Ratio and Proportion
(ii) Alternendo a b
a c c d
If b = d , then =
Proof a dc , bc ,
b
Here, = multiplying both sides by we get,
a × b = c × b or, a = b Proved
b c d c c d
Thus, the property of the proportion in which if a = c , then a = b is known as
alternendo. b d c d
Example: For 5 : 3 = 10 : 6 using alternendo property 5:10 = 1 : 2 = 1×3 : 2×3 = 3:6
? 5 : 3 = 10 : 6 implies 5 : 10 = 3 : 6
(iii) Componendo a b c
a c + + d
If b = d , then b = d
Proof a c
b d
Here, = , then adding 1 to both sides,
a +1 = c +1 or, a + b = c + d Proved
b d b d
Thus, the property of the proportion in which if a = c , then a + b = c + d is known
as componendo. b d b d
Example: For 4 : 6 = 10 : 15 using componendo property (4 + 6) : 6 = 10 : 6 = 5 : 3
And 5 : 3 = 5 × 5 : 3 × 5 = 25 : 15 = (10 + 15) : 15
? 4 : 6 = 10 : 15 implies (4 + 6) : 6 = (10 + 15) : 15
(iv) Dividendo
a c a–b c–d
If b = d , then b = d
Proof a c
b d
Here, = , subtracting 1 from both sides,
a –1 = c – 1 or, a –b = c – d Proved
b d proportion in b if da =
b c a–b c–d
Thus, the property of the which d , then b = d is known
as dividendo.
Example: 5 : 4 = 15 : 12 ; (5 – 4) : 4 = 1 : 4 = 3 : 12 = (15 – 12) : 12
? 5 : 4 = 15 : 12 implies (5 – 4) : 4 = (15 – 12) : 12
(v) Componendo and Dividendo
If a = c , then a+b = c+d
b d a–b c–d
Proof a c
b d
Here, = , then by componendo, we have, Also, by dividendo , we have
a+b = c + d ........ (i) or, a–b = c–d ........ (ii)
b d b d
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Ratio and Proportion
Dividing equation (i) by (ii), we get
a+b
b = c–d or, a+b = c+d proved
a–b a–b c–d
bd if a = c , then a+b = c+d
Thus, the property of the proportion in which b d a–b c–d
is known as componendo and dividendo.
Example: 5 : 3 = 10 : 6 then (5 + 3) : (5 – 3) = 8 : 2 = 4 : 1
(10 + 6) : (10 – 6) = 16 : 4 = 4 : 1
? 5 : 3 = 10 : 6 implies (5 + 3) : (5 – 3) = (10 + 6) : (10 – 6)
(vi) Addendo a c + c
a c b d a + d
If b = d , then = = b
Proof
Here, a = c , then by alternendo, we have, a = b
b d c d
and by componendo, we have = b+d
d
Again, by alternaendo, we have, a+c = c ? a = c = a+c proved.
b+d d b d b+d
Similarly, a = c = e = g = ... = a + c + e + g + ...
b d f h b + d + f + h + ...
Also, a = c = a– c
b d b– d
This property of proportion is known as addendo.
Example: 2 = 4 = 6 ; 2 + 4 + 6 = 12 = 32.
3 6 9 3 + 6 + 9 18
Worked-out examples
Example 1: If 12, 15, 20 are in proportion, find the fourth proportional.
Solution:
Let the fourth proportional be x.
Since 12, 15, 20 and x are in proportion,
12 = 20 or, 12x = 15 × 20 or, x = 25
15 x
Hence, the required fourth proportional is 25.
Example 2: Find the mean proportional between 20 and 45.
Solution:
Here, the first proportional is 20 and the third proportional is 45.
Now, mean proportional = 1st proportional × 3rd proportional
= 20 × 45 = 2 × 2 × 3 × 3 × 5 × 5 = 2 × 3 × 5 = 30
Hence, the required mean proportional is 30.
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Ratio and Proportion
Example 3: What should be added to each term 7, 10, 15 and 20 so that they are in
proportion?
Solution:
Let the required number to be added be x.
Now, 7+x = 15 + x
10 + x 20 + x
or, 140 + 7x + 20x + x2 = 150 + 10x + 15x + x2
or, x = 5
Hence, the required number to be added is 5.
3a + 4b 3a – 4b
Example 4: If a : b = c : d, prove that 3c + 4d = 3c – 4d .
Solution:
Here, a = c , multiplying both sides by 3 , 3a = 3c
b d 4 4b 4d
Now, by using componendo and dividendo, we get, 3a + 4b = 3c + 4d
3a – 4b 3c – 4d
Again, by using alternendo, we get, 3a + 4b = 3a – 4b proved.
3c + 4d 3c – 4d
Alternative process (k-method)
a c
Let, b = d = k , then a = bk and c= dk
Now, L.H.S. = 3a + 4b = 3bk + 4b = b (3k + 4) = b
3c + 4d 3dk + 4d d (3k + 4) d
R.H.S. = 3a – 4b = 3bk – 4b = b (3k – 4) = b
3c – 4d 3dk – 4d d (3k – 4) d
? L.H.S. = R.H.S. proved
Example 5: If x = y = z , prove that x3 + y3 + z3 = xyz .
Solution: a b c a3 + b3 + c3 abc
Let, x = y = z =k, then, x =k i.e. x = ak, y = k i.e. y = bk , z =k i.e. z = ck
a b c a b c
Now, L.H.S. = x3 + y3 + z3 = (ak)3 + (bk)3 + (ck)3 = a3k3 + b3k3 + c3k3 = k3(a3+ b3 + c3) = k3
a3 + b3 + c3 a3 + b3 + c3 a3 + b3 + c3 a3 + b3 + c3
ak.bk.ck
R.H.S. = xyz = abc = k3
abc
Hence, L.H.S. = R.H.S. Proved
Example 6: If a = b , prove that a2 + b2 = a + b
Solution: b c b2 + c2 b + c
Let a = b =k, then b = k i.e. b = ck, a = k i.e. a = bk = ck.k = ck2
b c c b
Now, L.H.S.= a2 + b2 = (ck2)2 + (ck)2 = c2k4 + c2k2 = c2k2(k2 + 1) = k2 = k
b2 + c2 (ck)2 + c2 c2k2 + c2 c2 (k2 +1)
Again, R.H.S. = a + b = ck2 + ck = ck(k + 1) = k
b + c ck + c c(k + 1)
Hence, L.H.S. = R.H.S. Proved.
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Ratio and Proportion
Example 7: If b is the mean proportional between a and c, prove that
a–b+c = a2 + b2 + c2
a+b+c (a + b + c)2
Solution:
Since b is the mean proportional between a and c. So, a : b : : b : c.
Let, a = b = k
b c
Then, b = k i.e, b = ck and a = k i.e, a = bk = ck.k = ck2
c b
a–b+c ck2 – ck + c c(k2 – k + 1) k2 – k + 1
Now, L.H.S.= a+b+c = ck2 + ck + c = c(k2 + k + 1) = k2 + k + 1
Again,
a2 + b2 + c2 (ck2)2 + (ck)2 + c2 c2k4 + c2k2 + c2 c2(k4 + k2 + 1)
R.H.S. = (a + b + c)2 = (ck2 + ck + c)2 = [c(k2 + k + 1)]2 = c2(k2 + k + 1)2
= (k2)2 + (1)2 + k2 = (k2 + 1)2 – 2.k2.1 +k2 = (k2 + 1)2 – k2
(k2 + k + 1)2 (k2 + k + 1)2 (k2 + k + 1)2
= (k2 + k + 1) (k2 – k + 1) = k2 – k + 1
(k2 + k + 1)2 k2 + k + 1
Hence, L.H.S. = R.H.S. Proved.
Example 8: If b is the mean proportional between a and c, prove that
Solution: abc(a + b + c)3 = (ab + bc + ca)3.
Here, b is the mean proportional between a and c. Then, b2 = ac.
L.H.S. = abc(a + b +c)3 = ac.b(a + b + c)3 = b2.b(a + b + c)3 = b3(a + b + c)3
R.H.S. = (ab + bc + ca)3 = (ab + bc + b2) = {b(a + b + c)}3 = b3(a + b + c)3
Hence, L.H.S. = R.H.S. Proved.
