AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.1

AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1

Question 1.
In ΔABC, ∠ABC = 90°; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of ΔADB
AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 1
Solution:
ΔADB = \(\frac { 1 }{ 2 }\) ΔABC [ ∵ AD is a median of ΔABC]
\(\frac { 1 }{ 2 }\) = [ \(\frac { 1 }{ 2 }\) AB x BC]
= \(\frac { 1 }{ 4 }\) x 12 x 6.5
= 19.5 cm2

AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1

Question 2.
Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR =17 cm.
[Hint: PQRS has two parts]
AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 2
Solution:
Area of ΔQPS = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 9 x 12
= 54cm2
In ΔQPS
QS2 = PQ2 + PS2
QS = \(\begin{aligned}
\sqrt{12^{2}+9^{2}} &=\sqrt{144+81} \\
&=\sqrt{225}=15
\end{aligned}\)
Area of ΔQSR =\(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 15 x 8 = 60 cm2
∴ □PQRS = ΔQPS + ΔQSR
= 54 + 60= 114 cm2

AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1

Question 3.
Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle.
[Hint: ABCD has two parts]
AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 3
Solution:
Area of trapezium 1
= \(\frac { 1 }{ 2 }\) (sum of parallel sides) x (distance between the parallel sides)
= \(\frac { 1 }{ 2 }\) (a + b) h
From the figure, a = 3 + 3 = 6 cm
b = 3 cm
(∵ Opp. sides of rectangle)
h = 8 cm
∴ A = \(\frac { 1 }{ 2 }\)(6 + 3)x8 = 36cm2

AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1

Question 4.
ABCD is a parallelogram. The diago-nals AC and BD intersect each other at O. Prove that ar (ΔAOD) = ar (ΔBOQ. [Hint: Congruent figures have equal area]
AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 4
Solution:
Given that □ABCD is a parallelogram.
Diagonals AC and BD meet at ‘O’.
In ΔAOD and ΔBOC
AD = BC [ ∵ Opp. sides of a ||gm]
AO = OC [ ∵ diagonals bisect each
OD = OB other]
ΔAOD = ΔBOC [S.S.S. congruence]
∴ ΔAOD = ΔBOC (i.e., have equal area)