

∵ Perimeter of a square = 4 * Side

∴ Perimeter of the given figure = 4 * 7 cm = 28 cm

Exercise 11.1

Question 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Solution:

(a) Side of the square = 60 m

 ∴ Its perimeter = 4 * Side

= 4 * 60 m = 240 m

 Area of the square = Side * Side

= 60 m * 60 m = 3600 m²

(b) ∵ Perimeter of the rectangle = Perimeter of the given square

∴ Perimeter of the rectangle = 240 m

or 2 * [Length + Breadth] = 240 m

or 2 * [80 m + Breadth] = 240 m

or 80 m + Breadth = 240/2 m = 120 m

∴ Breadth = (120 – 80) m = 40 m

Now, Area of the rectangle = Length* Breadth

= 80 m *40 m

= 3200 m²

Since, 3600 m² > 3200 m²

∴  Area of the square field (a) is greater.

Question 2. Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m².

Solution: ∵ The given plot is a square with side as 25 m.

∴ Area of the plot = Side * Side

= 25 m * 25 m = 625 m²

∵ The constructed portion is a rectangle having length = 20 m and breadth = 15 m.

∴ Area of the constructed portion = 20 m * 15 m = 300 m²

Now area of the garden = [Total plot area] – [Total constructed area]

= (625 – 300) m² = 325 m²

∴ Cost of developing the garden = Rs 55 * 325 = Rs 17,875

Question 3. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres.]

Solution: For the semi-circular part:

Diameter of the semi-circle = 7 m

∴ Radius of the semi-circle = 7/2 = 3.5 m

Area of the 2 semi-circles = 

= πr² = 22/7 *(3.5 m)²

Also, perimeter of the 2 semicircles

For rectangular part:

Length of the rectangle = 20 – (3.5 + 3.5) m = 13 m

Breadth of the rectangle = 7 m

∴ Area = Length * Breadth

= 13 m * 7 m = 91 m²

Perimeter = 2 * [Length + Breadth]

= 2 * [13 m + 0 m]

= 2 * 13 m = 26 m

Now, Area of the garden = (38.5 + 91) m²

= 129.5 m²

Perimeter of the garden = 22 m + 26 m = 48 m

Question 4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way want to fill up the corners.)

Solution: Area of a parallelograms = Base * Corresponding height

Area of a tile = 

∴ Area of the floor = 1080 m²

Now, number of tiles = Total area/Area of one tiles

= 1080/0.024 = 45000 tiles

Question 5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Solution: (a) Diameter = 2.8 cm fi Radius = 2.8/2 cm = 1.4 cm

Perimeter of a circle = 2pr ⇒ Perimeter of a semicircle = 2πR/2 = πr

∴  Perimeter of the figure = πr + Diameter

= 22/7 * 1.4 cm + 2.8 cm

= 22/7* 14/10 cm + 2.8 cm

= 4.4 cm + 2.8 cm = 7.2 cm

(b) Perimeter of the semi-circular part = πr

= 22/7 * 1.4 cm

= 4.4 cm

Perimeter of the remaining part = 1.5 cm + 2.8 cm + 1.5 cm= 5.8 cm

∴ Perimeter of the figure = 4.4 cm + 5.8 cm = 10.2 cm

(c) Perimeter of the semi-circular part = πd/z  =  22/7 * 2.8/2  = 4.4 cm

∴ Perimeter of the figure = 4.4 cm + 2 cm + 2 cm

= 8.4 cm

Since, 7.2 cm < 8.4 cm < 10.2 cm

∴ Perimeter of figure ‘b’ has the longest round.

Note: In figures ‘b’ and ‘c’, the diameters are not part of the figures.

Area of Trapezium

The area of a trapezium is the half of the sum of the length of parallel sides and then multiply the perpendicular distances between then.

Here ABCD is a trapezium.

Here AB and CD are parallel sides.

Also AE and BF are perpendicular distance between them.

∴  Area of trapezium of ABCD = 1/2  (AB + CD) * AE

Question 1. Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown. Show that the area of trapezium WXYZ =

Solution:

Area of Δ PWZ =1/2 * Base * Altitude

= 1/2 * c * h = 1/2 ch

Area of the rectangle PQYZ = b * h = bh

Area of Δ QXY = 1/2 * d * h = 1/2 dh

∴ Area of the trapezium XYZW = 1/2 ch + bh +1/2 dh                [∵ c + d + b = a]

= 1/2 (c+d)h + bh       [∵ c + d = a – b]

Question 2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separately and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression 

Solution: Area of Δ PWZ = 1/2 ch =1/2 * 6 * 10 cm² = 30 cm²

Area of Δ QXY = 1/2 dh = 1/2 * 4 * 10 cm² = 20 cm²

Area of rectangle PQYZ = Length * Breadth = b * h

= 12 * 10 cm² = 120 cm²

\ Area of trapezium WXYZ = Area of Δ PWZ + Area of Δ QXY + Area of rectangle PQYZ

= 30 cm² + 20 cm² + 120 cm² = 170 cm²

Also, area of the trapezium WXYZ =

Hence, the area of trapezium is verified.

Question: Find the area of the following trapeziums.

