Category: Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 1

Question 1.
Given below are some fiugres. Which of them are symmetric? Draw the axes of symmetry
for the symmetric figures.

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Solution:
A ball -Circle
A cylindrical pipe – Rectangle
A book – Rectangle

Question 2.
Here are the shadows of some 3D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be many answers for these!)

Solution:
i) A Circle – Spherical/circular objects
ii) A Square – Cube/square sheets
iii) A Triangle – Triangular/right prism with triangular base
iv) A Rectangle – Cuboid/rectangular shapes.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 3

Question 1.
Use an isometric dot paper and make an isometric sketch for each one of the given shapes.

Solution:

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Draw an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each of the given isometric shapes.

Solution:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique’?)
(b) A cube with an edge 4 cm long.
Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 2

Question 1.
Some nets are given below. Trace them and paste them on a thick paper. Try to make 3-D
shapes by suitably folding them and gluing together. Match the net with it’s 3-D shape.
Nets
3D shapes
(i)

Solution:

Question 2.
Three nets for each shape are given here. Match the net with its 3-D shape.
Solution:
Solution:


Solution:

Question 3.
A dice is a cube with dots on each face. The opposite faces of a dice always have a total of seven dots on them.
Here are two nets to make dice. Insert the suitable number of dots in blanks.

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 14th Lesson Understanding 3D and 2D Shapes Exercise 1

Question 1.
Given below are the pictures of some objects. Categorise and fill write their names according to their shape and fill the table with name of it.

Solution:

Question 2.
Write names of at least 2 objects from day-to-day life, which are in the shape of the basic 3D shapes given below:
(i) Cone – ……………
(ii) Cube – ……………
(iii) Cuboid – ……………
(iv) Sphere – ……………
(v) Cylinder – ……………
Solution:
i) Cone : Ice-cream cone, birthday caps
ii) Cube : Die, chalk piece box
iii) Cuboid : Eraser, brick
iv) Sphere : Marbles, balls
v) Cylinder: Pencil, candle, pipe

Question 3.
Identify and state the number of faces, edges and vertices of the figures given below:

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 6

Question 1.
A path 2.5 m wide is running around a square field whose side is 45 m. Determine the area of the path.
Solution:

Area of the path= (Area of outer figure) – (Area of inner figure)
= (50 × 50) – (45 × 45)
= 2500 – 2025 = 475 m²

Question 2.
The central hail ofa school is 18m long and 12.5 m wide. A carpet is to be laid on the floor leaving a strip 50 cm wide near the walls, uncovered. Find the area of the carpet and also the uncovered portion?
Solution:

Given
Length of the hail = 18m
Breadth of the hail 12.5m
Width of the strip = 50cm = \(\frac { 1 }{ 2 }\)m
Length of inner rectangle = 18 – (\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)) 17m
Breadth of inner rectangle = 12.5 – (\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)) = 11.5 m
Area of the strip = (Area of the outer fIgure) – (Area of the inner figure)
= 18 × 12.5 – 17 ×11.5
= 225 – 195.5 = 29.5 m²
∴ Area of the carpet = 195.5 m²

Question 3.
The length of the side of a grassy square plot is 80 m. Two walking paths each 4m wide
are constructed parallel to the sides of the plot such that they cut each other at the centre
of the plot. Determine the area of the paths.
Solution:

Given : A square grassary plot of length 80m
Two paths of width 4m each.
From the question
KL = 4m and KN = 80m
EH = 4m and EF 80m
PQ = 4m and PS 4m
Area of two paths = 🖾 KLMN + 🖾 EFGH – 🖾 FQRS
= KN x KL + EF × EH – PQ × PS
= 80 × 4 + 80 × 4 – 4 × 4
= 320 + 320 – 16
= 624sq.m.

