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GROUP TUITION
Chapter-7 SIMILARITY AND THE THEOREM OF PYTHAGORAS
Theorem 7.1
If an altitude is drawn to the hypotenuse of a right angled triangle, then the triangles formed in the different
closed semiplanes of the altitude are similar to the given triangle and they are similar to each other.
A
Thus if in ABCB is a right angle in ∆ABC. BD  AC , D  AC then
ADB  ABC is a similarity,
BDC  ABC is a similarity and
D
ADB  BDC is a similarity.
Geometric mean:
If b is the geometric mean of a and c then b2 = a  c.
C
B
Adjacent segments:
In ABC, if B and C are acute angles and AD  BC , D  BC then
BD is called adjacent to AB and CD is called adjacent to AC .
Corollary 1:
If an altitude is drawn to hypotenuse of a right angles triangle, then (1) length of altitude is the geometric mean
of lengths of segments of hypotenuse formed by the altitude (2) length of each side other than the hypotenuse is
the geometric mean of the length of hypotenuse and the segments of hypotenuse adjacent to the side.
Thus if in ABCB is a right angle in ∆ABC. BD  AC , D  AC then [Refer above figure]
AB2 = AD  AC, BC2 = CD  AC and BD2 = AD  CD
Theorem 7.2 [Pythagoras Theorem]
Square of the length of the hypotenuse of a right angled triangle is the sum of the squares of the lengths of other
two sides.
Theorem 7.3 [Converse of Pythagoras Theorem]
In a triangle, if the square of a side is equal to the sum of the squares of a side is equal to the sum of the squares
of other two sides, then the angle opposite to the first side is a right angle.
A
Apollonius Theorem:
If AD is a median of ABC then AB2 + AC2 = 2(AD2 + BD2).
B
D
C
Some Important result:
 In a right angled triangle the length of median of hypotenuse is half of the hypotenuse.
 In an isosceles right angled triangle the length of hypotenuse is 2 times of the sides forming right angle.
 In a right angled triangle the side opposite to 30 is half of the hypotenuse.
Exercise – 7.1
1. B is a right angle in ABC. BD  AC and D  AC . If AD = 4DC, prove that BD = 2DC.
Data: In ∆ABC, B is a right angle. BD  AC and D  AC . Moreover, AD = 4DC.
A
To prove: BD = 2DC.
Proof: In ∆ABC, B is a right angle. BD  AC , D  AC .
 BD2 = AD . DC
But AD = 4DC
 BD2 = 4DC.DC = 4DC2
B
 BD2 = (2DC)2  BD = 2DC
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AJAY PARMAR
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GROUP TUITION
2. 5, 12, 13 are the lengths of the sides of a triangle. Show that the triangle is right angled. Find the
length of altitude on the hypotenuse.
In ∆ABC, let AB = 5, BC = 12, and AC = 13. Here, in ∆ABC, the longest side is AC .
AC2 = 132 = 169
A
13
D
AB2 + BC2 = 52 + 122 = 25 + 144 = 169
2
2
2
Thus, in ∆ABC, AC = AB + BC
5
?
In ABC, B is a right angle [ Converse of Pythagoras Theorem]
B
C
12
Suppose BD is the altitude.
Area of right angled ∆ABC
½ × AB × BC = ½ × AB × BD
AB × BC = AB × BD
5 × 12 = 13 × BD BD =
5 12
13
=
60
13
.
Thus, the length of the altitude on the hypotenuse of the given triangle is
60
13
.
3. In PQR, Q M is the altitude to hypotenuse PR . If PM = 8, RM = 12, find PQ, QR and QM.
In ∆PQR, QM is the altitude to hypotenuse PR .
P 8
M
 P  M  R  PR = PM + RM = 8 + 12 = 20
2
2
2
PQ
= PM . PR;
QR
= RM . PR;
QM = PM . RM
12
?
?
= 8 × 20
= 12 × 20
= 8 × 12
= 16 × 10
= 16 × 15
= 96 = 16 × 6
Q
?
PQ = 4 10
 QR = 4 15
QM = 4 6
4. In ABC mB = 90, BM  AC , M  AC . If AM – MC = 7, AB2 – BC2 = 175, find AC.
A
In ∆ABC mB = 90, BM  AC , M  AC .
2
2
 AB = AM . AC and BC = MC . AC
 AB2  BC2 = AM . AC  MC . AC
 175 = AM . AC  MC . AC
 175 = AC (AM  MC)
 175 = AC (7)
B
 AC = 25
R
?
M
C
5. A is right angle in ABC. AD is an altitude of the triangle. If AB = 5 , BD = 2, find the length of the
B
hypotenuse of the triangle.
In ∆ABC, A is a right angle.
?
2
BC is the hypotenuse.
5
D
In ∆ABC, A is a right angle. AD is an altitude.
2
 AB = BD . BC
 ( 5 )2 = 2 . BC
C
A
V
 5 = 2BC
 BC = ½(5) = 2.5
Thus, the length of the hypotenuse of ∆ABC is 2.5.