Alternative process:
Here, b is the mean proportional between a and c. i.e. a : b : : b : c
a b
b = c = k (suppose), then, b = ck and a = bk = ck.k = ck2
L.H.S. = abc(a + b + c)3 = ck2.ck.c(ck2 + ck + c)3 = c3k3{c(k2 + k + 1)}3
= c3k3.c3(k2 + k + 1)3 = c6k3(k2 + k + 1)3
R.H.S. = (ab + bc + ca)3 = (ck2.ck + ck.c + c.ck2)3 = (c2k3 + c2k + c2k2)3
= {c2k(k2 + 1 + k)}3 = c6k3(k2 + k + 1)3
Hence, L.H.S. = R.H.S.
Example 9: If x, y, and z are in continuous proportion, prove that
Solution: x2y2z2 1 + 1 + 1 = (x3 + y3 + z3)
x3 y3 z3
Here, x = y , then, y2= xz, and (y2)3 = x3z3 , i.e. y6 = x3z3
y z
Now, L.H.S. = x2y2z2 1 + 1 + 1 = x2y2z2 y3z3 + x3z3 + x3y3 = y3z3 + y6 + x3y3
x3 y3 z3 x3y3z3 xyz
= y3 (z3 + y3 + x3) = x3 + y3 + z3 = R.H.S. proved.
y2.y
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Ratio and Proportion
Example 10: If a = b = c , prove that a3 + b3 + abc = a
Solution: b c d b3 + c3 + bcd d
Here, a = b = c = k
b c d
Then, c= dk, b = ck = dk.k = dk2, a = bk = dk2.k = dk3
Now, L.H.S.
= a3 + b3 + abc = (dk3)3 + (dk2)3 + dk3.dk2.dk
b3 + c3 + bcd (dk2)3 + (dk)3 + dk2.dk.d
= d3k9 + d3k6 + d3k6 = d3k9 + 2d3k6 = d3k6(k3 + 2) = k3
d3k6 + d3k3 + d3k3 d3k6 + 2d3k3 d3k3(k3 + 2)
R.H.S. = a = dk3 = k3
d d
Hence, L.H.S. = R.H.S. Proved
Example 11: If a = b = c prove that (a2 + c2) (b2 + d2) = (ab + cd)2
Solution: b c d
Here, a = b = c , then, b2= ac, c2 = bd and ad = bc
b c d
Now, L.H.S. = (a2 + c2) (b2 + d2)
= a2b2 + a2d2 + b2c2 + c2d2
= a2b2 + b2c2 + b2c2 + c2d2 (ad = bc and a2d2 = b2c2)
= a2b2 + 2b2c2 + c2d2
= a2b2 + 2bc.bc + c2d2
= a2b2 + 2ad.bc + c2d2 (bc = ad)
= (ab)2 + 2abcd + (cd)2 = (ab + cd)2 = R.H.S. Proved
Example 12: If a = b = c , prove that
b c d
ab – bc + cd = (a – b + c) (b – c + d)
Solution:
a b c
Let, b = c = d =k (suppose)
Then, c = dk, b = ck = dk2, a = bk = dk3
Now, L.H.S. = ab – bc + cd
= dk3.dk2 – dk2.dk + dk.d
= dk2 k – dk k + d k = d k (k2 – k + 1)
R.H.S. = (a – b + c) (b – c + d)
= (dk3 – dk2 + dk) (dk2 – dk + d)
= (k2 – k + 1). d (k2 – k + 1)
= d k (k2 – k + 1)2 = d k (k2 – k + 1)
? L.H.S. = R.H.S. proved.
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Ratio and Proportion
Example 13: If a : b : : b : c : : c : d, prove that (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2.
Solution:
Here, a = b = c , b2 = ac, c2 = bd and ad = bc
b c d
Now, L.H.S. = (b – c)2 + (c – a)2 + (d – b)2
= b2 – 2bc + c2 + c2 – 2ca + a2 + d2 – 2bd + b2
= 2b2 + 2c2 + a2 + d2 – 2bc – 2ca – 2bd
= 2ac + 2bd + a2 + d2 – 2ad – 2ac – 2bd
= a2 – 2ad + d2
= (a – d)2 = R.H.S. Proved
EXERCISE 10.2
General section
1. a) If 6, 9, 16, and x are in proportion, find the fourth proportional x.
b) If 8, 10, x, and 35 are in proportion, find the third proportional x.
c) If 15, x, 40, and 24 are in proportion, find the second proportional x.
d) If x, 20, 45, and 60 are in proportion, find the first proportional x.
e) If 10, 40, 64, are in proportion, find the second proportional.
f) If 12, 15, 60, are in proportion, find the fourth proportional.
g) Find the mean proportional between 9 and 25 in a continued proportion.
a2
h) If (a – b), b and (a + b) are in continuous proportion, show that b2 = 2.
i) If (x + 4) is the mean proportional between (x + 2) and (x + 7) find the value of x.
2. a) What number should be added to each term 7, 10, 16, and 22 so that they are in
proportion?
b) What number should be subtracted from each term 11, 10, 29, and 26 so that they
are in proportion?
c) What number should be added to each term 2, 8, and 17 so that they are in a
continued proportion?
d) What number should be subtracted from each term 7, 17 and 47 so that they are in
a continued proportion?
Creative section A
3. If a : b = c : d, prove that
a) a + b = c + d b) a – b = c –d c) a – b = c – d
b d b b a + b c + d
d) 3a + 5b = 3a – 5b e) ma + nb = a f) a2 + c2 = ac
3c – 5d 3c – 5d mc + nd c b2 + d2 bd
g) c2 = a2 + c2 h) ab – cd = a2 – c2
d2 b2 + d2 ab + cd a2 + c2
i) a2 – ac + c2 = b2 – bd + d2 j) (a2 – c2) (b2 – d2) = (ab – cd)2
a2 + ac + c2 b2 + bd + d2
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Ratio and Proportion
k) a = a2 + b2 l) xa – yc = a2 + c2
b c2 + d2 xb – yd b2 + d2
m) a2 – c2 = a4 – c4 n) ac = a3 + c3 o) a2 + c2 = a4 + c4
b2 – d2 b4 – d4 bd 3 b3 + d3 b2 + d2 4 b4 + d4
4. If a = c = e , prove that
b d f
a) a+c+e = c+e b) 2a – 3c + 5e = e c) a2 + c2 + e2 = ac
b+d+f d+f 2b – 3d + 5f f b2 + d2 + f2 bd
d) ac + ce + ae = a2 e) a3 + c3 + e3 = ace f) (a + c – e)3 = a(a – c + e)2
bd + df + bf b2 b3 + d3 + f3 bdf (b + d – f)3 b(b – d + f)2
g) la + mc + ne = ace h) a3c3 + c3e3 + a3e3 = ace
lb + md + nf 3 bdf b3d3 + d3f3 + b3f3 bdf
5. If x = y = z , prove that
a b c
a) (x + y + z)3 = x3 + y3 + z3 b) x3 + y3 + z3 = xyz
(a + b + c)2 a2 b2 c2 a3 + b3 + c3 abc
c) y–z 3 x3 – y3 d) x2 – yz = a2 – bc
b–c a3 – b3 y2 – zx b2 – ac
=
6. If a = b prove that
b c
a a–b a+b 2 a2 + b2 (a – b)2 (b – c)2
a) a+b = a–c b) b+c b2 + c2 c) a = c
=
d) a2 + ab = b2 + bc e) (a + b)2 = (b + c)2 f) a+b+c = (a + b + c)2
b2 c2 b2 c2 a–b+c a2 + b2 + c2
g) a2 + ab + b2 = a h) a3 + b3 = a (a – b) i) a+b = a2 (b – c)
b2 + bc + c2 c b3 + c3 c (b – c) b+c b2 (a – b)
7. a) If a, b, c are in continuous proportion, show that (a2 + b2) (b2 + c2) = b2 (a + c)2.