Question 4.
A verandah 2 m wide is constructed all around a room of dimensions 8 m X 5 m. Find the area of the verandah
Solution:

Length of outer rectangle = 8 + 2 + 2 =12m
Breadth of outer rectangle = 5 + 2 + 2 = 9m
Area of outer rectangle = l x b = 12 × 9 = 108m²
Length of inner rectangle = 8m
Breadth of inner rectangle = 5m
Area of inner rectangle = l x b = 8 × 5 = 40m²
:. Area of the verandah = (Area of outer figure) – (Area of inner figure)
= 108 – 40 = 68m²

Question 5.
The length of a rectangular park is 700 m and its breadth is 300 m. Two crossroads, each of width 10 m, cut the centre of a rectangular park and are parallel to its sides. Find the area of the roads. Also, find the area of the park excluding the area of the crossroads.
Solution:

Given
Length of the rectangular park = 700 m
Breadth of the rectangular park = 300 ni
From the question,
PS 10m. PQ = 700 ni
Area of 🖾 PQRS = 10 × 700 = 7000m²
AB = 10m, AD = 300m
Area of 🖾 ABCD = 300 x 10 3000m²
KL = 10m, KN = 10m
🖾 KLMN = 10 × 10 = 100m²
Area of the two paths = 🖾 IQRS + 🖾 AF3CD – 🖾 KLMN
= 7000 + 3000 – 100
= 9900m²
The remaining area of the park = Area of the whole park – Area of the two paths
= 700 × 300 – 9900
= 210000 – 9900
= 200100 q.m.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 5

Question 1
Find the circumference of a circle whose radius is
(i) 35cm (ii) 4.2cm (iii) 15.4 cm
Solution:
i) Given: r= 35 cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 35 = 220 cms.

ii) Given : r = 4.2 cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 4.2 26.4 cms

iii) Given: r = 15.4cm
Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 15.4 = 96.8 cms

Question 2.
Find the circumference of circle whose diameter is
(i) 17.5 cm (ii) 5.6 cm (iii) 4.9 cm
Note : take π = \(\frac{22}{7}\) in the above two questions.
Solution:
i) Given : d = 17.5 cm
Circumference of the circle = πd = \(\frac{22}{7}\) × 17.5 = 55 cm
ii) Given d = 5.6 cm
Circumference of the circle = πd = \(\frac{22}{7}\) × 5.6 = 17.6 cm
iii) Given d = 4.9 cm
Circumstances of the circle = πd = \(\frac{22}{7}\) × 4.9 = 15.4 cm

Question 3.
(i) Taking π = 3.14, find the circumference of a circle whose radius is
(a) 8 cm (b) 15 cm (c) 20cm
(ii) Calculate the radius of a circle whose circumference is 44cm?
Solution:
i) a) Given r = 8 cm π = 3.14
Circumference of circle = 2πr
= 2 × 3.14 × 8 = 50.24cm

b) Given r = 15cm
Circumference of circle = 2πr
= 2 × 3.14 × 15 = 942cm

c) Given r 20cm
Circumference of circle = 2πr
= 2 × 3.14 × 20 = 125.6cm

ii) Given,
Circumference = 44cm
Circumference of a circle = 2πr
2 × \(\frac{22}{7}\) x r =
2 × 22 × r = 44 x 7
r = \(\frac{44 \times 7}{2 \times 22}\) = 7cm
∴ radius of the circle = 7cm.

Question 4.
If the circumference of a circle is 264 cm, find its radius. Take π = \(\frac{22}{7}\)
Solution:
Given : Circumference = 264 cm
Circumference of a circle = 2πr = 2 × \(\frac{22}{7}\) x r = 264
2 × 22 × r = 264 × 7
r = \(\frac{264 \times 7}{2 \times 22}\) = 42
∴ radius of the circle = 42 cm.

