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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

Question 1.
Find the area of each of the following triangles.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 1
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.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 2
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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

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.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 3
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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

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?
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 4
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?
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 5
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?
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 6
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.

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

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.

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

Question 10.
In Figure ABCD find the area of the shaded region.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 7
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.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 8
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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

Question 12.
Find the area of a parallelogram PQRS, if PR =24 cm and QU = ST =8 cm.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 9
Solution:
Given : In PQRS, PR = 24cm, ST = 8cm, QU =8cm
Area of AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 10PQRS = 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

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3

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