Solution:

Note: Area of a trapezium = 1/2* [sum of the parallel sides]*[Perpendicular distance between the parallel sides]

(i) Area of the given trapezium

= 1/2 * (Sum of parallel sides) * (Perpendicular distance between the parallel sides)

= 1/2 * (9 + 7) cm * 3 cm

= 1/2 * 16 cm * 3 cm = 24 cm²

(ii) Area of the given trapezium

= 1/2 * (Sum of the parallel sides) * (Perpendicular distance between the parallel sides)

= 1/2 * (10 + 5) cm * 6 cm

= 1/2 * 15 cm * 6 cm = 45 cm² 

Area of a General Quadrilateral

Let ABCD be a quadrilateral. Join its vertices A and C such that AC is a diagonal.

Draw BP  AC and DQ AC.

Let BP = h₁ and DQ = h₂

Now, area of quadrilateral ABCD

[Area of a quadrilateral] = 1/2 * (One of the diagonals) *(Sum of the perpendicular drawn on the diagonal from the opposite vertices)

Example 1. Find the area the quadrilateral LMNO (as shown in the figure).

Solution: Diagonal LN = 8 cm

Perpendicular MX = 4.5 cm

Perpendicular OY = 3.5 cm

∵ Sum of the perpendiculars = MX + OY = 4.5 cm + 3.5 cm = 8 cm

∴  Area of the quadrilateral =1/2 * (A diagonal) * (Sum of the lengths of the perpendiculars on it from opposite vertices)

Question: We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?

Solution: The diagonal BD of quadrilateral ABCD is joined and it divides the quadrilateral into two triangles.

Now,

Area of quadrilateral ABCD = Area of D ABD + Area of D BCD

Infact ABCD is a parallelogram.

∴ Area of a parallelogram ABCD = b * h

Area of a parallelogram = Base * Height

We know that a parallelogram can also be a trapezium. We already know that

Area of trapezium ABCD = 1/2 (Sum of parallel sides) * [Perpendicular distance between the parallel sides]

or Area of parallelogram ABCD = bh.

Yes, the above relation agrees with formula that we know already.

Question: A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?

Solution: No, a trapezium cannot be divided into two congruent triangles.

Area of Special Quadrilaterals

Let ABCD be a rhombus. Therefore, its diagonals are perpendicular to each other.

Area of rhombus ABCD = Area of Δ ACD + Area of Δ ABC

where d₁ and d₂ are the diagonals of the rhombus.

Thus, the area of a rhombus =1/2 * The product of its diagonals

Note: Splitting a quadrilateral into triangles is called triangulation.

Example 2. Find the area of a rhombus whose diagonals are 12 cm and 9.2 cm.

Solution: Let d₁ and d₂ be the diagonals of the rhombus.

∴ d₁ = 12 cm and d₂ = 9.2 cm

∵ Area of rhombus =1/2  * d₁ * d₂

∴ Area of the given rhombus =1/2 * 12 * 9.2 cm²

= 6 * 9.2 cm² = 55.2 cm²

Question: Find the area of these quadrilaterals:

Solution:

(i) Area of quadrilateral ABCD

= 1/2 * AC * [Sum of perpendiculars on AC from opposite vertices]

= 1/2 * 6 cm * [3 cm + 5 cm]

= 1/2 * 6 cm * 8 cm = 3 cm * 8 cm = 24 cm²

(ii) The given figure is a rhombus having d₁ = 7 cm and d₂ = 6 cm.

∴  Area of the given rhombus =1/2 * Product of diagonals

= 1/2 * d₁ * d₂

= 1/2 *7 cm * 6 cm

= 7 cm * 3 cm = 21 cm²

(iii) The given figure is a parallelogram. Its diagonal divides it in totwo congruent triangles.

∴ Area of the parallelogram = 2 * [Area of one of the triangles]

Area of a Polygon

To find area of polygons, we divide them into shapes, for which we have a formula for the area. First we find the areas of various parts and then add them to get the area of given polygon.

Question 1. Divide the following polygons into parts (triangles and trapezium) to find out its area.

Solution:

(a) We draw perpendiculars from opposite vertices on FI, i.e. GL ⊥ FI, HM ⊥ FI and EN ⊥ FI

Area of the polygon EFGHI

= ar (Δ GFL) + ar (trapezium GLMH) + ar (Δ HMI) + ar (Δ NEI) + ar (Δ EFN)

(b) NQ is a diagonal. Draw OA ⊥ NQ, MB ⊥ NQ, PC ⊥ NQ and RD ⊥ NQ

∴ Area of polygon OPQRMN = ar (Δ OAN) + ar (trap. CPOA) + ar (Δ PCQ) + ar (Δ RDQ)+ ar (trap. MBDR) + ar (Δ MBN)

Question 2. Fill in the blanks.

Polygon ABCDE is divided into parts as shown below. Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Area of polygon ABCDE = Area of ΔAFB + …

Area of Δ AFB =1/2 * AF * BF = 1/2 * 3 * 2 = …

Area of trapezium FBCH =

So, the area of polygon ABCDE = …

Solution: Area of polygon ABCDE = Area of Δ AFB + Area of trapezium FBCH + Area of Δ CHD + Area of Δ ADE

Area of D AFB = 1/2 * AF * BF

= 1/2 * 3 * 2 = 3 cm²

Area of trapezium FBCH 

So, the area of polygon ABCD = 3 cm² + 7.5 cm² + 3 cm² + 10 cm² = 23.5 cm²

Question 3. Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm MC = 6 cm, MB = 4 cm, MA = 2 cm. NA, OC, QD and RB are perpendiculars to diagonal MP.

Solution: Area of polygon MNOPQR = ar (Δ MAN) + ar (trap. ACON) + ar (Δ OCP) + ar (Δ PDQ) + ar (trap. DBRQ) + ar Δ RBM)

 ∵            

∴ Area of polygon MNOPQR