Question 5.
If the circumference of a circle is 33 cm, fmd its diameter.m
Solution:
Given circumference = 33cm
Circumference of a circle = πd
\(\frac{22}{7}\) × d = 33
22d = 33 × 7
d = \(\frac{33 \times 7}{22}=\frac{21}{2}\) = 10.5cm
∴ diameter of the circle = 10.5 cm

Question 6.
How many times will a wheel of radius 35cm be rotated to travel 660 m?
(Take π = \(\frac{22}{7}\)).
Solution:
Given radius = 35cm.
Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × 35
= 220cm.
∴ The number of times the wheel will he rotated = \(\frac{660}{220}\) = 3 times.

Question 7.
The ratio of the diameters of two circles is 3 :4. Find the ratio of their circumferences.
Solution:
Given: Ratio of diameter = d₁ : d₂ = 3 : 4
Ratio of their circumferences C₁ : C₂ = πd₁ : πd₁
= \(\frac{22}{7}\) × 3: \(\frac{22}{7}\) : 4 = 3 : 4

Question 8.
A road roller makes 200 rotations in covering 2200 m.
Find the radius of the roller.
Solution:
Given : Road roller makes 200 rotations covering a distance = 2200m
(i.e.,) 200 × circumference = 2200
200 × 2 × π × r = 2200
200 × 2 \(\frac{22}{7}\) × r = 2200
r = 2200 × \(\frac{7}{22} \times \frac{1}{200} \times \frac{1}{2}=\frac{7}{4}\) = 1.75m

Question 9.
The minute hand of a circular clock is 15 cm.
How far does the tip of the minute hand move in 1 hour?
(Take π = 3.14)
Solution:
Given : The length of minute hand = 15cm
In one hour it completes a rotation.
i.e. distance covered by its tip = circumference of the circle with a radius 15cm
22 660
= 2 × \(\frac{22}{7}\) × 15 = \(\frac{660}{7}\) cm = 94.28cm

Question 10.
A wire is bent in the form of a circle with radius 25 cm. It is straightened and made ¡rito a square. What is the length of the side of the square?
Solution:
Given: Radius of the circle 25 cm .
Circumference of the circle = Perimeter of the square
2πr = 4 × side
4 × side=2 × \(\frac{22}{7}\) × 25
Side = 2 × \(\frac{22}{7}\) × 25 × \(\frac{1}{4}\) = \(\frac{275}{7}\) = 39.28
∴ Side of square = 39.28 cm

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 4

Question 1.
Find the area of the following rhombuses.

Solution:
Area = \(\frac { 1 }{ 2 }\)d₁d₂
d₁ = 5 + 5 = 10cm
d₂ = 2 + 2 = 4cm
A = \(\frac { 1 }{ 2 }\) × 10 × 4 = 20 cm²

d₁ = 3 + 3 = 6 cm
d₂ = 4 + 4 = 8cm
Area = \(\frac { 1 }{ 2 }\)d₁d₂
= \(\frac { 1 }{ 2 }\) × 6 × 8
= 24 cm²

Question 2.
Find the missing values.

Diagonal – 1 (d1)	Diagonal – 2 (d2)	Area of the rhombus
12cm	16cm
27mm		2025mm2
14m	57.6m

Solution:

Diagonal – 1 (d1)	Diagonal – 2 (d2)	Area of the rhombus
12cm	16cm	\(\frac{1}{2}\) × 12 × 16 = 96cm2
27mm	\(\frac{2025 \times 2}{27}\) = 150mm	2025mm2
14m	57.6m	\(\frac{1}{2}\) × 24 × 57.6 = 691.2 m2

Question 3.
If length of diagonal of a rhombus whose area 216 sq. cm. is 24 cm. when find the length of second diagonal.
Solution:
Given: Length of one diagonal d₁ = 24cm, d₂ =?
Area = \(\frac { 1 }{ 2 }\)d₁d₂ = 216
\(\frac { 1 }{ 2 }\) × d₁ × d₂ = 216
\(\frac { 1 }{ 2 }\) × 24 × d₂ = 216
d₂ = \(\frac{216}{12}\) = 18