[Hint: a = b , i.e. b2 = ac]
b c
b) If a, b, c are in continuous proportion, show that (ab + bc + ca)3 = abc (a + b + c)3
[Hint: a = b , i.e. b2 = ac. Then, L.H.S. = (ab + bc + b2)3]
b c
c) If x, y, z are in continuous proportion, prove that
x2y2z2 1 + 1 + 1 = x3 + y3 + z3
x3 y3 z3
p q
d) If q = r , show that p3q3 + q3r3 + r3p3 = pqr (p3 + q3 + r3)
[Hint: p = q , i.e. q2 = pr, then L.H.S. = p3.pqr + r3.pqr + q3.pqr]
q r
e) If b is the mean proportional between a and c, prove that
(a + b – c) (a + b + c) = a2 + b2 + c2.
f) If y is the mean proportional between x and z, prove that (x – y)2 = (y – z)2.
x z
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Ratio and Proportion
8. If a = b = c , prove that
b c d
a) a2 + b2+ ca = a+b b) a–b = a2 + b2 + c2
b2 + c2 + bd c+d c–d b2 + c2 + d2
c) ab – cd = b2 – d2 d) a2 – ab = b2 – bc = c2 – cd
ab + cd b2 + d2 b2 c2 d2
e) a3 + b3+ c3 = a f) a3 + b3+ abc = a
b3 + c3 + d3 d b3 + c3 + bcd d
g) a3 – b2c+ c3 = a3 h) (ax – by – cz)3 = bc
abc – bd2 + d3 b3 (bx – cy + dz)3 d2
i) (a + b + c) (b + c + d) = ab + bc + cd
j) ab – bc + cd = (a – b + c) (b – c + d)
Creative section B
9. If a : b : : b : c : : c : d, prove that
a) (b + c) (b + d) = (c + a) (c + d)
b) (a + d) (b + c) – (a + c) (b + d) = (b – c)2
c) (a – c)2 + (b – c)2 + (b – d)2 = (a – d)2
d) (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2
10. a) If a, b, c, and d are in continues proportion, prove that (a2 – b2) (c2 – d2) = (b2 – c2)
b) If p, q, r and s are in continuous proportion show:
(p2 + q2 + r2) : (q2 + r2 + s2) = (p2 + qr) : (q2 + rs)
c) If a, b, c, d, and e are in continuous proportion, prove that a : e = a4 : b4.
Project work
11. a) Take any two ratios of positive integers and find the compounded ratio.
b) Take a ratio of any two positive integers and find:
(i) duplicate and sub-duplicate ratios (ii) triplicate and sub-triplicate ratios
c) How many students are there in your class? Write the ratio of the number of boys
and girls. Then, make an equation to find the number of boys and girls.
d) How many teachers are there in your school? Write the ratio of the number of
male and female teachers. Then, make an equation to find the number of male and
female teachers.
12. a) Take any four positive integers which are in proportion. Then, verify the following
properties of proportion.
(i) Invertendo (ii) Alternendo (iii) Componendo
(iv) Dividendo (v) Componendo and dividendo (vi) Addendo
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Ratio and Proportion
Objective Questions
1. Which one is the equal expression to a2 + b2?
(A) (a + b)2 – 2ab (B) (a – b)2 + 2ab
(C) (a + b) (a – b) (D) Both (A) ands (B)
2. The factorization of x4 + x2y2 + y4 is
(A) (x2 – xy + y2) (x2 + xy + y2) (B) (x2 – xy + y2) (x2– xy + y2)
(C) (x2 + xy – y2) (x2 + xy + y2) (D) (x2 + xy – y2) (x2 + xy – y2)
3. What is the HCF of p2 – 1 and p3 + 1?
(A) p + 1 (B) p – 1 (C) (p – 1) (p3 +1) (D) (p + 1) (p3 – 1)
4. If m + n = m, what is the value of mn?
(A) 0 (B) 1 (C) m (D) n
5. What is the value of a0 – 2 ? (A) 4 (B) 16 (C) 64 (D) 512
64 3
6. The value of 5m+2 – 5m is
5m+1 – 5m
(A) 2 (B) 4 (C) 6 (D) 8
7. For what value of x, will the equation 2x + 2x + 1 = 12 be satisfied?
(A) 0 (B) 1 (C) 2 (D) 3
8. If 23x – 5 × ax – 2 = 2x – 2 × a1 – x, what is the value of 4x?
(A) 1 (B) 1.5 (C) 4 (D) 64
9. The sum of two numbers is 30. If twice the bigger number is thrice the smaller, the
smaller is
(A) 6 (B) 8 (C) 12 (D) 18
10. According to marriage ritual, a groom takes the bride to his house. If Ram’s daughter gets
married to Sita’s son, both the family will have equal members. But, if Ram’s son gets
married with Sita’s daughter, Ram’s family will be twice as large as Sita’s family. How
many members are there in Ram and Sita’s family?
(A) 4, 3 (B) 5, 3 (C) 7, 5 (D) 9, 5
11. The equation ax2 + bx + c = 0, a ≠ 0 becomes pure quadratic if -
(A) b < 0 (B) b > 0 (C) b = 0 (D) c = 0
12. What are the roots of the quadratic equation ax2 + bx + c = 0, a ≠ 0?
(A) – b ± b2 – 4ac (B) – a ± b2 – 4ac (C) – b ± b2 + 4ac (D) – b ± b2 – ac
2a 2b 2a a
13. What are the roots of the equation x2 + 1 = 5 (x – 1) ?
(A) 1, 2 (B) 2, 3 (C) 1, 5 (D) 1, 6
14. The duplicate ratio of 4:9 is
(A) 2:3 (B) 4:9 (C) 16:81 (D) 64:729
15. The number of girls and boys in a school are in the ratio 8:9. If 24 new girls join and
7 boys leave the school; the ratio becomes 9:8. How many students are there in the
school?
(A) 120 (B) 135 (C) 255 (D) 272
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Unit Geometry - Triangle
11
11.1 Various types of angles B
A
In the given figure, straight line segments AO and Bo meet at the O
point O to from AOB. Here, O is the vertex, AO and BO are the
arms of AOB.
Types of angles on the basis of their sizes
According to the size of angles, they are classified in the following ways.
Acute angle Right angle Obtuse angle Straight angle Reflex angle Complete turn
B B B O A OA
OA B Exactly 360°
OA OA QO P
Less than 90° Exactly 90° Greater than 90°
but less than 180° Exactly 180° Greater than
180° but less
than 360°
Types of pairs of angles
Following are the different types of pairs of angles according to their construction and
properties.
Adjacent angles Vertically opposite Complementary Supplementary
angles angles angles
DC C B
B
C B O B 180°
B A OA
O AC O A AOB + BOC = 90° CO A
AOB and BOC
AOB and BOC are a AOB and COD are complementary AOB + BOC = 180°
pair of adjacent angles. AOD and BOC are angles. AOB and BOC are
two pairs of vertically supplementary angles.
opposite angles.
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Geometry - Triangle
Pairs of angles between two line segments intersected by a transversal
When a transversal intersects two straight line segments at two distinct points, following
pairs of angles are formed.
Co-interior angles Alternate angles Corresponding angles
ac ac ac ac ac ae
bd bd db db gc
b
bf
d hd
Fig. 1
Fig. 2
Fig. 1 Fig. 2 Fig. 1 Fig. 2 a and b, c and d are
a and b, c and d are two pairs a and b, c and d are two pairs of corresponding
of co-interior angles. two pairs of alternate angles. angles.
The pair of corresponding
The sum of a pair of co-interior The pair of alternate angles angles between parallel line
segments are equal.
angles between parallel line between parallel lines
segments is 180°. segments are equal.
In figure 2, In figure 2, In figure 2
a + b = 180° and c + d = 180° a = b and c = d a = b and c = d
e = f and g = h
11.2 Axioms and postulates
Axioms
Axioms and postulates are the most important building blocks of geometric proofs.
An axiom is a statement that is taken to be true, to serve as a premise or starting point
for further reasoning and arguments. In other words, an axiom is a self-evident truth.
Following are a few examples of axioms.
1. The reflective axiom: It states that any quantity is equal to itself. For example,
lines segments, angles, and polygons are always equal to themselves.
2. The transitive axiom: It states that if two quantities are both equal to a third
quantity, then they are equal to each other. For example,
If A = B and B = C, then, A = B = C
3. The substitution axiom: It states that if two quantities are equal, then one can
be replaced by the other in any expression, and the result won’t be changed. For
example, If A = B and B + C = 180° , then, A + C = 180°.
4. The partition axiom: It states that a quantity is equal to the B P
sum of its parts. It is well known as ‘whole part axiom’ in
geometry. OA
For example, in the given figure, AOB = AOP + POB
5. The addition, subtraction, multiplication, and division axioms:
(i) Axiom of equality of addition
When the equal quantities are added to both sides of equal quantities, the sum
is also equal.
If a = b, then a + c = b + c
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(ii) Axiom of equality of subtraction
When the equal quantities are subtracted from both sides of equal quantities,
the difference is also equal.