6. mB = 90 in ABC. BM is altitude to AC .
1) If AM = BM = 8, find AC.
In ∆ABC, mB = 90 and BM is altitude to AC .
 A  M  C and BM2 =AM . CM
BM2 =AM . CM
 82 = 8 . CM
 CM = 8
Now, since A  M  C, AC = AM + CM = 8 + 8 = 16.
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2) If BM = 15, AC = 34, find AB.
In ∆ABC, mB = 90 and BM is altitude to AC .
 A  M  C and BM2 =AM . CM and AB2 = AM . AC
Suppose, AM = x
 CM = AC  AM = 34  x
BM2 =AM . CM
 152 = x . (34  x)
 225 = 34x  x2
 x2  34x + 225 = 0
 (x  25) (x  9) = 0
 x = 25
OR
x=9
 AM = 25
OR
AM = 9
If AM = 25, then
If AM = 9, then
AB2 = AM . AC
AB2 = AM . AC
= 25 × 34
= 9 × 34
 AB = 5 34
 AB = 3 34
A
?
M
34
15
B
3) If BM = 2 30 , MC = 6, find AC.
In ∆ABC, mB = 90 and BM is altitude to AC .
 A  M  C and BM2 =AM . CM
BM2 =AM . MC
 (2 30 )2 = 6 . AM
 120 = 6 . AM
 AM = 20
Now, since A  M  C,  AC = AM + CM = 20 + 6 = 26.
4) If AB = 10 , AM = 2.5, find MC.
In ∆ABC, mB = 90 and BM is altitude to AC .
 A  M  C and AB2 = AM . AC
AB2 = AM . AC
 ( 10 )2 = 2.5 . AC
10 = 2.5 . AC
 AC = 4
Now, since A  M  C, MC = AC  AM = 4  2.5 = 1.5.
C
A
?
M
2 30
6
C
V
B
A
2.5
10
M
?
B
C
V
7. In PQR, mQ = 90, PQ = x, QR = y Q D  PR , D  PR . Find PD, QD, RD in terms of x and y.
P
In ∆PQR, mQ = 90.
?
 PR2 = PQ2 + QR2
M
x
 PR2 = x2 + y2  PR = x 2  y 2
?
?
In ∆PQR, mQ = 90 and QD is an altitude.
 PQ2 = PD . PR
QR2 = RD . PR
Q
R
QR 2
 RD =
=
PR
x2
PQ 2
 PD =
=
PR
x2  y2
y2
y
x2  y2
2
Also QD = PD . RD
 QD2 =
Thus, PD =
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x2
x y
2
2
.
x2
x y
2
2
y2
x y
2
, RD =
2
x 2 .y 2
=
x y
2
y2
x y
2
2
2
 QD =
and QD =
x. y
x y
x. y
x y
.
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GROUP TUITION
8. Q is a right angle in PQR and Q M  PR , M  PR . If PQ = 4QR, then prove that PM = 16RM.
Data: In ∆PQR, Q is a right angle. QM  PR , M  PR , PQ = 4QR.
P
To Prove: PM = 16RM.
Proof: In ∆PQR, Q is a right angle. QM  PR , M  PR .
 PQ2 = PM . PR and QR2 = RM . PR …(i)
Also PQ = 4 QR
M
PQ
PQ 2
= 4  2 = 16
QR
QR
PM  PR
by (i)
= 16
RM  PR
PM

= 16  PM = 16RM
RM

Q
R
V
9. PQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of PQRS.
In rectangle PQRS
S
PR = QS, also PR + QS = 10
 PR + PR = 10  2PR = 10 PR = 5
5
PQ + QR = 7, suppose, PQ = x,  QR = 7  PQ = 7  x
In ∆PQR, Q is a right angle.
x
P
 PR2 = PQ2 + QR2
 52 = x2 + (7  x)2
 25 = x2 + 49  14x + x2
 2x2 + 49  14x  25 = 0
 2x2  14x + 24 = 0
 x2  7x + 12 = 0
 (x  4) (x  3) = 0
x=4
OR
x=3
 PQ = 4
OR
PQ = 3
If PQ = 4, then QR = 3 and if PQ = 3, then QR = 4
Thus, under both the situations, the lengths of two adjacent sides of rectangle PQRS are 3 and 4.
Area of rectangle PQRS = PQ  QR = 3 × 4 = 12 Thus, the area of rectangle PQRS is 12.
R
7–x
Q
10. The diagonals of a convex ABCD intersect at right angles. Prove that AB2 + CD2 = AD2 + BC2.
A
Data: The diagonals of a convex ABCD intersect at right angles at M.
2
2
2
2
To Prove: AB + CD = AD + BC .
D
B
Proof: In ∆AMB, CMD, AMD and BMC mM = 90.
M
 by Pythagoras theorem in all the triangles
AB2 = AM2 + BM2 …(i)
CD2 = CM2 + DM2 …(ii)
C
AD2 = AM2 + DM2 …(iii)
BC2 = BM2 + CM2 …(iv)
Adding (i) and (ii)
 AB2 + CD2 = AM2 + BM2 + CM2 + DM2.