Question 4.
The floor of a building consists of3 000 tiles which are rhombus shaped. The diagonals of each of the tiles are 45 cm and 30 cm. Find the total cost of polishing the floor, if cost per m² is Rs.2.50.
Solution:
Diagonals of each (shape / rhombus) tiles
d₁ = 45cm, d₂ = 30cm
Total tiles = 3000
Total area = 3000 × Area of each tile = 3000 × \(\frac { 1 }{ 2 }\) × d₁ × d₂
= 3000 x \(\frac { 1 }{ 2 }\) × 45 × 30 = 2025000 cm²
= \(\frac{2025000}{100 \times 100}\) m² = 202.5m²
∴ Cost of polishing the floor at the rate of Rs. 2.50 per a square metre 202.5 × 2.50 = Rs. 506.25

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 3

Question 1.
Find the area of each of the following triangles.

Solution:
î) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 5 x 8 = 20 cm²
ii) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 6 x 4 = 12 cm²
iii) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 5.4 x 7.5 = 20.25 cm²
iv)Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 6 x 4 = 12 cm²

Question 2.
In ΔPQR, PQ = 4 cm, PR =8 cm and RT = 6cm. Find (i) the area of ΔPQR (ii) the length of QS.

Solution:
Given PQ 4cm, PR = 8cm, RT = 6cm
1) Area of ΔPQR = \(\frac { 1 }{ 2 }\) × base × height = \(\frac { 1 }{ 2 }\) × PQ × RT
= \(\frac { 1 }{ 2 }\) × 4 × 6 = 12cm²

ii) Also area of ΔPQR = \(\frac { 1 }{ 2 }\) × base x height = \(\frac { 1 }{ 2 }\) × PR × QS
12 = \(\frac { 1 }{ 2 }\) × 8 × QS
QR = \(\frac{12 \times 2}{8}\) = 3cm

Question 3.
ΔABC is right-angled at A. AD is perpendicular to BC, AB =5 cm, BC = 13 cm and AC = 12 cm. Find the area of ΔABC. Also, find the length of AD.

Solution:
Given.
In ΔABC, ∠A = 90°
AB = 5cm; AC = 12cm; AD ⊥ BC; BC = 13cm.
Now area of ΔABC = \(\frac { 1 }{ 2 }\) × base × height
= \(\frac { 1 }{ 2 }\) x AB x AC = \(\frac { 1 }{ 2 }\) × 5 × 12 = 30cm²
Also area of MBC = \(\frac { 1 }{ 2 }\) × BC × AD
30 = \(\frac { 1 }{ 2 }\) × 13 × AD
∴ \(\frac{30 \times 2}{13}=\frac{60}{13}=4 \frac{8}{13}\) cm

Question 4.
ΔPQR is isosceles with PQ = PR = 7.5 cm and QR = 9 cm. The height PS from P to QR, is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?

Solution:
Given: PQ = PR = 7.5cm, QR = 9cm PS = 6cm
Area of ΔPQR = \(\frac { 1 }{ 2 }\) bh = latex]\frac { 1 }{ 2 }[/latex] x 9 x 6 = 27cm²
Also area of ΔPQR = \(\frac { 1 }{ 2 }\) × PQ × RT
27 = \(\frac { 1 }{ 2 }\) × 7.5 × RT [PQ = PR]
∴ RT = \(\frac{2 \times 27}{7.5}=\frac{2 \times 27 \times 2}{15}=\frac{36}{5}\) = 7.2 cm

Question 5.
ABCD rectangle with AB =8 cm, BC = 16 cm and AE = 4 cm. Find the area of ABCE. Is the area of ΔBEC equal to the sum of the area of ΔBAE and ΔCDE. Why?

