Well-designed AP 8th Class Maths Textbook Solutions Chapter 11 Mensuration InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

## AP 8th Class Maths 11th Chapter Mensuration InText Questions

Try These

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

Answer:

b) Write the perimeter of each shape.

Try These

Question 1.

Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown (Fig.). Show that the area of , (a + b) trapezium WXYZ = h \(\frac{a+b}{2}\).

Solution:

Shape of the PLOT is trapezium ‘WXYZ’

Let y1, z’ are two points on WX.

Such that Zz’, Yy’ are two perpendiculars then Yz = y’z’ = b ; let Wz’ = c, Xy’ = d ; Zz’ = Yy’ = h

then area of trapezium

= area of ∆Wz’Z + area of rectangle ZYy’z’ + area of ∆Yy’x

formula for area of triangle = \(\frac{1}{2}\) × b × h

then area of ∆Wz’Z = \(\frac{1}{2}\) × Wz’ × Zz’

= \(\frac{1}{2}\) × c × h = \(\frac{ch}{2}\) …………..(1)

area of ∆Yy’X = \(\frac{1}{2}\) × (Yy’) × (Xy’)

= \(\frac{1}{2}\) dh = \(\frac{dh}{2}\) …………….(2)

Formula for area of retangle

= length × breadth

∴ area of y’z’ZY = y’z’ x Zz’ = b × h ………….. (3)

∴ area of trapezium = (1) + (2) + (3)

= \(\frac{ch}{2}\) + \(\frac{dh}{2}\) +bh = 2(c + d) + bh

We can write it as = \(\frac{h}{2}\) (c + d) + \(\frac{2bh}{2}\)

⇒ \(\frac{h}{2}\) (c + d + b + d) ⇒ \(\frac{h}{2}\) (a + b)

(∵ b + c + d = a from figure)

Area of trapezium = \(\frac{h}{2}\) (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 separetely and add to find the area WXYZ.

Verify it by putting the values of h, a and b in the expression \(\frac{h(a + b)}{2}\).

Solution:

Given h = 10 cm; c = 6 cm, b = 12 cm, d = 4 cm

∴ area of ∆Wz’Z = \(\frac{1}{2}\) × c × h = \(\frac{1}{2}\) × 6 × 10 = 30 cm² ………….(1)

∴ area of ∆XYy’ = \(\frac{1}{2}\) × d × h = \(\frac{1}{2}\) × 4 × 10 = 20 cm² …………..(2)

area of z’y’YZ = b × h = 12 × 10= 120 cm² …………. (3)

Total area of Trapezium = (1) + (2) + (3) = 30 + 20 + 120 = 170 cm² ………. (4)

Verification from formula,

Putting these values in formula \(\frac{h}{2}\) (a + b)

where h = 10 ; .

a = b + c + d = 12 + 6 + 4 = 22; b = 12 cm

then area = \(\frac{10}{2}\) (22 + 12) = 5(34) = 170 cm² ……………(5)

(4) = (5) Hence verified.

Do This

Question 1.

Draw any trapezium WXYZ on a piece of graph paper as shown in the figure and cut it out (Fig).

Answer:

Question 2.

Find the mid point of XY by folding the side and name it A (Fig).

Answer:

Bisect each non-parallel side compare ZC and YD with CE and DF by folding away AB.

Hence, we get mid-point of XZ i.e. A.

Question 3.

Cut trapezium WXYZ into two pieces by cutting along ZA. Place AZYA as shown in Fig., where AY is placed on AX.

What is the length of the base of the larger triangle ? Write an expression for the area of this triangle.

Answer:

WX + ZY = WB is base of larger triangle.

Here XB = ZY

Area of this larger triangle

= \(\frac{1}{2}\) (WX + ZY)h.

Question 4.

The area of this triangle and the area of the trapezium WXYZ are same (How?). Get the expression for the area of trapezium by using the expression for the area of triangle.

Answer:

Yes, area of trapezium WXYZ = \(\frac{h}{2}\) (WX + ZY) and

area of larger ∆ = \(\frac{h}{2}\) (WX + ZY)

Thus area of trapezium

= (\(\frac{1}{2}\) × a × h) + (\(\frac{1}{2}\) × b × h)

Try These

Find the area of the following trapeziums (Fig).

i)

Solution:

Height of trapezium = h = 3 cm

Lengths of two parallel sides of trapezium = a= 7cm, p = 9 cm.

then formula for area of trapezium

= \(\frac{h}{2}\) (a + b).