= AM2 + BM2 + CM2 + DM2.
= AD2 + BC2 [ (iii) and (iv)]
AB2 + CD2 = AD2 + BC2.
11. In PQR, mQ = 90, M  Q R and N  PQ . Prove that PM2 + RN2=PR2 + MN2.
Given: Q is a right angle in PQR. N  PQ and M  QR .
To prove: PM2 + RN2 = PR2 + MN2
Proof: mQ = 90 in PQM, NQR, PQR and NQM
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By Pythagoras theorem
PM2 = PQ2 + QM2
…(i)
RN2 = NQ2 + QR2
…(ii)
2
2
2
PR = PQ + QR
…(iii)
MN2 = NQ2 + QM2 …(iv)
adding (i) and (ii)
PM2 + RN2
= PQ2 + QM2 + NQ2 + QR2
= PQ2 + QR2 + NQ2 + QM2
= PR2 + MN2 [ (iii) and (iv)]
 PM2 + RN2 = PR2 + MN2
P
N
Q
M
R
12. The sides of a triangle have lengths a2 + b2, 2ab, a2 – b2, where a > b and a,b, R+. Prove that the angle
opposite to the side having length a2 + b2 is a right angle.
Given: ABC such that AB = 2ab; BC = a2 – b2; AC = a2 + b2, a > b > 0.
A
To prove: ABC is a right angled triangle
Proof: in ABC
a2 + b2
2ab
AB2 = (2ab)2 = 4a2b2;
BC2 = (a2 – b2)2 = a4 – 2a2b2 + b2;
AC2 = (a2 + b2)2 = a4 + 2a2b2 + b2
B
a2 – b2
C
 AB2 + BC2 = 4a2b2 + a4 – 2a2b2 + b4
4
2 2
4
= a + 2a b + b
= AC2
by converse of Pythagoras theorem mB = 90
ABC is a right angled triangle.
13. In ABC, mB = 90 and BE is a median. Prove that AB2 + BC2 + AC2 = 8BE2.
Given: mB = 90 and BE is a median of ABC.
A
To prove: AB2 + BC2 + AC2 = 8 BE2.
Proof: BE is a median of ABC and mB = 90
BE = ½ AC
AC = 2BE
………(i)
B
mB = 90 in ABC
 AB2 + BC2 = AC2
……...(ii)
 AB2 + BC2 + AC2 = AC2 + AC2 [Adding AC2 on both sides]
 AB2 + BC2 + AC2 = 2AC2 = 2(2BE)2 …...by (i)
 AB2 + BC2 + AC2 = 2(4AE2) = 8BE2
 AB2 + BC2 + AC2 = 8BE2
E
C
14. AB = AC and A is right angle in ABC. If BC = 2 a, then find the area of the triangle. (aR, a > 0).
In ∆ ABC AB = AC and A is right angle.
B
 AB2 + AC2 =BC2
 AB2 + AB2 =( 2 a)2
(∵AB = AC)
2a
2
2
x
 2AB =2a
 AB2 =a2
 AB =a
A
C
x
Area of right angled ∆ ABC = ½ × AB × AC = ½ × a × a = ½ a2
Exercise – 7.2
1. In rectangle ABCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.
 ABCD is a rectangle.
 AC = BD and AC + BD = 34  AC = ½ (34) = 17
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AB + BC = 23
D
Suppose, AB = x
 BC = 23  x
17
In ∆ABC , mB = 90
 AB2 + BC2 = AC2
x
A
 x2 + (23  x)2 = 172
2
2
 x + 529  46x + x = 289
 2x2  46x + 240 = 0
 x2  23x + 120 = 0
 (x  15) (x  8) = 0
 x = 15
OR
x=8
 AB = 15
OR
AB = 8
If AB = 15, then BC = 23  15 = 8
If AB = 8, then BC = 23  8 = 15
Thus, both the situations, the lengths of two adjacent sides of rectangle ABCD are 15 and 8.
Area of rectangle ABCD = Product of two adjacent sides = 15 × 8 = 120
Thus, the area of rectangle ABCD is 120.
2. In ABC mA = mB + mC, AB = 7, BC = 25. Find the perimeter of ∆ ABC.
In ∆ABC mA = mB + mC
But mA + mB + mC = 180
B
 mA + mA = 180
 2mA = 180
7
 mA = 90
A
In ∆ABC mA = 90
2
2
2
 AB + AC = BC
 72 + AC2 = 252
 49 + AC2 = 625
 AC2 = 625  49 = 576 = (24)2
 AC = 24
Perimeter of ∆ABC = AB + BC + AC = 7 + 25 + 24 = 56
Thus, the perimeter of ∆ABC is 56.
C
23 – x
B
25
C
?
3. A staircase of length 6.5 meters touches a wall at height of 6 meter. Find the distance of base of the
staircase from the wall.
In the figure, AB represents the part of the wall and AC represents the staircase.