Solution:
Given : In rectangle ABCD,
AB = 8cm, BC = 16 cm, AE = 4cm
Now Area of ΔBAE = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) × \(\overline{\mathrm{BC}} \times \overline{\mathrm{AB}}\)
= \(\frac { 1 }{ 2 }\) × 16 x 8 = 64cm²

Area of ΔCDE = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) × \(\overline{\mathrm{ED}} \times \overline{\mathrm{DC}}\)
= \(\frac { 1 }{ 2 }\) × 12 × 8 = 48cm²
(∵ \(\overline{\mathrm{ED}}\) = AD – AE = 16 – 4 = 12cm and \(\overline{\mathrm{DC}}=\overline{\mathrm{AB}}\))
Now ΔBAE + ΔCDE = 16 + 48 = 64 = ΔBCE.

Question 6.
Ramu says that the area of ΔPQR is, A = \(\frac{1}{2}\) × 7 × 5 cm².
Gopi says that it is, A = \(\frac{1}{2}\) × 8 × 5 cm². Who is correct’? Why?

Solution:
Ramu is correct.
Since area of a triangle = \(\frac { 1 }{ 2 }\) × base × corresponding height
= \(\frac { 1 }{ 2 }\) × 7 × 5cm²
But height 5cm is not corresponding to side &m.
∴ Gopi is not correct.

Question 7.
Find the base of a triangle whose area is 220 cm² and height is 11cm.
Solution:
Given Area = 220 cm², h = 11 cm
Area of a triangle = \(\frac { 1 }{ 2 }\)b.h
= \(\frac { 1 }{ 2 }\) × b × 11 = 220cm² (given)
b × 11 = 220 × 2
b = \(\frac{220 \times 2}{11}\) = 40
∴ base of the triangle = 40 cm.

Question 8.
In a triangle the height is double the base and the area is 400 cm². Find the length of the base and height.
Solution:
Area = 400 cm²
Let the base of the triangle be x.
height = 2x
Area of the triangle = \(\frac { 1 }{ 2 }\)b.h
\(\frac { 1 }{ 2 }\) × x × 2x = 400 (given)
x² = 400
x = \(\sqrt{400}\) = 20
∴ base of the triangle = 20 cm
height = 2(base)= 2(20) = 40 cm.

Question 9.
The area of triangle is equal to the area of a rectangle whose length and breadth are 20 cm and 15 cm respectively. Calculate the height of the triangle if its base measures 30 cm.
Solution:
Given :Length of the rectangle = 20 cm
Breadth of the rectangle = 15 cm
Base of the triangle = 30 cm
Area of the rectangle = Area of the triangle
Length x breadth = \(\frac { 1 }{ 2 }\) × base × height
20 × 15 = \(\frac { 1 }{ 2 }\) × 30 × height
∴ height = \(\frac{20 \times 15}{15}\) = 20cm

Question 10.
In Figure ABCD find the area of the shaded region.

Solution:
Given : Side of the square ABCD = 40cm
Area of the shaded rgion = (Area of the square) — (Area of the unshaded region)
= (side × side) – \(\frac { 1 }{ 2 }\) × base × height
= 40 × 40 – \(\frac { 1 }{ 2 }\) × 40 × 40
= 1600 – 800 = 800cm²

Question 11.
In Figure ABCD, find the area of the shaded region.

Solution:
Given: Side of the square = 20 cm
Area of the shade region
= (Area of the square) – (Area of unshaded region)
= (Square ABCD) – (ΔEAF + ΔFBC + ΔEDC)
= side × side – (\(\frac { 1 }{ 2 }\)AF × AE + \(\frac { 1 }{ 2 }\) FB × BC + \(\frac { 1 }{ 2 }\) DE × DC)
= 20 × 20 – (\(\frac { 1 }{ 2 }\) × 10 × 12 + \(\frac { 1 }{ 2 }\) × 10 × 20 + \(\frac { 1 }{ 2 }\) × 8 × 20)
= 400 – [60 + 100 + 80] = 400 – 240 = 160cm²

Question 12.
Find the area of a parallelogram PQRS, if PR =24 cm and QU = ST =8 cm.

