**Telangana**** SCERT Solution Class IX (9) Math Chapter 11 Areas Exercise 11.1**

**(1) In ΔABC, ****∠****ABC = 90 ^{o} , AD = DC, AB = 12cm and BC = 6.5 cm. Find the area of ΔADB.**

**Solution:** In △ABC,

Given,

∠ABC = 90

AB = 12cm

BC = 6.5 cm

AD = DC

∴ BD is a median of △ABC

∴ BD divides △ABC into two equal triangles △ADB & △BDC

∴ Area of △ABC = ½ x b x h

= ½ x AD x BC

= ½ x 12 x 6.5

= 39 cm^{2}

∴ Area of △ADB = ½ of △ABC

= ½ x 39

= 19.5 cm^{2}

** (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:** From the adjacent figure,

In △PQS,

**SQ is the hypotenuse**

∴ SQ^{2} = PQ^{2} + PS^{2}

Or, SQ = √12^{2} + 9^{2}

= √(144+81)

= √225

= 15cm

∴ Area of △PQS = ½ x b x h

= ½ x PQ x PS

= ½ x 12 x 9

= 54 cm^{2}

In △QPS

Height = SQ = 15 cm

∴ Area of △QRS = ½ x b x h

= ½ x QR x SQ

= ½ x 8 x 15

= 60 cm^{2}

∴ Total area of equal PQRS = area of △PQS + area of △QPS

= 54 + 60

= 114 cm^{2}

**(3) Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle. (Hint: ABCD has two parts)**

Given, ADCE is a rectangle,

∴ EC = AD & EC ∥ AD

Or, EC = 8 cm; BE = 3 cm, AE = 3 cm

∠BEC = ∠EAD = 90^{o} [corresponding angle]

∴ △BEC is right angle triangle at ∠BEC = 90^{o}

∴ Area of △BEC = ½ x b x h

= ½ x EC x BE

= ½ x 8 x 3

= 12 cm^{2}

∴ Area of rectangle ADCE = l x b

= AD x AE

= 8 x 3

= 24 cm

∴ Area of trapezium ABCD = area of △BEC + area of rectangle ADCE

= 12 + 24

= 36 cm^{2}

** (4) ABCD is a parallelogram. The diagonals AC and BD intersect each other at ‘O’. Prove that ar(ΔAOD) = ar(ΔBOC). (Hint: Congruent figures have equal area)**

**Solution:** Given,

ABCD is a parallelogram

AC and BD are diagonals that intersect each other at point O.

∴ In △AOD & △BOC

(i) DO = BO [∵ diagonals of a parallelogram bisect each other]

(ii) AO = OC [∵ diagonals of a parallelogram bisect each, other]

(iii) AD = BC [∵opposite sided of a parallelogram are equal and parallel to each other]

∴ △ABD ≅ △BOC by SSS

∴ Area of △ AOD = area of △BOC [area of congruent triangles are equal]