= \(\frac{3}{2}\) (7 + 9) = \(\frac{3}{2}\) × 16 = 3 × 8 = 24 cm²

ii)

Height of trapezium = h = 6 cm Parallel sides

a, b of trapezium = 10 cm, 5 cm

∴ h = 6 cm, a = 10 cm, b = 5cm

then area of trapezium = \(\frac{h}{2}\) (a + b)

= \(\frac{6}{2}\) (10 + 5) = 3 × 15 = 45 cm²

Do This

In Class VII we learnt to draw parallelograms of equal areas with different perimeters. Can it be done for trapezium? Check if the following trapeziums are of equal areas but have different perimeters (Fig).

Answer:

1)

Area = \(\frac{h}{2}\)(a + b) = \(\frac{4}{2}\) (10 + 14) = 48 sq. units.

Perimeter = 5 + 10 + 4+14 = 33 units

2)

Area of this trapezium = \(\frac{h}{2}\)(a + b)

= \(\frac{8}{2}\) (8 + 4) = 48 sq. units and

Perimeter = 8 + 8 + 8 + 4 = 28 units.

3)

Area of this trapezium = \(\frac{h}{2}\) (a + b)

= \(\frac{6}{2}\) (6 + 10) = 48 sq. units

Perimeter = 6+10+ 7 + 6 = 29 units.

Think, Discuss And Write

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

Answer:

We can divide a parallelogram into two congruent triangles by drawing a diagonal.

Because the opposite sides are parallel and equal but in the case of trapezium, the opposite sides are not equal then AB ≠ CD.

But height is same for the both AABD and ABCD, then areas

\(\frac{1}{2}\) × AB × h = \(\frac{1}{2}\) × CD × h (∵ AB ≠ CD)

So they are not congruent.

So we cannot divide a trapezium into two congruent triangles.

Try These

Find the area of these quadrilaterals

(Fig).

i)

Solution:

In quadrilateral ABCD

diagonal AC = d = 6 cm

heights h_{1} = 3 cm, h_{2} = 5 cm

∴ area of quadrilateral ABCD = \(\frac{1}{2}\) d(h_{1} + h_{2})

= \(\frac{1}{2}\) × 6 × (3 + 5) = \(\frac{1}{2}\) × 6 × 8 = 24 cm²

ii)

Solution:

In the given quadrilateral all sides are equal (4 cm each) and two diagonals are different.

So it is Rhombus.

∴ Area of Rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

= \(\frac{1}{2}\) × 7 × 6 = 21 cm²

iii)

Solution:

In the given quadrilateral, opposite sides are equal (5 cm, 4 cm). So it is aparaller logram.

In a parallelogram a diagonal divides it into two congruent triangles.

So area of first triangle = \(\frac{1}{2}\) bh

= \(\frac{1}{2}\)× 8 × 2 = 8 cm²

then area of second triangle = 8 cm²

∴ area of quadrilateral = 8 + 8 = 16 cm²

Try These

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

FI is a diagonal of polygon EFGHI

NQ is a diagonal of polygon MNOPQR

Answer:

Dividing into parts

Think, Discuss And Write

Why is it incorrect to call the solid shown here a cylinder ?

Solution:

Yes, it is incorrect to call above as a cylinder.

By definition, cylinder is a solid geometrical figure with straight parallel sides and circular / oval cross section.

But in the given figure, there are no parallel sides. Hence we can’t call it cylinder.

Try These

Find the total surface area of the following cuboids (Fig):

Solution:

Here (l) = 6 cm, b = 4 cm, h = 2 cm

formula for total surface area of a cuboid = 2(lb + bh + lh) now putting values in it

We get T.S.A = 2[(6 × 4) + (4 × 2) + (2 × 6)]

= 2[24 + 8 + 12] = 2{44]

= 88 sq.cm.

In this cuboid

length = 4 cm ; b = 4 cm, h = 10 cm

∴ TSA = 2[(4 × 4) + (4 × 10) + (10 × 4)]

= 2[16 + 40 + 40] = 2[96]

= 196 cm

Do This

i) Cover the lateral surface of a cuboidal duster (which your teacher uses in the class room) using a strip of brown sheet of paper, such that it just fits around the surface. Remove the paper. Measure the area of the paper. Is it the lateral surface area of the duster ?

