Telangana SCERT Solution Class IX (9) Math Chapter 11 Areas Exercise 11.1
(1) In ΔABC, ∠ABC = 90o , 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 cm2
∴ Area of △ADB = ½ of △ABC
= ½ x 39
= 19.5 cm2
(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
∴ SQ2 = PQ2 + PS2
Or, SQ = √122 + 92
= √(144+81)
= √225
= 15cm
∴ Area of △PQS = ½ x b x h
= ½ x PQ x PS
= ½ x 12 x 9
= 54 cm2
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 cm2
∴ Total area of equal PQRS = area of △PQS + area of △QPS
= 54 + 60
= 114 cm2
(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 = 90o [corresponding angle]
∴ △BEC is right angle triangle at ∠BEC = 90o
∴ Area of △BEC = ½ x b x h
= ½ x EC x BE
= ½ x 8 x 3
= 12 cm2
∴ 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 cm2
(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]