Telangana SCERT Solution Class X (10) Maths Chapter 8 Similar Triangles Exercise 8.4
Exercise 8.4
(Q1) Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
=> Given ABCD is a rhombus
AB = BC = CD = DA and AC ⊥ BD
OA = OC & OB = OD
To prove:
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof: in △AOB by Pythagoras theorem,
DA2 + OB2 = AB2
(AC/2)2 + (BD/2)2 = AB2
AC2/4 + BD2/4 = AB2
AC2+BD2/4 = AB2
AC2 + BD2 = 4AB2
AC2 + BD2 = AB2 + AB2 + AB2 + AB2
We know that,
All the sides are equal in a rhombus.
AB = BC = CD = DA
=> AC2 + BD2 = AB2 + BC2 + CD2 + DA2 (∵ Given)
∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
(Q2) ABC is a right triangle A right angled at B. Let D and E be only points on AB and BC respectively prove that AE2 + CD2 = AC2 + DE2
=> Given: In △ABC
∠B = 900
D is the point on AB & E is the point on BC
To prove: AE2 + CD2 = AC2 + DE2
Proof: In △ABC
AB2 + BC2 = AC2 (By Pythagorous Theorem) ——– (1)
In △DBE
DB2 + BE2 = DE2 (By Pythagorous theorem) ——— (2)
Add equation (1) and (2),
AB2 + BC2 + DB2 + BE2 = AC2 + DE2
AE2 + CD2 = AC2 + DE2
[∵ In △ABE AE2 = AB2 + BE2 In △DBE CD2 = DB2 + BC2]
(Q3) Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Given: ABC is an equilateral triangle of side ‘a’ and altitude ‘h’
To prove:
3a2 = 4h2
Proof: In △ADB, by Pythagorous theorem,
AB2 = BD2 + AD2
a2 – (a/2)2 = h2
a2 – a2/4 = h2
4a2–a2/4 = h2
3a2/4 = h2
3a2 = 4h2
(Q4) PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR show that PM2 = QM.MR.
=> Given: In △PQR ∠P = 900
PM ⊥ QR
To prove: PM2 = QM.MR
Proof: In △RPQ & △RMP
∠R = ∠R (Common)
∠RPQ = ∠RMP = 900
BY A – A Similarity
△RPQ ~ △RMP —– (1)
Similarity
△RPQ ~ △QMP ——- (2)
From (1) & (2)
△RMP ~ △QMP
If two triangles are similar then their corresponding sides are proportional.
∴ MR/PM = PM/QM
QM.MR = PM2
∴ PM2 = QM.MR
(Q5) ABD is a triangle right angled at A and AC ⊥ BD show that
(i) AB2 = BC.BD
(ii) AC2 = BC.DC
(iii) AD2 = BD.CD
=> Given: ABP is a right angled triangle.
∠A = 900 & AC ⊥ BD
To prove: (i) AB2 = BC.BD
(ii) AC2 = BC.DC
(iii) AD2 = BD.CD
Proof: In △BAD and △BCA
∠B = ∠B (Common)
∠BAD = ∠BCA = 900
By A – A Similarity,
△BAD ~ △BCA ——- (1)
AB/BC = BD/AB
=> AB2 = BC.BD
In △BAD and △DCA
∠D = ∠D (Common)
∠BAD = ∠DCA = 900
By A – A similarity,
△BAD ~ △DCA —— (2)
=> AD/CD = BD/AD
=> AD2 = BD.CD
From equation (1) & (2)
△BCA ~ △DCA
=> AC/DC = BC/AC
=> AC2 = BC.DC
(Q6) ABC is an isosceles triangle right angled at C prove that AB2 = 2AC2
=> Given: ABC is an isosceles right angled triangle (PHOTO)
∠C = 900
AC =BC
To prove: AB2 = 2AC2
Proof: In △ACB, by Pythagorous theorem,
AB2 = AC2 + BC2
= AC2 + AC2 (∵ AC = BC)
AB2 = 2AC2
(Q7) ‘O’ is any point in the interior of a triangle ABC.
If OD ⊥ BC, OF
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
=> Given: ‘O’ is any point in the interior of a triangle ABC.