If a = b, then a – c = b – c
(iii) Axiom of equality of multiplication
When both sides of the equal quantities are multiplied by equal quantities, the
product is also equal.
If a = b, then a × c = b × c
(iv) Axiom of equality of division
When both sides of the equal quantities are divided by equal quantities, the
quotient is also equal.
If a = b, then a = b
c c
Postulates
A statement that is accepted as true without any proof is called a postulate. A postulate
forms the basis of a theory. A few geometrical postulates which are used as the basis
to prove different theorems are given below.
(i) Only one straight line can be drawn by joining any two points.
(ii) Infinite number of straight lines can be drawn from a point.
(iii) Two straight lines can intersect only at a point.
(iv) A straight line can be extended infinitely to both directions.
(v) A line segment can have only one mid-point.
(vi) Only one perpendicular can be drawn from a point to a line.
(vii) Only one line can be drawn from a point parallel to the given straight line.
(viii) The distance between a point and a straight line is the length of the perpendicular
drawn from the point to the line.
(ix) Only one straight line can be the bisector of an angle.
EXERCISE 11.1
General section
1. Identify the types of these angles on the basis of their sizes. Then, list them separately:
a) b) R c) d)
B A Y X
G
QP
C Z FE
e) f) A BX R
QO P g) h)
Q
Y O W O
CD Z P
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2. Answer the following questions:
a) How many straight lines can be drawn by joining two given points?
b) At how many points can two straight lines be intersected?
c) If x = y and y = z, is z = x true or false? What type of axiom is it?
d) If an angle is half of its complement, find the angles.
e) If an angle is three times of its supplement, find the angles.
3. a) In the adjoining figure, QPS = QRS. Show that S RT
QPS + QRT = SRT mentioning necessary axioms.
=fig32ureCaOloDn.gsSidheo,wtAhaOtBC=OD21 P Q
b) In the BOC and C
BOC B
= 1 AOD
2 OA
mentioning necessary axioms. D
c) In the given figure, w = x and x = y. Show that zy
w + z = y + z mentioning necessary axioms. wx
11.3 Types of triangle - review
Types of triangles by angles
On the basis of the sizes of angles, there are three types of triangles:
(i) Acute angled triangle (ii) Obtuse angled triangle (iii) Right angled triangle
AA A
BC BC BC
It has three acute angles. It has one obtuse angle. It has one right angle.
Types of triangles by sides
On the basis of the length of the sides of triangles, there are three types of triangles.
(i) Scalene triangle (ii) Isosceles triangle (iii) Equilateral triangle
AA A
BC BC BC
It has none of the It has two It has all three
sides equal. sides equal. sides equal.
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11.4 Median and altitude of a triangle
A median of a triangle is a straight line that joins a vertex of a triangle to the
mid-point of its opposite side. A
AD, BE, and CF are the medians of ∆ ABC. All three F E
medians of a triangle are concurrent and the meeting
point of medians is called the centroid of the triangle. In G
DC
the figure, G is the centroid of ∆ ABC. The centroid of a B
triangle divides each median in the ratio of 2:1.
A perpendicular drawn from a vertex to the opposite side of a triangle is known as the
altitude (or height) of the triangle. A
AD, BE, and CF are the altitudes of ∆ABC. The altitudes F E
O C
of a triangle are also concurrent and the meeting point of D
altitudes of a triangle is called Orthocentre of the triangle. In
the figure, O is the orthocentre of the triangle ABC. B
Following are the important facts about the medians and altitudes of different types
of triangles.
(i) In an equilateral triangle, its medians are also the altitudes of the triangle or
vice versa.
(ii) In an isosceles triangle, the median drawn from its vertex to the base is also its
altitude.
(iii) In a right-angled triangle, the median drawn to its hypotenuse is half of the
hypotenuse in length.
(iv) Each median of a triangle divides the triangle in two triangles having equal
area.
11.5 Properties of triangles
Following are a few important properties of triangles. We can verify these properties
experimentally as well as theoretically.
(i) The sum of the angles of a triangle is equal to two right angles (180°).
(ii) The exterior angle of a triangle is equal to the sum of two opposite interior
angles.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) In any triangle, the angle opposite to longer side is greater than the angle opposite
to shorter side.
(v) The base angles of an isosceles triangle are equal.
(vi) In an isosceles triangle, perpendicular drawn from the vertical angle to the base
bisects the base.
(vii) A straight line that joins the mid-points of any two sides of a triangle is parallel
to the third side of the triangle, ... and so on.
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Theorem 1
The sum of the angles of any triangle is equal to two right angles.
Experimental verification
Step 1: Three triangles with different shapes and sizes are drawn. Suppose the name of
each triangle is ABC.
Step 2: All three angles of each triangle are measured and the measurements are tabulated
in the table.
Figure A B C A + B + C Result
(i) A + B + C = 180°
(ii) A + B + C = 180°
(iii) A + B + C = 180°
Conclusion: The sum of the angles of any triangle is equal to two right angles.
Theoretical proof
Given: ABC, BCA and BAC are the angles of ∆ABC.
To prove: ABC + BCA + BAC = 2 rt. angles
Construction: Through the vertex A, a straight line XY parallel to BC is
drawn.
Proof:
Statements Reasons
1. XAB = ABC 1. XY // BC and alternate angles
2. YAC = BCA 2. XY // BC and alternate angles
3. XAB + BAC + YAC = XAY 3. Whole part axiom
4. XAB + BAC + YAC = 2 rt. angles 4. XAY is a straight angle
5. ABC + BAC + BCA = 2 rt. angles 5. From statements (1), (2) and (4)
Proved
Theorem 2
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn. A side BC is
produced to D in each triangle.
Step 2: The exterior angle ACD and the opposite interior angles ABC and BAC are measured
in each triangle. The measurements are tabulated in the table.
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Figure ABC BAC ABC + BAC ACD Result
(i) ABC + BAC = ACD
(ii) ABC + BAC = ACD
(iii) ABC + BAC = ACD
Conclusion: The exterior angle of a triangle is equal to the sum of the two opposite interior
angles.
Theoretical proof A
CD
Given: In ∆ ABC, the side BC is produced to D. ACD is the
exterior angle so formed. ABC and BAC are the
opposite interior angles. B
To prove: ACD = ABC + BAC
Proof
Statements Reasons
1. BCA + ACD = BCD 1. Whole part axiom
2. BCA + ACD = 180° 2. BCD is a straight angle
3. ABC + BCA + BAC = 180° 3. Sum of the angles of a triangle
4. BCA + ACD = ABC + BCA + BAC 4. From statements (2) and (3)
5. ACD = ABC + BAC 5. Cancelling BCA from both
sides of statement (4).
Worked-out examples Proved
Example 1: In the figure given alongside, find the values of x and y. A
Solution: 40° x
(i) In 'ABD, BAD + ABD + ADB = 180° 100° D
60°
or, 40° + y° + 100° = 180° yx C
? y = 180° – 140° = 40° B
(ii) In 'ABC, BAC + ABC + ACB = 180°
(40° + x) + (y + x) + 60° = 180°
or, 2x + 40° + 40° + 60° = 180°
or, 2x = 40° A
56°
? x = 20° D
x
Hence, x = 20° and y = 40°.
Example 2: In the figure BD and CD are the bisectors of B C E
ABC and ACE respectively. If BAC = 56°, A D
find the value of x. 56° x
E
Solution:
(i) Here, suppose ABD = DBC = a and ACD = DCE = b.
In ABC, BAC + ABC = ACE ab
or, 56° + (a + a) = (b + b) Ba b
C
[Being exterior angles of triangle equal
to the sum of opposite interior angles]
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or, 56° + 2a = 2b
or, 2(28° + a) = 2b
? b = 28° + a .......... (i)
(ii) In 'BCD, DBC + BDC = DCE
a+x =b
or, a + x = 28° + a [From (i), b = 28° + a]
? x = 28°
Hence, x = 28° D 50° C x E
z B
Example 3: In the figure given alongside, find the values of x,
P R
y ,and z. y O
Solution: A 30°
(i) x = 30° [Being DE // AB and alternate angles]
(ii) 30° + y = 50° [Being DE // AB and alternate angles]
or, 30° + y = 50°
or, y = 50° – 30° = 20°
(iii) z = 90° – (30° + y) [In rt. angled ∆ ABC, ABC = 90° – CAB]
or, z = 90° – 50°
= 40°
So, x = 30°, y = 20° and z = 40°
Example 4: In the adjoining figure, the bisectors of PQR and
Solution: PRQ meet at O. Prove that QOR = 90° + 2P . Q
Given: In 'PQR, QO bisects PQR and RO bisects PRQ.
i.e. PQR = 2OQR and PRQ = 2ORQ.