Answer:

After measuring the area of paper, we observe it is equal to its lateral surface area of the duster.

ii) Measure length, width and height of your classroom and find .

a) the total surface area of the room, ignoring the area of windows and doors.

b) the lateral surface area of this room.

c) the total area of the room which is to be white washed.

Answer:

Student Activity.

Think, Discuss And Write

Question 1.

Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base ?

Answer:

Yes, we can

∵ Lateral Surface Area = 2h(l + b) = 2lh + 2bh

2 × area of base = 2 lb

∴ Lateral Surface Area + 2 • area of base

2lh + 2bh + 2lb = 2(lb + bh + lh)

= total surface area.

Question 2.

If we interchange the lengths of the base and the height of a cuboid (Fig(i)) to get another cuboid (Fig (ii)), will its lateral surface area change

Answer:

No. If we change ‘l’ and ‘b’ vice-versa, then the lateral surface area wont change.

But, If we change

l → b, b → h and h → l then the lateral surface area changes, but volume won’t change.

Do This

Draw the pattern shown on a squared paper and cut it out [Fig (i)]. You know that this pattern is a net of a cube. Fold it along the lines [Fig (ii)] and tape the edges to form a cube [Fig (iii)].

a) What is the length, width and height of the cube? Observe that all the faces of a cube are square in shape. This makes length, height and width of a cube equal (Fig (i)).

Solution:

i) Length = 3 units

ii) Width = 3 units

iii) Height = 3 units

We observe that all the faces of cube are square in shape.

b) Write the area of each of the faces. Are they equal.

Solution:

Area of each face = l × l = l²

(∵ face is square in shape)

Yes. they are equal.

c) Write the total surface area of this cube.

Solution:

Total surface area = l² + l² + l² + l² + l² + l²

(∵ 6 faces)

= 6l²

d) If each side of the cube is Z, what will be the area of each face? (Fig (ii)). Can we say that the total surface area of a cube of side l is 6l²?

Solution:

Yes, we can say T.S.A = ‘6/2

Try These

Find the surface area of cube A and lateral surface area of cube B (Fig).

Solution:

i) l = 10, b = 10, h = 10

total surface area = 2(lb + bh + lh)

= 2(l.l + l.l + l.l) (Y l = b = h)

= 2(3l²) = 6l² = 6 × 10 × 10 = 600 cm²

ii) lateral surface area of cube ‘B’

l = b = h = 8

lateral surface area = 2h(l + b)

= 2l(l + l) = 2l . (2l) =4l² = 4 × 10 × 10 = 400 cm²

Think, Discuss And Write

Two cubes each with side b are joined to form a cuboid (Fig). What is the surface area of this cuboid ? Is it 12b^{2}? Is the surface area of cuboid formed by joining three such cubes, 18b^{2}? Why ?

Solution:

Two cubes each with side “b” are joined to form a cuboid.

then lengh (l) = 2b, breadth (b) = b; height (h) = b

then surface area = 2(lb + bh + lh)

= 2(2b.b + b.b + 2b.b)

= 2(2b^{2} + b^{2} + 2b^{2})

= 2(5b^{2}) = 10b^{2}

but not 12b^{2}.

• If three such cubes joined, then l = 3b, b = b, h = b

then surface area = 2(3b.b + b.b + 3b.b)

= 2(3b^{2} + b^{2} + 3b^{2})

= 2 (7b^{2}) = 14b^{2}

but not 18b^{2}

ii) How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area ?

Solution:

We can join in a row to form a cuboid of smallest surface i.e 12 × 1 × 1.

(iii) After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions (Fig).

How many have no face painted ? 1 face painted ? 2 faces painted ? 3 faces painted ?

Solution:

The cube is cut into 64 smaller cubes. So the first cube would be 4 × 4 × 4.

How many have no face painted ? = 8

Formula = (n – 2)^{3} = (4 – 2)^{3} = 2^{3} = 8

How many cubes have 1 face painted = 6(n – 2)^{2} = 6(4 – 2)^{2} = 6 × 4 = 24

How many cubes have 2 face painted = 12(n – 2) = 12(4 – 2) = 12 × 2 = 24

How many cubes have 3 face painted = 64 – (8 + 24 + 24) = 8

• Cylinders :

Formula for curved (lateral) surface area of a cylinder (CSA) = 2πrh.

(∵ h – height of cylinder; r – radius of cylinder)

Total Surface area of cylinder = C.S.A + area of top + area of bottom

= 2πrh + πr^{2} + πr^{2} = 2 πrh + 2π²

= 2πr(r + h)