Solution:
Given : In PQRS, PR = 24cm, ST = 8cm, QU =8cm
Area of PQRS = ΔPQR + ΔPRS
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\)base x height + x base x height
= \(\frac { 1 }{ 2 }\) × 24 x 8 + \(\frac { 1 }{ 2 }\) × 24 × 8 = 96 + 96 = 192 cm²

Question 13.
The base and height of the triangle are in the ratio 3:2 and its area is 108 cm². Find its base
and height.
Solution:
Given : Area = 108 cm2
Let the base of the triangle be 3x and height of the triangle be 2x.
Area of the triangle = \(\frac { 1 }{ 2 }\) × base × height = \(\frac { 1 }{ 2 }\) x (3x) (2x) = 108 cm² (given)
3x² = 108
x² = 108/3 = 36
x = \(\sqrt{36}\) = 6
∴ base 3x = 3(6) = 18 cm
height = 2x = 2(6) = 12cm

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 2

Question 1.
Find the area of the each of the following parallelograms.

Solution:

Fig	Base	Height	A = Base x Height
(i)	7	4	7 x 4 = 28cm2
(ii)	5	3	5 x 3  = 15 cm2
(iii)	5.1	7.6	5.1 x 7.6 = 38.76 cm2
(iv)	4	6	4 x 6 = 24cm2

Question 2.
PQRS is a parallelogram. PM is the height from P to \(\overline{\mathrm{SR}}\) and PN is the height from P to \(\overline{\mathrm{QR}}\). If SR = 12cm and PM = 7.6 cm.
(i) Find the area of the parallelogram PQRS
(ii) Find PN, if QR = 8 cm.
Solution:

i) Given:SR = 12cm
PM = 7.6cm
∴ Area = base × height
= 12 × 7.6 = 91.2 cm²

ii) Also area = base × height
91.2 = QR × PN
91.2 = 8 × PN
∴ PN = \(\frac{91.2}{8}\) = 11.4cm

Question 3.
DF and BE are the height on sides AB and AD respectively in parallelogram ABCD. Ifthe area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BE and DF.

Solution:
Given : AB = 35 cm, AD = 49 cm
area of 🖾ABCD = 1470 cm²
Area of a parallelogram = base × height
AB × DF = area of 🖾ABCD
35 × DF = 1470
DF = \(\frac{1470}{35}\) = 42cm
Also AD × BE = 1470
49 × BE = 1470
∴ BE= \(\frac{1470}{49}\) = 30cm

Question 4.
The height of a parallelogram is one third of its base. If the area of the parallelogram is 192cm², find its height and base.
Solution:
Area of the parallelogram = 192 cm²
Area of parallelogram = base × height = 192 cm² (given)
Let the base of the parallelogram be x.
height = \(\frac{x}{3}\)
Area = base × height = x x \(\frac{x}{3}=\frac{x^{2}}{3}\) = 192
⇒ x² = 192 × 3 = 576
x = \(\sqrt{576}\) = 24
∴ base = 24 cm
height = \(\frac{x}{3}=\frac{24}{3}\) = 8cm.

Question 5.
In a parallelogram the base and height are is in the ratio of 5:2. If the area of the parallelogram is 360m², find its base and height.
Solution:
Let the base of the parallelogram be 5x and height of the parallelogram be 2x.
Area = height x base = 5x × 2x = 10x² = 360 cm² (given)
10x² = 360
x² = \(\frac{360}{10}\) = 36
x = √36 = 6
base = 5x = 5(6) = 30 cm
height = 2x = 2(6) = 12 cm

Question 6.
A square and a parallelogram have the same area. If a side of the square is 40m and the height of the parallelogram is 20m, find the base of the parallelogram.
Solution:
Given: Side of a square = 40m
Height of a parallelogram = 20m
Area of parallelogram = Area of the square
base × height side × side
base × 20 40 × 40
∴ base = \(\frac{40 \times 40}{20}\) = 80 m.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Area and Perimeter Exercise 1

Question 1.

Solution:

Question 2.
The measurements of some squares are given in the table below. However, they are incomplete. Find the missing information.