To prove: QOR = 90° + 2P .
Statements Reasons
1. PQR = 2OQR and PRQ = 2ORQ 1. Given
2. PQR + PRQ + QPR = 180° 2. Being sum of angles of 'PQR
2OQR + 2ORQ + P = 180°
3. or, 2(OQR + ORQ) = 180° – P 3. From statements (1) and (2)
?or,OOQQRR++OORRQQ==9210(°1–80°2–P P) 4. Being sum of angles of 'QOR.
5. From statements (3) and (4).
4. OQR + ORQ + QOR = 180°
Proved
90° – P + QOR = 180°
2
5. P
or, QOR = 180° – 90° + 2
? QOR = 90° + 2P
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EXERCISE 11.2
General section B A C
1. a) In 'ABC, M is the mid-point of side BC and AP A BC. PM
Then, name the altitude and median of the triangle.
b) In 'XYZ, A, B and C are the mid-points of sides X
XY, YZ and XZ respectively. Then, name the medians A GC
of the triangle. What is the point G called?
Y BZ R
c) In 'PQR. PX A QR, QY A PR and RZ A PQ. P
Then, name the altitudes of the triangle.
What is the point O called? ZY
O
2. Draw the rough sketches for the following:
a) In 'ABC, BD is a median and AE is an altitude. QX
b) In 'LMN, LQ and MR are altitudes.
c) In 'PQR, PT is an altitude in the exterior of triangle. A
3. a) In the figure, ABC = 90°. Name a right angled triangle and an
obtuse angled triangle.
C DB
P
b) In the figure, PQ = PR and 'QSR = 90°. Write the name of an
isosceles triangle and a right angled triangle. S
4. Answer the following questions. Q R
a) If x°, 40, and 85° are the angles of a triangle, find the size of x°.
b) If y°, 2y°, and 30° are the angles of a triangle, find the unknown sizes of angles of
the triangle.
c) If one of the acute angles of a right angled triangle is 55°, calculate the size of the
other acute angle.
d) One of the acute angles of an obtuse angled triangle is one–third of the second one.
If the obtuse angle of the triangle is 120°, find the sizes of the two acute angles of
the triangle.
e) If the angles of a triangle are in the ratio 1: 2 : 3, find them.
f) In 'ABC, 2A = 3B = 6C, find A, B, and C.
g) In 'PQR, P –Q = 20° and Q – R = 50°, find P, Q, and R.
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5. Find the unknown sizes of angles:
a) A b) A c) A
2x b 2x
50°
B 3x xC D 120° a 70° B B 3x+5°
C CD
S e) Px Q f) 3x Q
A C
d) R y
140°
p 2p P 120° a bU A 2x P
2 Rx B C R 4x B
T
Q S
A P
x
g) 64° h) 60° i)
O O 50° R
x y 20° Q yx
Bx
20° C
j) k) l) A
60° C
B D
O
m) P n) 36° o)
A
S R xx
65° y 35°
B DC
QT
p) q) F r)
E
P By A E
x C x
T x 110° 145° 56° 35° S
100° AD CG B R
30° Q D
SR
Creative section - A P
6. a) In the given figure, PQ = PR, and QS is the bisector of PQR.
If PQS = 30°, find PSQ.
Q
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P
b) In the figure alongside, 'ABC is an equilateral triangle B
and 'PBC is an isosceles triangle. If ABP = 20°, find
BCP and BPC.
c) In the adjoining figure, GP bisects BGH and HP bisects
GHD. Calculate the size of a.
d) In the figure alongside, AP and BP A B
P R
are the angular bisectors of BAD
R
and ABC respectively. Show that
APB = 1 rt. angle. D C
e) In the adjoining figure, QO and RO are the angular P
bisectors of PQR and PRQ. Find the size of QOR. 80°
Creative section -B O
Q
7. a) In the adjoining figure, the bisectors of ABC and ACB
of ∆ ABC meet at O. Prove that BOC = 90° + A .
2
b) In the figure alongside, BP and CP are the angular bisectors
of the exterior angles CBE and BCD of ∆ABC. Prove that
BPC = 90° – A .
2
c) In ∆PQR, RX A PQ and QY A PR, RX and QY intersect P Y
at O. Prove that QOR = 180° – P. XO
Q
8. a) In the adjoining figure, ABC is an isosceles triangle. BO and
CO are the bisectors of ABC and ACB respectively. Prove
that BOC is an isosceles triangle.
A
b) In the given figure, AM = BM = CM. M
Prove that 'ABC is a right angled triangle. BC
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D
9. a) In the given figure ABCD, prove that C
BCD = BAD + ABC + ADC.
AB
b) In the adjoining star shaped figure, prove that
A + B + C + D + E = 180° A E
T
PS
B QR D
C
AE
10. a) In the figure alongside, BE and CE are the angular
bisectors of ABC and ACD respectively.
Prove that BAC = 2BEC. B CD
T
b) In the given figure, the bisector of ACU meets AU at O.
U
O
Prove that COT = 1 (CAT + CUT). C A
2 A
c) In the adjoining 'ABC, AY is the bisector of BAC and
AX A BC. Prove that XAY = 1 (B – C). B XY C
2
Theorem 3
The sum of any two sides of a triangle is greater than the third side.
Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn.
Step 2: All three sides of each of ∆ABC are measured and the lengths are tabulated in the
table.
Figure AB BC CA AB + BC BC + CA AB + CA Result
(i) AB + BC > CA,
BC + CA > AB, AB + CA > BC
(ii) AB + BC > CA,
BC + CA > AB, AB + CA > BC
(iii) AB + BC > CA,
BC + CA > AB, AB + CA > BC
Conclusion: The sum of any two sides of a triangle is greater than the third side.
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Theorem 4
In any triangle, the angle opposite to the longer side is greater than the angle opposite to
the shorter side.
Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that
BC is the longest and CA is shortest sides in each triangle.
Step 2: The angle opposite to the greatest side BC and the angle opposite to the shortest side
CA are measured and tabulated in the table.
Figure Angle opposite to BC: A Angle opposite to CA: B Result
(i) A > B
(ii) A > B
(iii) A > B
Conclusion: The angle opposite to the longer side of any triangle is greater than the angle
opposite to the shorter side.
Converse of Theorem 4
In any triangle, the side opposite to the greater angle is longer than the side opposite to the
smaller angle.
Experimental verification
Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that
A is the greatest and B is the smallest angles in each triangle.
Step 2: The side opposite to the greatest angle A and the side opposite to the smallest angle
B are measured and tabulated in the table.
Figure Side opposite to A: BC Side opposite to B: CA Result.
(i) BC > CA
(ii) BC > CA
(iii) BC > CA
Conclusion: The side opposite to the greater angle of any triangle is longer than the side
opposite to the smaller angle.
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Theorem 5
Of all straight line segments drawn to a given line from a point outside it, the perpendicular
is the shortest one.
Experimental verification
Step 1: Three straight line segments XY with different lengths and position are drawn.
A point P is marked outside the each line segment. Three line segments PA, PB, PC
and a perpendicular PQ are drawn from P to XY.
(i) (ii) (iii)
Step 2: The lengths of the line segments PA, PB, PC, and PQ are measured and tabulated in
the table.
Figure Length of the line segments Result
(i) PA PB PC PQ
(ii)
(iii) Perpendicular PQ is the shortest one.
Perpendicular PQ is the shortest one.
Perpendicular PQ is the shortest one.
Conclusion: Of all line segments drawn to a given line from a point outside it, the
perpendicular is the shortest one.
11.6 Congruent triangles
Two or more triangles are said to congruent if they have the same shape and size. In this
case, when one triangle is placed over another triangle, their corresponding parts exactly
coincide to each other. The congruent triangles are always similar and they have equal area.
11.7 Conditions of congruency of triangles
There are 3 sides and 3 angles in a triangle. Two triangles can be congruent if 3 parts (out
of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle as per the
following conditions. We use these conditions as axioms.