 In ∆ABC, mB = 90, AB = 6 m and AC = 6.5 m
A
2
2
2
 AC = AB + BC
 (6.5)2 = (6)2 + BC2
6.5
6
 42.25 = 36 + BC2
 BC2 = 42.25  36 = 6.25 = (2.5)2
C
B
?
 BC = 2.5
Thus, the distance of base of the staircase from the wall is 2.5 m.
4. In ABC, AB = 7, AC = 5, AD = 5. Find BC, if the mid – point of BC is D.
In ∆ABC, D is the mid – point of BC
 AD is a median and BC = 2BD
 AB2 + AC2 = 2(AD2 + BD2)
 (7)2 + (5)2 = 2(52 + BD2)
 49 + 25 = 2(25 + BD2)
 74 = 2(25 + BD2)
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 25 + BD2 = 37
 BD2 = 12
 BD = 12 = 4  3 = 2 3
Now, BC = 2BD = 2(2 3 ) = 4 3 .
5. In equilateral ABC, D  BC such that BD : DC = 1 : 2. Prove that 3AD = 7 AB.
Given: In equilateral ∆ABC, D  BC and BD : DC = 1 : 2.
To prove: 3AD = 7 AB.
Construction: Draw AM  BC , M  BC .
Proof: ∆ABC is an equilateral triangle.
Suppose, AB = BC = AC = 6a
In equilateral ∆ABC, AM is an altitude.
 AM is a median.
 BM = ½ BC = ½ (6a) = 3a
B
B  D  C and BD : DC = 1:2
But BD + DC = BC  x + 2x = BC  3x = BC  x = 1/3  BC
 BD = 1/3  BC = 1/3  (6a) = 2a
DM = BM  BD = 3a  2a = a
In ∆ AMB, mM = 90
 AB2 = AM2 + BM2
 36a2 = AM2 + 9a 2
 AM2 = 27a 2
In ∆ AMD, mM = 90
 AD2 = AM2 + DM2 = 27a 2 + a2 = 28a2
 AD2 = 28a2 = 4  (7a2)
 AD = 2 7 a
 3AD = 6 7 a = 7  6a
 3AD = 7 AB
A
D
M
C
6. In ABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest side.
A
In ∆ABC, AB = 17, BC = 15, AC = 8.
Thus, in ∆ABC, the longest side is AB .
17
2
2
D
AB = 17 = 289
8
BC2 + AC2 = 152 + 82 = 225 + 64 = 289
 AB2 = BC2 + AC2
C
15
B
 ∆ABC is a right angled triangle in which AB is hypotenuse and C is right angle. In right angled triangle
ABC, the length of the median on the longest side hypotenuse AB is half the length of AB .
Let CD be the median on hypotenuse AB .
 CD = ½ AB = ½ (17) = 8.5
Thus, in ∆ABC, the length of the median on the longest side AB is 8.5.
7. AD is a median of ABC. AB2 + AC2 = 148 and AD = 7. Find BC.
In ∆ABC, AD is a median.
 AB2 + AC2 = 2(AD2 + BD2) [ Apollonius theorem]
 148 = 2(72 + BD2)
 49 + BD2 = 74
 BD2 = 25
 BD = 5
Also as AD is a median, D is the midpoint of AC .
 BC = 2 BD = 2(5) = 10.
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8. In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.
B
 ABCD is a rectangle.
 mD = 90
 AC2 = AD2 + CD2
 252 = AD2 + 72
 AD2 + 49 = 625
A
 AD2 = 576 = 242
 AD = 24
Perimeter of rectangle ABCD = 2(AD + CD) = 2(24 + 7) = 2(31) = 62
Thus, the perimeter of rectangle ABCD is 62.
9. In rhombus XYZW, XZ= 14 and YW = 48. Find XY.
Let, XZ  YW = {M}
XM = ½XZ = ½ (14) = 7, and YM = ½ YW = ½ (48) = 24
In ∆XMY, mM = 90
[Diagonals of rhombus bisect at right angles]
 XY2 = XM2 + YM2 = 72 + 242 = 49 + 576 = 625 = 252
 XY = 25
C
25
7
?
D
X
?
Y
7
24
W
M
Z
10. In PQR, mQ : mR : mP = 1 : 2 : 1. If PQ = 2 6 , find PR.
P
In ∆PQR, mQ : mR : mP = 1 : 2 : 1. If PQ = 2 6 , find PR.
If mQ = x, then mR = 2x and mP = x
In ∆PQR, mP + mQ + mR = 180
2 6
x
 x + 2x + x = 180
 4x = 180  x = 45
mP = x = 45, mQ = x = 45 and mR = 2x = 2(45) = 90
Q
R
x
In ∆PQR mP = mQ
 QR = PR = x
In ∆PQR, mR = 90.
 PR2 + QR2 = PQ2
 x2 + x2 = (2 6 )2
 2x2 = 24
 x2 = 12  x = 2 3
 PR = 2 3
Exercise – 7
1. AD , BE , CF are the medians of ABC. If BE = 12, CF = 9 and AB 2 + BC2 + AC2 = 600, BC = 10, find
A
AD.