Solution:
i) Perimeter = 4a = 4(15) = 60cm
ii) Perimeter = 4a = 88
a = \(\frac{88}{4}\)= 22
Side = 22cm
Area = a² = (22)² = 484 cm²

Question 3.
The measurements of some rectangles are given in the table below. However, they are incomplete. Find the missing information.

Solution:
Given:
i) Length = 20 cm
breadth = 14cm
Area = l x b = 20 x 14 =280 cm²
Perimeter = 2(l + b) = 2(20 + 14)
= 2(34) = 68cm

ii) Given breadth = 12cm, perimeter = 60cm
Perimeter 2(l + b) 60 cm
= 2(l + 12) =60cm
2l + 24 =60cm
2l = 60 – 24 = 36
2l = \(\frac{36}{2}\) = 18cm
Area = l x b = 18 x 12 = 216cm²

iii) Given length = 15cm, Area = 150 cm
Area = l x b = 15 x b = 150
⇒ b = \(\frac{150}{15}\) = 10cm
Perimeter = 2(l + b) = 2(15 + 10)
= 2(25) =50cm

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Quadrilaterals Exercise 2

Question 1.
State whether true or false
(i) All rectangles are squares ( )
(ii) All rhombuses are parallelogram ( )
(iii) All squares arc rhombuses and also rectangles ( )
(iv) All squares are not parallelograms ( )
(v) All kites arc rhombuses ( )
(vi) All rhombuses are kites ( )
(vii) All parallelograms are trapeziums ( )
(viii) All squares are trapeziums ( )
Solution:
(i) All rectangles are squares ( False)
(ii) All rhombuses are parallelogram ( True)
(iii) All squares arc rhombuses and also rectangles ( True)
(iv) All squares are not parallelograms (False )
(v) All kites arc rhombuses ( False)
(vi) All rhombuses are kites (True )
(vii) All parallelograms are trapeziums ( True)
(viii) All squares are trapeziums ( True)

Question 2.
Explain how a square is a
(i) quadrilateral
(ii) parallelogram
(iii) rhombus
(iv) rectangle.
Solution:
i) quadrilateral : A square is a closed figure bounded by four line segments and hence it is a
quadrilateral.
ii) Parallelogram : In a square both pairs of opposite sides are parallel and hence it is a parallelogram.
iii) Rhombus : All four sides of a square are equal, thus it is a Rhombus.
iv) Rectangle : In a square each angle ¡s a right angle and hence it is a rectangle.

Question 3.
In a rhombus ABCD, ∠CBA = 40°.
Find the other angles.

Given that ∠CBA , ∠B = 40°
We know that rhombus is a parallelogram and
adjacent angles are supplementary.
Thus ∠A + ∠B = 180°
∠A + 40° = 180°
∠A = 180° – 40° = 140°
Also opposite angles are equal.
∴ ∠A =∠C = 140°
∠D = ∠B = 40°

Question 4.
The adjacent angels of a parallelogram arex° and (2x + 30)°.
Find all the angles of the parallelogram.
Solution:

Given that the adjacent angles of a parallelogram are x° ; (2x + 30)°
Adjacent angles of a parallelogram are supplementary.
∴ x + (2x + 30) = 180°
3x + 30 = 180°
3x = 180° – 30° = 150°
x = \(\frac{150^{\circ}}{3}\) = 50°
∴ The adjacent angles are x = 50°
2x + 30°= 2 x 50 + 30 = 130°
∴ The other two angles are 50°, 130°
∴ The four angles are 50°, 130°, 50°, 130°

Question 5.
Explain how DEAR is a trapezium. Which of its two sides are parallel?
Solution:

In quadrilateral DEAR
∠D = 80° and ∠E = 100°
Also ∠D + ∠E = 80°+ 100° = 180°
(∵ D, E are interior angles on the same side of \(\overline{\mathrm{DE}}\))
∴ DR //EA
Also ∠A = ∠R = 90° and ∠A + ∠R = 180°
∴ DEAR is a trapezium.

