Triangles EX 6.5 Solutions
Question - 1 : - Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
Answer - 1 : -
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution
Question - 2 : - PQR is a triangle right angled at P and M is apoint on QR such that PM ⊥QR. Show that PM2 = QM × MR.
Answer - 2 : -
Given, ΔPQR is right angled at P is a point onQR such that PM ⊥QR
We have to prove, PM2 =QM × MR
In ΔPQM, by Pythagoras theorem
PQ2 = PM2 + QM2
Or, PM2 = PQ2 –QM2 ……………………………..(i)
In ΔPMR, by Pythagoras theorem
PR2 = PM2 + MR2
Or, PM2 = PR2 –MR2 ………………………………………..(ii)
Adding equation, (i) and (ii), we get,
2PM2 = (PQ2 +PM2) – (QM2 + MR2)
= QR2 –QM2 – MR2 [∴ QR2 = PQ2 + PR2]
=(QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
Question - 3 : - In Figure, ABD is atriangle right angled at A and AC ⊥ BD. Show that
Answer - 3 : - (i) AB_2 = BC × BD
(ii) AC_2 = BC × DC
(iii) AD_2 = BD × CD
Solution:
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD
Solution:
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
Similarly, in ΔCAD
∠CAD = 90° – ∠CBA
= 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)
∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC
Solution:
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
ABC is an isoscelestriangle right angled at C. Prove that AB2 = 2AC2 .
Answer - 4 : -
Given, ΔABC is an isosceles triangle rightangled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB2 = AC2 + BC2 [ByPythagoras theorem]
= AC2 + AC2 [Since,AC = BC]
AB2 = 2AC2
Question - 5 : - ABC is an isoscelestriangle with AC = BC. If AB2 = 2AC2, provethat ABC is a right triangle.
Answer - 5 : -
Given, ΔABC is an isosceles triangle having AC= BC and AB2 = 2AC2
In ΔACB,
AC = BC
AB2 = 2AC2
AB2 = AC2 + AC2
= AC2 +BC2 [Since, AC = BC]
Hence, by Pythagoras theorem ΔABC isright angle triangle.
Question - 6 : - ABC is anequilateral triangle of side 2a. Find each of its altitudes.
Answer - 6 : -
Given, ABC is an equilateral triangle of side2a.
Draw, AD ⊥ BC
In ΔADB and ΔADC,
AB = AC
AD = AD
∠ADB = ∠ADC [Both are 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB,
AB2 = AD2 + BD2
(2a)2 = AD2 + a2
⇒ AD2 = 4a2 – a2
⇒ AD2 = 3a2
⇒ AD = √3a
Question - 7 : - Prove that the sumof the squares of the sides of rhombus is equal to the sum of the squares ofits diagonals.
Answer - 7 : -
Given, ABCDis a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB2 + BC2 + CD2 +AD2 = AC2 + BD2
Since, the diagonals of a rhombus bisect eachother at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 …………………….. (i) [By Pythagoras theorem]
Similarly,
AD2 = AO2 + DO2 …………………….. (ii)
DC2 = DO2 + CO2 …………………….. (iii)
BC2 = CO2 + BO2 …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB2 + AD2 + DC2 + BC2 =2(AO2 + BO2 + DO2 + CO2)
= 4AO2 + 4BO2 [Since, AO =CO and BO =DO]
= (2AO)2 +(2BO)2 = AC2 + BD2
AB2 + AD2 + DC2 + BC2 =AC2 + BD2
Hence, proved.
Question - 8 : - In the given figure, O is a point in theinterior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
Answer - 8 : - (i) OA_2 + OB_2 + OC_2 – OD_2 – OE_2 – OF_2 = AF_2 + BD_2 + CE_2
(ii) AF_2 + BD_2 + CE_2 = AE_2 + CD_2 + BF_2
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Question - 9 : - A ladder 10 m long reaches a window 8 m abovethe ground. Find the distance of the foot of the ladder from base of the wall.
Answer - 9 : -
Given, a ladder 10 m long reaches a window 8 mabove the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC2 = AB2 +BC2
102 = 82 + BC2
BC2 = 100 – 64
BC2 = 36
BC = 6m
Therefore, the distance of the foot of theladder from the base of the wall is 6 m.
Question - 10 : - A guy wire attachedto a vertical pole of height 18 m is 24 m long and has a stake attached to theother end. How far from the base of the pole should the stake be driven so thatthe wire will be taut?
Answer - 10 : -
Given, a guy wire attached to a vertical poleof height 18 m is 24 m long and has a stake attached to the other end.
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2 = AB2 +BC2
242 = 182 + BC2
BC2 = 576 – 324
BC2 = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.