In ∆ABC, AD , BE , CF are the medians.
Hence, by Apollonius theorem, it is proved that
F
E
3(AB2 + BC2 + AC2) = 4(AD2 + BE2 + CF2)
2
2
2
 3(600) = 4(AD + 12 + 9 )
 AD2 + 144 + 81 = 3 × 150
 AD2 = 450  144 81
B
D
C
 AD2 = 225 = 152
 AD = 15
2. AD is altitude of ABC such that B – D – C. If AD2 = BD × DC, prove that BAC is right angle.
Given: In ∆ABC, AD is altitude such that B  D  C. Moreover, AD2 = BD × DC.
To prove: BAC is right angle.
Proof: In ∆ABC, AD is altitude such that B  D  C.
 mADB = mADC = 90
B
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In ∆ADB, mD = 90
 AB2 = AD2 + BD2 = BD . DC + BD2
 AB2 = BD (DC + BD)
 AB2 = BD . BC
(∵ B  D  C) … (1)
In ∆ADC, mD = 90
 AC2 = AD2 + DC2 = BD . DC + DC2
 AC2 = DC (BD + DC)
 AC2 = DC . BC
(∵ B  D  C) … (2)
Adding (1) and (2),
 AB2 + AC2 = BD . BC + DC . BC
= BC (BD + DC)
= BC . BC
(∵ B  D  C)
2
2
2
 AB + AC = BC
Hence, ∆ABC is a right angled triangle in which BC is the hypotenuse and BAC is right angle.
3. In ABC, AD  BC , B – D – C. If AB2 = BD × BC, prove that BAC is a right angle.
Given: In ∆ABC, AD  BC , B – D – C. Moreover, AB2 = BD × BC.
B
To prove: BAC is right angle.
Proof: In ∆ABC, AD  BC , B – D – C.
 mADB = mADC = 90
D
In ∆ADB, mD = 90
and
In ∆ADC, mD = 90
 AB2 = AD2 + BD2
 AC2 = AD2 + DC2
 BD . BC = AD2 + BD2
 AC2 = BD . DC + DC2
A
2
2
2
 AD = BD . BC  BD
 AC = DC (BD + DC)
2
 AD = BD (BC  BD)
 AC2 = DC . BC
( B  D  C)
 AD2 = BD . DC (∵ B  D  C)
Now,
 AB2 + AC2 = BD . BC + DC . BC
= BC (BD + DC)
= BC . BC
(∵ B  D  C)
 AB2 + AC2 = BC2
Hence, ∆ABC is a right angled triangle in which BC is the hypotenuse and BAC is right angle.
4. In ABC, AD  BC , B – D – C. If AC2 = CD × BC, prove that BAC is a right angle.
Given: In ∆ABC, AD  BC , B – D – C. Moreover, AC2 = CD × BC.
B
To prove: BAC is right angle.
Proof: In ∆ABC, AD  BC , B – D – C.
 mADC = mADB = 90
In ∆ADC, mD = 90
and
In ∆ADB, mD = 90
 AC2 = AD2 + CD2
 AB2 = AD2 + BD2 = CD . BD + BD2
A
 CD . BC = AD2 + CD2
 AB2 = BD (CD + BD)
2
2
2
 AD = CD . BC  CD = CD (BC  CD)
 AB = BD . BC
(∵ B  D  C)
2
 AD = CD . BD
(∵ B  D  C)
Now,
 AB2 + AC2 = BD . BC + CD . BC
= BC (BD + DC)
= BC . BC
(∵ B  D  C)
 AB2 + AC2 = BC2
Hence, ∆ABC is a right angled triangle in which BC is the hypotenuse and BAC is right angle.
C
D
C
5. AD is a median of ABC. If BD = AD, prove that A is a right angle in ∆ABC.
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B
Given: In ∆ABC, AD is a median and BD = AD.
To prove: A is a right angle in ∆ABC.
Proof: In ∆ABC, AD is a median.
 AB2 + AC2 = 2(AD2 + BD2) [Apollonius theorem]
= 2(BD2 + BD2) ( AD = BD)
A
= 2(2BD2) = 4BD2 = (2BD)2
= BC2
(D is the midpoint of BC )
Hence, ∆ABC is a right angled triangle in which BC is the hypotenuse and A is right angle.
D
C
6. In figure, AC is length of a pole standing vertical on the ground. The pole is bent a point B, so that the
top of the pole touches the ground at a point 15 meters away from the base of the pole. If the length of
the pole is 25, find the length of the upper part of the pole.
When the pole is bent at point B, its top touches the ground at point Cʹ.
C
 BC = BCʹ
Here, ACʹ = 15 m and AC = 25 m.
25
AC = AB + BCʹ
B
Suppose, AB = x m
A
C' 15
 25 = x + BCʹ
 BCʹ = (25 – x) m
In ABC, mA = 90
 BC2 = AB2 + AC2
 (25 – x)2 = x2 + 152
 625 – 50x – x2 = x2 + 225
 625 – 225 = 50x
 50x = 400
x=8
 AB = 8 m
Length of the upper part of the pole = BC = BCʹ = 25 – x = 25 – 8 = 17 m.