(i) S.S.S. axiom
When three sides of a triangle are respectively equal to three corresponding sides of
another triangle, they are said to be congruent.
In ∆s ABC and PQR,
a) AB = PQ (S)
b) BC = QR (S)
c) CA = RP (S)
d) ? ∆ ABC # ∆ PQR [S.S.S. axiom]
Corresponding parts of congruent triangles are also equal.
? A = P, B = Q and C = R.
Note: The parts opposite to the equal parts of triangles are called the corresponding
parts.
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(ii) S.A.S. axiom
When two sides of one triangle and angle made by them are respectively equal to the
corresponding sides and angle of another triangle they are said to be congruent triangles.
In ∆s ABC and PQR [S.A.S. axiom]
a) AB = PQ (S)
b) B = Q (A)
c) BC = QR (S)
d) ?∆ABC # ∆PQR
Now, C = R and A = P [Corresponding angles of congruent triangles]
CA = RP [Corresponding sides of congruent triangles]
(iii) A.S.A. axiom
When two angles and their adjacent side of one triangle are respectively equal to
the corresponding angles and sides of another triangle, they are said to be congruent
triangles.
In ∆s ABC and PQR,
a) B = Q (A)
b) BC = QR (S)
c) C = R (A)
d) ? ∆ABC # ∆PQR [A.S.A. axiom]
Now, AC = PR and AB = PQ [Corresponding sides of congruent triangles]
A = P [Corresponding angles of congruent triangles]
(iv) R.H.S. axiom
In two right angled triangles, when the hypotenuse and one of the two remaining sides
are respectively equal, they are said to congruent triangles.
In right angled ∆s ABC and PQR,
a) B = Q (R)
b) AC = PR (H)
c) BC = QR (S)
d) ? ∆ABC # ∆PQR [R.H.S. axiom]
Now, C = R and A = P [Corresponding angles of congruent triangles]
AB = PQ [Corresponding sides of congruent triangles]
(v) A.A.S. axiom
When two angles and a side of one triangle are respectively equal to the corresponding
angles and sides of another triangle, they are said to be congruent triangles.
This axiom can be verified by using A.S.A. axiom.
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Here, [Given]
a) A = P [Given]
b) B = Q
c) A + B + C = P + Q + R [Sum of the angles of any triangle is 180°]
d) ? C = R [From (a), (b) and (c)]
Now, in ∆s ABC and PQR,
e) B = Q (A) [Given]
f) BC = QR (S) [Given]
g) C = R (A) [from (d)]
h) ? ∆ABC # ∆PQR [A.S.A axiom]
Now, AB = PQ and AC = PR [Corresponding sides of congruent triangles]
EXERCISE 11.3
General section
1. a) Identify and name the longest and the shortest sides of the following triangles.
FR
60°
35° 85° DE PQ
b) Identify and name the greatest and the smallest angles of the following triangles.
B
X E
G
Y CA F
Z
c) In ∆ABC, if A = 85°, B = 65° and C = 30°, name the longest and the shortest
sides of the triangle.
d) In ∆PQR, if Q = 75°, and R = 25°, name the longest and the shortest sides of the
triangle.
e) In ∆XYZ, if XY = 5.6 cm, YZ = 4.7 cm and ZX = 6.5, name the greatest and the
smallest angles of the triangle.
Creative section - A
2. Use the necessary axioms to verify that the following pairs of triangles are congruent.
Also, write the equal corresponding parts of the triangles. Y
a) A P b) D X
B CQ R E FZ
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c) C
A BQ P E FT
3. a) In the adjoining figure, show that
∆ABC # ∆PQR. Also find the unknown
sizes of angles of ∆ABC and ∆PQR.
b) In the figure alongside, show that
∆RST # ∆XYZ. Also find the
unknown sizes of x cm, y cm and
x°.
Creative section - B P
4. a) In ∆PQR, O is the interior point. Prove that O
OP + OQ + OR > 1 (PQ + QR + RP) . Q R
2 A C
O 8 cm
b) In the figure alongside, O is the interior point of 'ABC. 7 cm
If AB = 7 cm, BC = 11 cm and CA = 8 cm, show that B 11 cm
OA + OB + OC > 13 cm.
5. a) In the given figure, AB // PQ and AB = PQ.
Prove that ∆ABO # ∆POQ. Also show that AO = OQ and
BO = OP
b) In the given figure, PQ // SR and PS // QR.
Prove that ∆PQR # ∆PRS.
Also show that PQ = RS and PS = QR.
Project work!
6. a) Take three sticks of different lengths such that the total length of two shorter sticks
is greater than longer stick. Now, make a triangle by joining these sticks.
b) Take three sticks of different lengths such that the total length of two shorter
sticks is less than the longer stick. Can you make a triangle by joining these sticks?
Discuss in your class.
7. a) Fold a rectangular sheet of paper diagonally and cut into two halves.
(i) What type of triangles did you get?
(ii) Are these triangles congruent? Discuss in the class.
(iii) Which are the corresponding sides and angles in these two triangles? Discuss
in the class.
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Theorem 6
If any two sides of a triangle are equal, the angles opposite to them are equal.
Or
Base angles of an isosceles triangle are equal.
Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn.
Step 2: The angles opposite to the equal sides of each triangle are measured and tabulated
in the table.
Figure Angles opposite to the equal sides (Base angles) Result
(i) ABC ACB ABC = ACB
(ii) ABC = ACB
(iii) ABC = ACB
Conclusion: Base angles of an isosceles triangles are equal.
Theoretical proof
Given: In isosceles ∆ABC, AB = AC
To prove: ABC = ACB
Construction: From the vertex A, AD A BC is drawn.
Proof
Statements Reasons
1. In ∆s ABD and ACD 1.
(i) By construction, AD A BC
(i) ADB = ADC (R) (ii) Given
(iii) Common side
(ii) AB = AC (H) (iv) R.H.S. axiom
(iii) AD = AD (S) 2. Corresponding angles of congruent triangle.
(iv) ? ∆ABD # ∆ACD
2. ABD = ACD
i.e. ABC = ACB
Proved
Converse of Theorem 6
If two angles of a triangle are equal, the sides opposite to them are also equal.
Or
It two angles of a triangle are equal, it is an isosceles triangle.
Experimental verification
Step 1: Three line segments AB of different lengths are drawn.
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Step 2: At the points B and C of each line segment, angles of equal sizes are drawn. Let A be
the point at which the arms of the angles meet. Now, 's ABC are formed.
Step 3: The lengths of the sides opposite to the equal angles of each triangle are measured
and tabulated in the table.
Figure Lengths of sides opposite to equal angles Result
AB AC
(i) AB = AC
(ii) AB = AC
(iii) AB = AC
Conclusion: The sides opposite to the equal angles of a triangle are equal, i.e. each of triangle
ABC are isosceles triangles.
Theoretical proof
Given: In ∆ABC, B = C
To prove: AB = AC.
Hence ∆ABC is an isosceles triangle.
Construction: From the vertex A, AD A BC is drawn.
Proof
Statements Reasons
1. In ∆s ABD and ACD
1.
(i) ABD = ACD (A) (i) Given
(ii) By construction, AD A BC
(ii) ADB = ADC (A) (iii) Common side
(iv) A.A.S. axiom
(iii) AD = AD (S)
2. Corresponding sides of congruent triangles
(iv) ? ∆ BD # ∆ACD
2. AB = AC
i.e. ∆ABC is an isosceles triangle.
Proved
Theorem 7
The bisector of the vertical angle of an isosceles triangle is the perpendicular bisector of
the base.
Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a
way that AB = AC in each triangle.
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Step 2: The bisector of A is drawn in each triangle. Suppose it is AD and it meets the
opposite side BC at D.
Step 3: The lengths of BD and DC, and the angles ADB and ADC are measured and tabulated
in the table.
Figure BD DC Result ADB ADC Result
(i) ADB = ADC, AD A BC
(ii) BD = DC ADB = ADC, AD A BC
(iii) ADB = ADC, AD A BC
BD = DC
BD = DC
Conclusion: The angular bisector of the vertical angle of an isosceles triangle is the
perpendicular bisector of its base.
Theoretical proof
Given: ∆ABC is an isosceles triangle in which AB = AC.
AD is the bisector of BAC.
To prove: AD A BC and BD = DC.
