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

Question 1.

In ΔABC, ∠ABC = 90°; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of ΔADB

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 cm^{2}

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]

Solution:

Area of ΔQPS = \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x 9 x 12

= 54cm^{2}

In ΔQPS

QS^{2} = PQ^{2} + PS^{2}

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 cm^{2}

∴ □PQRS = ΔQPS + ΔQSR

= 54 + 60= 114 cm^{2}

Question 3.

Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle.

[Hint: ABCD has two parts]

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 = 36cm^{2}

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]

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)