7. In ABC, AB > AC, D is the mid – point of BC . AM  BC such that B – M – C. prove that
AB2 – AC2 = 2BC × DM.

Given: In ∆ABC, AB > AC, D is the mid – point of BC . AM  BC such that B – M – C.
To prove: AB2 – AC2 = 2BC × DM.
Proof: In ∆ABC, AB > AC, D is the mid – point of BC and M is the foot of perpendicular from A to BC
such that B – M – C.
 B and C are both acute angles.
Since AB > AC, point D lies between B and M. Hence, we get B – D – M – C.
A
In AMB mM = 90
 AB2 = AM2 + BM2
In AMC mM = 90
 AC2 = AM2 + CM2
 AB2 – AC2 = (AM2 + BM2) – (AM2 + CM2)
= AM2 + BM2 – AM2 – CM2
B
D
C
M
= BM2 – CM2
= (BD + DM)2 – (CD – DM)2
( B – D – M – C)
2
2
= (BD + DM) – (BD – DM)
(Since D is the midpoint of BC , CD = BD)
= (BD + DM + BD – DM) (BD + DM – BD + DM)
=(2BD) (2DM)
=(BC) (2DM)
( D is the point midpoint of BC )
2
2
 AB – AC = 2BC × DM.
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8. In ABC, BD  AC , D  AC and B is right angle. If AC = 5CD, prove that BD = 2CD.
Given: In ∆ABC, BD  AC , D  AC and B is right angle. Moreover AC = 5CD.
A
To prove: BD = 2CD.
Proof: D  AC  A – D – C
 AC = AD + CD
 5CD = AD + CD
 AD = 4CD.
In  ABC, B is a right angle and BD is an altitude.
B
 BD2 = AD . CD
= (4CD) . CD = 4CD2
= 4 CD2
= (2CD)2
 BD = 2CD
D
C
9. Select a proper (a), (b), (c), or (d) from given options and write in the box given on the right so that
the statement becomes correct : –
1. In PQR, if mP + mQ = mR. PR = 7, QR = 24, then PQ = ____.
(a) 31
(b) 25
(c) 17
(d) 15
In ∆PQR, mP + mQ = mR.
mR = 90
 PQ2 = PR2 + QR2 = 72 + 242 = 49 + 576 = 625 =252
 PQ = 25
2. In ABC, AD is an altitude and A is right angle. If AB = 20 , BD = 4, then CD = ____.
(a) 5
(b) 3
(c) 5
(d) 1
B
In ∆ABC, mA = 90 and AD is an altitude.
2
 AB = BD . BC
4
 ( 20 )2= 4 . BC
20
D
 4BC= 20
?
 BC = 5
Since B – D – C,
C
A
BC = BD + CD
 CD = BC – BD = 5 – 4 = 1
3. In ABC, AB2 + AC2 = 50. The length of the median AD = 3. So, BC = ____.
A
(a) 4
(b) 24
(c) 8
(d) 16
In ∆ABC, AD is a median.
 AB2 + AC2 = 2(AD2 + BD2)
3
50 = 2(32 + BD2)
9 + BD2 = 25
B
D
C
BD2 = 16  BD = 4 BC = 2BD = 2(4) = 8
4. In ABC mB = 90, AB = BC. Then AB : AC = _____.
(a) 1 : 3
(b) 1 : 2
(c) 1 : 2
(d) 2 : 1
In ∆ABC, mB = 90
 AB2 + BC2 = AC2
 AB2 + AB2 = AC2
( AB = AC)
 2AB2 = AC2
 AB2 : AC2 = 1 : 2
 AB : AC = 1 : 2
5. In ABC, mB = 90, AC = 10. The length of the median BM = _____.
(a) 5
(b) 5 2
(c) 6
(d) 8
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In ∆ABC, mB = 90.  AC is hypotenuse.
BM
is the median on hypotenuse AC ,  BM =
AC
6. In ABC, AB = BC =
(a) is acute
(b) is obtuse
In ∆ABC, AB = BC =
2
 AC 
2
AC =
1
2
(10) = 5
. mB = ____.
2
AC
1
2
(c) is right angle (d) cannot be obtained
.
 AC 
2
AC 2
AC 2
 AB2 + BC2 =   +   =
+
= AC2
2
2
2
2




Hence, ABC is right angled triangle in which AC is the hypotenuse and B is a right angle.
7. In ABC,
AB
1
=
(a) 90
AC
2
BC
=
3
, then mC = _____.
(b) 30
In ∆ABC,
AB
1
=
AC
2
=
(c) 60
BC
3
(d) 45
= k.