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB
To prove: (i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Proof: In △AFO,
AF2 + OF2 = OA2
=> AF2 = OA2 – OF2 ——- (1)
In △BDO
BD2 + OD2 = OB2
BD2 = OB2 – OD2 —— (2)
In △CEO,
CE2 + OE2 = OC2
=> CE2 = OC2 – OE2 ——- (3)
Add equation (1), (2) and (3)
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
=> OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) From (i)
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
= AE2 + CD2 + BF2
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2
(Q8) A 24m long wire is attached to a vertical pole of height 18m. And it has a stake attached to the end. How far from the base of the pole should the stake be driven so that the wire will be that?
=> Given: Height of the pole AB = 18m
Length of the wire BC = 24m
Distance between wire and base of the pole on earth is AC.
In △ABC
By Pythagorous theorem.
AB2 + AC2 = BC2
AC2 = BC2 – AB2
AC2 = (24)2 – (18)2
AC2 = 576 – 324
AC2 = 252
AC = √252
AC = √36X7
AC = 6√7m
The distance between pole and wire on earth = 6√7m
(Q9) Two pole of heights 6m and 11m stand on a plane ground. If the distance between the feet of the poles is 12m find the distance between their tops.
=> Given:
Height of the first pole AB = 6m
Height of the second 6m pole CD = 11m
Distance between the feet of the poles AC = 12m
The distance between top of the poles is BD
Here BACE is a rectangle.
AB = CE = 6m &
AC = BE = 12m
DE = CD – CE
= 11m – 6m
DE = 5m
In △BED by Pythagorous theorem,
BD2 = BE2 + DE2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
√BD2 = √169
BD = 13m
∴ The distance between top of the poles = 13m
(Q10) In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC prove that 9AD2 = 7AB2
=> Given: In equilateral triangle ABC
D is a point on BC such that BD = 1/3 BC.
To prove: 9AD2 = 7AB2
Proof: Draw an altitude on
BC => AE ⊥ BC
BE = 1/2 BC
DE = BE – BD
= BC/2 – BC/3
= 3BC-2BC/6
DE = BC/6
In △AEB
AB2 = AE2 + BE2 ——- (1)
In △AED,
AD2 = AE2 + DE2 ——- (2)
Subtracting equn (1) – (2) gives,
AB2 – AD2 = AE2 + BE2 + AE2 – DE2
= BE2 – DE2
= (BC/2)2 – (BC/6)2
AB2 – AD2 = BC2/4 – BC2/36
AB2 – AD2 = 9BC2-BC2/36
= 8BC2/36
= 2BC2/9
AB2 – AD2 = 2BC2/9
9AB2 – 9AD2 = 2BC2
=> 9AD2 = 9AB2 – 2BC2
= 9AB2 – 2AB2 (BC = AB)
9AD2 = 7AB2
(Q11) In the given figure, ABC is a triangle right angled at B, D and E are points on BC trisect it. Prove that,
8AE2 = 3AC2 + AD2
=> Given: In △ABC
∠B = 900
D and E are points on BC trisect it
=> BD = DE = EC
To prove: 8AE2 = 3AC2 + 5AD2
Proof: Let BD = DE = EC = ‘a’
=> BE = 2a and BC = 3a
In △ABD,
AD2 = AB2 + BD2
= AB2 + a2
In △ABE,
AE2 = AB2 + BE2
= AB2 + (2a)2
AE2 = AB2 + 4a2
In △ABC,
AC2 = AB2 + BC2
= AB2 + (3a)2
AC2 = AB2 + 9a2
Take 3AC2 + 5AD2
= 3 (AB2 + 9a2) +5 (AB2 + a2)
= 3AB2 + 27a2 + 5AB2 + 5a2 [∵ AC2 = AB2 + 9a2]
= 8AB2 + 32a2
= 8 (AB2 + 4a2)
= 8 (AB2 + (2a)2)
= 8 (AB2 + BE2) (∵ BE = 2a)
= 8 AE2 (by p.g. Theorem)
∴ 8AE2 = 3AC2 + 5AD2
(Q12) ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of △ABE and △ACD
Given: In isosceles △ABC
AB = BC & ∠B = 900
△ABE and △ACD are equilateral triangles
In △ABC,
By Pythagorous Theorem,
AC2 = AB2 + BC2
= AB2 + AB2
AC2 = 2AB2
And also given,
△ABE ~ △ACD
Area of △ABE/Area of △ACD = AB2/AC2
= AB2/2AB2
= 1/2
∴ Area of △ABE : Area of △ACD = 1:2
Here is your solution of Telangana SCERT Class 10 Math Chapter 8 Similar Triangles Exercise 8.4
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