Proof
Statements Reasons
1. In ∆ABD and ∆ACD 1.
(i) Given
(i) AB = AC (S) (ii) Given
(iii) Common side
(ii) BAD = CAD (A) (iv) S.A.S. axiom
(iii) AD = AD (S)
(iv) ? ∆ABD # ∆ACD
2. ADB = ADC 2. Corresponding angles of congruent
triangles
3. AD A BC 3. Adjacent angles in linear pair are equal.
4. BD = DC 4. Corresponding sides of congruent
triangles.
5. AD is the perpendicular bisector 5. From statements 3 and 4
of BC.
Proved
Converse of Theorem 7
The straight line joining the vertex and the mid–point of the base of an isosceles triangle is
perpendicular to the base and bisects the vertical angle.
Experimental verification
Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a
way that AB = AC in each triangle.
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Step 2: The mid–point of BC is marked as D. A and D are joined in each triangle.
Step 3: ADB, ADC, BAD, and DAC are measured and tabulated in the table.
Figure ADB ADC Result BAD DAC Result
(i) ADB = ADC, AD A BC BAD = DAC
(ii) ADB = ADC, AD A BC BAD = DAC
(iii) ADB = ADC, AD A BC BAD = DAC
Conclusion: The straight line joining the vertex and the mid-point of the base of an
isosceles triangle is perpendicular to the base and bisects the vertical angle.
Theoretical proof
Given: ∆ABC is an isosceles triangle in which AB = AC.
AD joins the vertex A and mid-point D of the base BC.
To prove: AD A BC and BAD = CAD
Proof
Statements Reasons
1. In ∆ABD and ∆ACD 1.
(i) Given
(i) AB = AC (S) (ii) Common side
(ii) AD = AD (S) Given
S.S.S. axiom
(iii) BD = DC (S) (iii)
(iv)
(iv) ? ∆ABD # ∆ACD
2. (i) ADB = ADC 2. (i) Corresponding angles of congruent
(ii) AD A BC
triangles
(ii) Adjacent angles in linear pair are equal.
3. BAD = CAD 3. Corresponding angles of congruent
triangles
Proved
Worked-out examples
Example 1: Find the values of x and y from the figure given A
alongside. y
Solution: Bx C D
Here, BAC = 90° and AB = AC = CD.
(i) ABC = ACB = x [Base angle of isosceles ∆ ABC]
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(ii) ABC + ACB + BAC = 180° [ Sum of angles of 'ABC]
or, x + x + 90° = 180°
or, 2x = 90q
? x = 45°
(iii) CAD + ADC = y [Base angles of isosceles ∆ ACD]
(iv) CAD + ADC = ACB [Being exterior angle of ∆ACD equal to the
sum of two non-adjacent interior angles]
or, y + y = 45°
or, 2y = 45°
? y = 22.5°
Example 2: In the adjoining figure, find the unknown sizes of angles. 8
Solution:
(i) a = 60° [∆ ABC is an equilateral triangle]
(ii) b + a + 58° = 180° [Size of a straight angle PAQ]
or, b + 60° + 58° = 180°
or, b = 180q – 118q
or, b = 62°
(iii) b + c + ABP = 180° [Sum of the angles of ∆ ABP]
or, 62° + c + 62°
or, c = 180° [PA = PB, so, ABP = b = 62°]
or, c
= 180° – 124
= 56°
(iv) d + 58° + ACQ = 180° [Sum of the angles of ∆ ACQ]
or, d + 58° + 58°
or, d = 180° [AQ = QC, So, ACQ = CAQ = 58°]
or, d
= 180° – 116°
= 64°
Hence, a = 60°, b = 62°, c = 56° and d = 64°. S T
Example 3: In the adjoining figure, find the values of a and b. ab R
Solution: Here, PQ
(i) SPQ = PSQ = a [ PQ = QS]
(ii) SQR = SPQ + PSQ = a + a = 2a [Exterior angle of 'PQS]
(iii) SQR = SRQ = 2a [ SQ = SR]
(iv) SPR + PRS = TSR
or, a + 2a = 90°
? a = 30° [Being parts of straight angle]
(v) PSQ + QSR + RST = 180°
or, a + b + 90° = 180°
or, 30° + b + 90° = 180°
? b = 60°
Hence, a = 30° and b = 60°.
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Example 4: In 'ABC, AB = AC and AP A BC. If AB = (2x + 3) cm, A (3y – 1)cm
AC = (3y – 1) cm, BP = (y + 1) cm and PC = (x + 2) cm, (2x + 3)cm
find the values of x and y.
Solution: 2x – 3y = –4 ... (i) B (y+1)cm P (x+2)cm C
(i) AB = AC or, 2x + 3 = 3y – 1
(ii) BP = PC [ AB = AC and AP A BC]
or, y+1 =x+2 y = x + 1 ... (i)
Now,
Putting the value of y in equation (i) from equation (ii), we get,
2x – 3(x + 1) = –4
or, 2x – 3x – 3 = –4
? x = 1
Again, putting the value of x in equation (ii), we get,
y =1+1=2
Hence, x = 1 and y = 2.
Example 5: In the adjoining figure, ∆CDE is an equilateral A D
E
triangle and ABCD is a square. Prove that
(i) AE = BE (ii) DAE = 15°
Solution: ∆CDE is an equilateral triangle and ABCD is B C
Given: a square.
To prove: (i) AE = BE (ii) DAE = 15°
Proof
Statements Reasons
1. In ∆ADE and ∆BCE 1.
(i) Sides of the equilateral triangle
(i) DE = CE (S) (ii) ADE = 60° + 90° and BCE = 60° + 90°
(iii) Sides of a square
(ii) ADE = DCE (A) (iv) S.A.S. axiom
(iii) AD = BC (S) 2. Corresponding sides of congruent triangles
(iv) ? ∆ADE # ∆BCE 3. DE = DC and AD = DC
2. AE = BE 4. Angles opposite to the equal sides of
∆ADE
3. AD = DE
5. Sum of the angles of ∆AED
4. DAE = AED
5. DAE + AED + ADE = 180°
or, DAE + AED + 150° = 180°
or, DAE = 15°
Proved
A
Example 6: In the given figure, AB = AC and BD = EC. Prove
that ∆ ADE is an isiscels triangle.
Solution:
Given: AB = AC and BD = EC B D EC
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To prove: (i) ∆ADE is an isosceles triangle.
Proof
Statements Reasons
1. In ∆ABD and ∆ACE 1.
(i) Given
(i) AB = AC (S) (ii) Base angles of an isosceles ∆ABC.
(iii) Given
(ii) ABD = ACE (A)
2. By S.A.S. axiom
(iii) BD = EC (S)
3. Corresponding sides of congruent triangles
2. ? ∆ABD # ∆ACE
4. From statement (3)
3. AD = AE
4. 'ADE is an isosceles triangle
Proved
Example 7: If the perpendiculars drawn from any two vertices to
their opposite sides of a triangle are equal, prove that
the triangle is an isosceles triangle.
Solution:
Given: In ∆ABC, CP A AB, BQ A AC and CP = BQ
To prove: (i) ∆ABC is an isosceles triangle.
Proof
Statements Reasons
1.
1. In ∆PBC and ∆QBC (i) Both are right angles
(ii) Common sides
(i) BPC = BQC (R) (iii) Given
2. By R.H.S. axiom
(ii) BC = BC (A) 3. Corresponding angles of congruent triangles
(iii) CP = BQ (S) 4. Base angles are equal.
2. ? ∆PBC # ∆BQC
3. PBC = QCB
i.e. ABC = ACB
4. 'ABC is an isosceles triangle
Proved
EXERCISE 11.4 A C
General section P R
1. a) In the given figure, AB = AC, write down the relation between X
ABC and ACB.
B
b) In the figure alongside, PQR = PRQ, what is the relation
between PQ and PR?
Q
c) In the adjoining figure, if XY = XZ and YA = AZ, write down
the relation between XA and YZ. YAZ
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L
d) In the given figure, if LM = LN and LP A MN, write down the
relation between MP and PN, and LMP and LNP.
MP N
O
e) In the figure alongside, ON = OE and NOT = TOE, write
down the relation between, NT and TE, OT and NE.
NT E
2. Find the unknown sizes of angles in the following figures:
P A E
W
Q RS X YZ B CD F G H
D K A
C
A P
B CL M BD
i) A j) T N E l) X
D
B xx Py k)
xy
69° y 40° x V
D B 330°0° U y
x 30° RA CY Z
CQ S
Creative section - A
3. a) In the given figure, find the sizes of x, y, and z.
b) In the adjoining figure, calculate the sizes of 50
p, q, r, and s.
c) In the given figure, AB = AC and
ACD = 110°, what will be the measure of
BAC + ACB?