AB = k, AC = 2k and BC = 3 k
 AB2 + BC2 = k 2 + ( 3 k)2 = k2 + 3k2 = 4k2 = (2k)2 = AC2
mB = 90
Now in ABC, sin C =
AB
AC
=
k
1
1
= but sin 30 = mC = 30
2
2
2k
8. In XYZ, mX : mY : mZ = 1 : 2 : 3. If XY = 15, YZ = _____.
(a)
15 3
2
(b) 17
(c) 8
(d) 7.5
In ∆XYZ, mX : mY : mZ = 1 : 2 : 3.
 mX = 30, mY = 60 and mZ = 90.
Thus, ∆XYZ is a right angled triangle with hypotenuse XY and mZ = 90
 YZ =
1
2
XY =
1
2
(15) = 7.5
9. In ABC, B is a right angle and BD is an altitude. If AD = BD = 5, then DC = ____.
(a) 1
(b) 5
(c) 5
(d) 2.5
A
In ∆ABC, B is a right angle,
5
BD is an altitude and AD = BD = 5.
2
 BD = AD . DC
5
 52 = 5 . DC
 DC = 5
B
10. In ABC, AD is a median. If AB2 + AC2 = 130 and AD = 7, then BD = _____.
(a) 4
(b) 8
(c) 16
(d) 32
In ∆ABC, AD is a median.
 AB2 + AC2 = 2(AD2 + BD2) [Apollonius theorem]
 130 = 2(72 + BD2)
 65 = 49 + BD2
 49 + BD2 =65
 BD2 = 16
 BD = 4
11. The diagonal of a square is 5 2 . The length of the side of the square is _____.
(a) 10
(b) 5
(c) 3 2
(d) 2 2
For any square, length of the diagonal = 2  Length of the side.
 5 2 = 2  Length of the side.
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 Length of the side = 5.
12. The length of a diagonal of a rectangle is 13. If one of the side of the rectangle is 5, the perimeter of
the rectangle is ____.
(a) 36
(b) 34
(c) 48
(d) 52
For any rectangle,
(Diagonal)2 = (Length)2 + (Breadth)2
 (13)2 = (Length)2 + (5)2
 169 = (Length)2 + 25
 (Length)2 = 169 – 25 = 144 = 122
 Length = 12.
Perimeter of rectangle = 2(Length + Breadth) = 2(12 + 5) = 2(17) = 34
13. The length of a median of an equilateral triangle is 3 . Length of the side of the triangle is _____.
(a) 1
(b) 2 3
(c) 2
(d) 3 3
The length of the median (altitude) of equilateral triangle
3
 Length of the side
2
3
 3 =
 Length of the side
2
2
 Length of the side = 3 
3
=
 Length of the side = 2
14. The perimeter of an equilateral triangle is 6. The length of the altitude of the triangle is _____.
(a)
3
2
(b) 2 3
(c) 2
Length of the side of an equilateral triangle =
(d) 3
Perimeter
3
Length of an altitude of an equilateral triangle =
=
6
3
=2
3
3
 Length of the side =
2=
2
2
15. In ABC, mA = 90. AD is a median. If AD = 6, AB = 10, then AC = _____.
(a) 8
(b) 7.5
(c) 16
(d) 2 11
In ABC, mA = 90 and AD is a median.
 AD = ½ BC
 6 = ½ BC
 BC = 12
In ABC mA = 90
 AB2 + AC2 = BC2
 102 + AC2 = 122
 100+ AC2 = 144
 AC2 = 44 = 4  11
 AC = 2 11
16. In PRQ, mQ = 90 and PQ = QR. Q M  PR . M  PR . If QM = 2, PQ = _____.
P
(a) 4
(b) 2 2
(c) 8
(d) 2
In PRQ, mQ = 90 and PQ = QR. QM is an altitude.
 QM is a median.
?
In PRQ, QM is a median on hypotenuse PR .
 QR = ½ PR
Q
 2 = ½ PR PR = 4
2
2
2
In PRQ, mQ = 90, PQ + QR = PR
 PQ2 + PQ2 = 42  2PQ2 = 16  PQ2 = 8 = 4  2
 PQ = 2 2
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B
C
V
10
M
2
10
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GROUP TUITION
17. In ABC, mA = 90. AD is an altitude. So, AB2 = _____.
(a) BD . BC
(b) BD . DC
(c)
BD
BC
(d) BC . DC
In ABC, mA = 90. AD is an altitude.
 By corollary 1 of theorem 7.1, AB2 = BD . BC
18. In ABC, mA = 90. AD is an altitude. Therefore, BD . BC = _____.
(a) AB2
(b) BC2
(c) AC2
(d) AD2
In ABC, mA = 90. AD is an altitude.
 By corollary 1 of theorem 7.1, AD2 = BD . BC
L–7
Section – 1
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Select proper option and write correct answer (1 mark each):
1. In PQR, if mP + mQ = mR. PR = 7, QR = 24, then PQ = ____.
(a) 31
(b) 25
(c) 17
(d) 15
2. In ABC, AD is an altitude and A is right angle. If AB = 20 , BD = 4, then CD = ____.
(a) 5
(b) 3
(c) 5
(d) 1
3. In ABC mB = 90, AB = BC. Then AB : AC = _____.
(a) 4
(b) 24
(c) 8
(d) 16
4. In ABC, AB = BC =
AC
2
25
. mB = ____.