Question 6.
BASE is a rectangle. Its diagonals intersect atO. Findx, if OB = 5x+1 and OE = 2x + 4.
Solution:

Given BASE is a rectangle.
OB = 5x + 1 and OA = 2x + 4
In a rectangle diagonals are equal.
BS = EA
(i.e.) 2 x BO = 2 x OA
2(5x + 1) = 2(2x + 4 )
10x + 2 = 4x + 8
10x – 4x = 8 – 2
6x = 6
x = 1

Question 7.
Is quadrilateral ABCD a parallelogram, if ∠A = 70° and ∠C = 65° ? Give reason.
Solution:
Given that in quadrilateral ABCD, ∠A = 70° and ∠C = 65°
∠A and ∠C are opposite angles and are not equal.
Hence quadrilateral ABCD is not a parallelogram.

Question 8.
Two adjacent sides of a parallelogram are in the ratio 5:3 the perimeter of the parallelogram is 48cm. Find the length of each of its sides.
Solution:
Given that the ratio of two adjacent sides of a parallelogram = 5 : 3
Sum of the terms of the ratio = 5 + 3 = 8
Sum of two sides = half of the perimeter = \(\frac { 1 }{ 2 }\) x 48 = 24 cm
∴ one side = \(\frac { 5 }{ 8 }\) x 24 = 15 cm
other side = \(\frac { 3 }{ 8 }\) x 24 = 9 cm
∴ The four sides are 15 cm, 9 cm, 15 cm, 9 cm ( opposite sides are equal)

Question 9.
The diagonals of the quadrilateral are perpendicular to each other. Is such a quadrilateral
always a rhombus? Draw a rough figure to justify your answer.
Solution:

Given : Diagonals are perpendicular.
If the diagonals are perpendicular then the
quadrilateral need not necessarily be a rhombus.
Look at the figure,
the diagonals are perpendicular to each other.
But all the sides are not equal.
PR ⊥ QS but PQ ≠ QR ≠ RS ≠ SP

Question 10.
ABCD is a trapezium in which \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) If ∠A = ∠B =30°, what are the measures of the other two angles?
Solution:

Given : ABCD is a trapezium.
∠A = ∠B = 30°
Since AB//CD,
∠A + ∠D = ∠B + ∠C = 180° (interior angles on the same side of a
transversal are supplementary)
30° + ∠D = 180°
∠D = 180° – 300°
= 150°
and 30° + ∠C = 180°
∠C = 180° – 30’
= 150°

Question 11.
Fill in the blanks.
(i) A parallelogram in which two adjacent sides are equal is a ………………….
(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a ………………….
(iii) IntrapeziurnABCD,\(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) . If ∠D = x° then ∠A = ………………
(iv) Every diagonal in a parallelogram divides it into …………………. triangles.
(v) In parallelogram ABCD, its diagonals \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BD}}\) intersect atO. IfAO = 5cm then AC = …………….. cm.
(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = …………. degrees.
(vii) ABCD is a parallelogram then ∠A — ∠C = ………………… degrees.
(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = …………… cm.
(ix) In a square ABCD, the diagonal \(\overline{\mathrm{AC}}\) is drawn. Then ∠BAC = …………
Solution:
(i) A parallelogram in which two adjacent sides are equal is a rhombus
(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a square
(iii) IntrapeziurnABCD,\(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) . If ∠D = x° then ∠A = 180° – x
(iv) Every diagonal in a parallelogram divides it into two congruent triangles.
(v) In parallelogram ABCD, its diagonals \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BD}}\) intersect atO. If AO = 5cm then AC = 10 cm.
(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = 90°
(vii) ABCD is a parallelogram then ∠A — ∠C = zero degrees.
(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = 10 cm.
(ix) In a square ABCD, the diagonal \(\overline{\mathrm{AC}}\) is drawn. Then ∠BAC = 45°