(a) is acute
(b) is obtuse
(c) is right angle (d) cannot be obtained
5. In XYZ, mX : mY : mZ = 1 : 2 : 3. If XY = 15, YZ = _____.
(a)
15 3
2
(b) 17
(c) 8
(d) 7.5
6. In ABC, AD is a median. If AB2 + AC2 = 130 and AD = 7, then BD = _____.
(a) 4
(b) 8
(c) 16
(d) 32
7. The diagonal of a square is 5 2 . The length of the side of the square is _____.
(a) 10
(b) 5
(c) 3 2
(d) 2 2
8. The length of a diagonal of a rectangle is 13. If one of the sides of the rectangle is 5, the perimeter of the
rectangle is ____.
(a) 36
(b) 34
(c) 48
(d) 52
9. The length of a median of an equilateral triangle is 3 . Length of the side of the triangle is _____.
(a) 1
(b) 2 3
(c) 2
(d) 3 3
10. In PRQ, mQ = 90 and PQ = QR. QM  PR . M  PR . If QM = 2, PQ = _____.
(a) 4
(b) 2 2
(c) 8
(d) 2
11. In ABC, mA = 90. AD is an altitude. So, AB2 = _____.
(a) BD . BC
(b) BD . DC
(c)
BD
BC
(d) BC . DC
12. In ABC, mA = 90. AD is an altitude. Therefore, BD . BC = _____.
(a) AB2
(b) BC2
(c) AC2
(d) AD2
13. If in PQR, Q is a right angle. QD is an altitude. If PD = 9DR, then PQ = ____QR.
(a) 9
(b) 3
(c) 6
(d) 81
m
14. In  PQRS, PQ = QR = 10 and PR = 10 2 , then PQRS is a ____.
(a) Square
(b) Rectangle
(c) Rhombus
(d) Trapezium
15. In a recangle HIJK HJ = 13, HI = 5, then the perimeter of rectangle HIJK is ____.
(a) 34
(b) 20
(c) 18
(d) 36
16. In PQR mP = mQ + mR. If PR = 2.1, PQ = 2.0 then the perimeter of PQR is ____.
(a) 2.9
(b) 7.0
(c) 8.81
(d) 6.10
17. A 15m long ladder leans on a wall. Its top end reaches the wall at the height of 12m. Then distance of the
lower end of the ladder from the wall is ____m.
(a) 27
(b) 12
(c) 9
(d) 15
18. In PQR, PM is a median. PQ2 + PR2 = 148 and PM = 7. Then QR = ____.
(a) 10
(b) 9
(c) 28
(d) 49
19. In ABC mB = 90. If AB = 6, BC = 8, then the length of the altitude BM is ____.
(a) 10
(b) 14
(c) 48
(d) 4.8
20. PQR, PQ = 8, QR = 6 and PR = 13. PQR is a/an ____ triangle.
(a) right angled (b) acute angled (c) obtuse angled (d) None of these
21. Perimeter of an equilateral triangle is 18. Length of its altitude = ____.
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(a) 2 3
(b) 3 3
(c) 4 3
(d) 4 3
2
2
2
22. In ABC, mB = 90, AB + BC + CA = 128.  AC = ____.
(a) 64
(b) 8
(c) 144
(d) 4
23. ABC, mA = 90, b = 12, c = 35 then a = ____.
(a) 37
(b) 47
(c) 12
(d) 4 7
24. Lengths of diagonals of rhombus are 10 and 24. Its area = ____.
(a) 10
(b) 24
(c) 240
(d) 120
25. In ABC, AB = 4, BC = 2 3 , AC = 2 7 . The length of the median on the longest side is ____.
(a) 8 3
(b) 4 28
(c)
(d)
7
3
Section – 2
A Solve the following (2 marks each)
10
1. In ∆ABC mB = 90, BM  AC , M  AC . If AM – MC = 7, AB2 – BC2 = 175, find AC.
2. The perimeter of a rhombus is 100. If the length of one of the diagonals is 48 then find the length of the
other diagonal.
3. In ∆PQR, Q is a right angle. QM  PR , M  PR , PQ = 4QR. If RM = 2 then find PR.
4. In ABC mA = mB + mC, AB = 7, BC = 25. Find the perimeter of ∆ ABC.
5. mP = 90 in  PQR. PM is a median. If PQ = 12, PR = 9, then find PM.
B Solve the following (3 marks each)
6
6. In ∆ABC, mB = 90, BM  AC , M  AC . If AM = x, BM = y, find AB, BC and CM in terms of x and y.
(x > 0, y > 0)
7. In PQR, mQ = 90, PQ = x, QR = y and QD is an altitude. Find PD, QD, RD in terms of x and y.
C Solve the following (4 marks each)
4
8. State and prove Pythagoras Theorem.
D Solve the following (5 marks each)
5
9. State and prove Converse of Pythagoras Theorem.
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