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A
4. a) In the given figure, AB = AC, BAC = 44° and ACD = 24°, D 44°
show that BC = CD. 24°
B C
P
b) In the figure alongside, SR = QR, QPR = 48° and S 48°
PRS = 18°, show that PQ = PR.
18° (3y+1)cm (5y–2)cm
Q R
P
(x+8)cm
5. a) In the adjoining figure, find the values of x and y.
(3x+1)cm
Q(x+4)cm T (y+3)cmR
A
b) In 'ABC, AD bisects A and it is perpendicular to base BC.
If AB = (3x + 1) cm, AC = (5y – 2) cm, BD = (x + 1) cm and
DC = (y + 2) cm, find the values of x and y.
B(x+1)cm D (y+2)cmC
Creative section - B
6. a) In the given figure, AB = AC and BD = EC. Prove that AD = AE.
b) In the given figure, AB = AC, BD = EC and DAE = 30°. 30°
Prove that ∆ADE is an isosceles triangle. Also calculate the P
size of ADE.
7. a) In the given figure, PQ = PR, QO bisects PQR and RO bisects O
A
PRQ. Prove that OQR is an isosceles triangle. Q R
b) In the figure alongside, BD and CD are bisectors of ABC D
C
and ACB respectively. If BD = CD, prove that 'ABC is an B
isosceles triangle. Q
T
R
P
c) In the given figure PQRS, SP = RQ
and RP = SQ. Prove that RT = ST.
S
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8. a) In the given figure, X is the mid-point of QR, XA A PQ,
XB A PR and XA = XB. Prove that 'PQR is an isosceles triangle.
A
b) In the given figure, BN A AC, CM A AB and BN = CM. Prove that M N
'ABC is an isosceles triangle. C
B
c) In the given triangle ABC, AB = AC, BP A AC and CQ A AB.
Prove that (i) BP = CQ (ii) OP = OQ.
9. a) In the figure alongside, APB and AQC are equilateral X Y
triangles. Prove that PC = BQ. A
(Hint: ∆ APC # ∆ AQB, then PC = BQ)
b) In the figure alongside PABQ and ACYX are squares. P
Prove that PC = BX.
QB D
A
E
c) In the adjoining figure ABC is an equilateral triangle and BCDE is a
square. Prove that AE = AD.
B C
S R
d) In the figure alongside PQRS is a square in which PA and A
SB intersect at O. If PA = SB, prove that PA and SB are
perpendicular to each other at O. O Q
B
(Hint: ∆ PBS # ∆ PQA by RHS, then show that SOA = SPO + OPB = 90°) P
10. a) In the figure alongside, AB = AC and AD bisects CAE.
Prove that AD // BC.
b) In the triangle given alongside, PQ = PR. The bisector of
PQR meets PR at S. Prove that PSQ = 3PQS.
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c) In the given figure, BE = EC and CE is the bisector of
ACB. Prove that BEC = ACD.
11. a) In the adjoining figure, AD is the bisector of
BAC and AD // EC. Prove that AC = AE.
A M
b) In 'ABC, ABC = 90° and M is the mid-point of AC.
Prove that AM = BM = CM.
BC
12. a) If the perpendiculars drawn from any two vertices to their opposite sides of a
triangle are equal, prove that the triangle is an isosceles triangle.
b) If the perpendiculars drawn from the mid-point of any side to the other two sides
of a triangle are equal, prove that the triangle is an isosceles triangle.
c) Prove that the line joining the point of intersection of two angular bisectors of the
base angles of an isosceles triangle to the vertex bisects the vertical angle.
d) If the angular bisector of an angle of a triangle bisects the opposite side, prove that
the triangle is an isosceles triangle.
Project work
13. a) Take a square sheet of paper and fold it through its one diagonal. Cut out it through
the diagonal and get two right-angled triangles. Now, fold each right-angled triangle
bisecting the vertical angle.
(i) Is the folding perpendicular to the base in each triangle?
(ii) Does the folding bisect the base in each triangle?
(iii) What conclusion can you make from these activities? Discuss in your class.
b) Draw a triangle of your own. Produce its all sides as shown in the
diagram and get three exterior angle of triangle. Show that sum of
three exterior angles of a triangle is 360°.
11.8 Mid-point theorems
Theorem 8
A straight line segment joining the mid–points of any two sides of a triangle is parallel to
the third side and it is equal to half of the length of the third side.
Experimental verification
Step 1: Three triangles ABC of different shapes and sizes are drawn.
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Step 2: The mid-points P of the side AB and Q of the side AC are marked in each triangle
and P, Q are joined.
Step 3: In each triangle the corresponding angles APQ and ABC are measured. Also,
the lengths of PQ and BC are measured and tabulated the measurements.
Figure APQ ABC Result PQ BC Result
(i) APQ = ABC PQ = BC
? PQ // BC
(ii) APQ = ABC PQ = BC
? PQ // BC
(iii) APQ = ABC PQ = BC
? PQ // BC
Conclusion: A straight line segment joining the mid–points of any two sides of a triangle is
parallel to the third side and it is equal to half of the length of the third side.
Theoretical proof
Given: P and Q are the mid–points of the sides AB and
AC respectively of the triangle ABC.
To prove: PQ // BC and PQ = BC
Construction: PQ is produced to R. R and C are joined such that
CR // BP.
Proof
Statements Reasons
1. In ∆s APQ and CRQ
1.
(i) AQP = CQR (A) (i) Vertically opposite angles
(ii) Q is the mid–point of AC (Given).
(ii) AQ = QC (S)
(iii) PAQ = QCR (A) (iii) CR // BA and alternate angles
(iv) ? ∆APQ # ∆CRQ (iv) A.S.A. axiom
2. AP = RC 2. Corresponding sides of congruent triangles
3. AP = PB 3. P is the mid–point of AB (Given).
4. ? PB = RC 4. From statements (2) and (3)
5. BP // CR 5. By construction
6. ? PR // BC, i.e. PQ // BC 6. Being PB = RC and BP // CR.
and PR = BC 7. Corresponding sides of congruent triangles
7. PQ = QR
8. PQ = PR 8. From statement (7)
9. PQ = BC 9. From statement (6)
Proved
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Converse of Theorem 8
A straight line segment drawn through the mid–point of one side of a triangle and parallel
to another side bisects the third sides.
Experimental verification
Step 1: Three triangles ABC of different shapes and sizes are drawn.
Step 2: The mid–point P of side AB in each triangle is marked and PQ // BC is drawn.
Step 3: AQ and QC are measured and tabulated in the table.
Figure AQ QC Result
(i) AQ = QC
(ii) AQ = QC
(iii) AQ = QC
Conclusion: A straight line segment drawn through the mid–point of one side of a triangle
and parallel to another side bisects the third side.
Theoretical proof
Given: P is the mid–point of the side AB of ∆ ABC, PQ // BC.
To prove: Q is the mid–point of AC, i.e. AQ = QC.
Construction: PQ is produced to R. R and C are joined such that
CR // BP.
Proof
Statements Reasons
1. BCRP is a parallelogram 1. BP // CR and BC // PR
2. BP = CR 2. Opposite sides of the parallelogram
3. BP = PA 3. P is the mid–point of AB (Given).
4. ? PA = CR 4. From the statements (2) and (3)
5. In ∆s APQ and CRQ 5.
(i) AQP = CQR (A) (i) Vertically opposite angles
(ii) APQ = CRQ (A) (ii) BA // CR and alternate angles
(iii) PA = CR (S) (iii) From statement 4
(iv) ? ∆APQ # ∆CRQ (iv) A.A.S. axiom
6. AQ = QC 6. Corresponding sides of congruent triangles
7. PQ bisects AC at Q 7. From statement (6) Proved
Worked-out examples
AD
Example 1: In the given figure, AD // PR // BC and P is the mid-point P QR
of AB. If PQ = 3 cm and AD = 4 cm, find the lengths of
BC and PR. BC
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Solution:
(i) Q is the mid-point of AC. [In 'ABC, P is the mid-point of AB and PQ // BC]
Noro, w3,cPmQ==1212BBCC?[PBaCn=d Q are mid-points of AB and AC is 'ABC]
6 cm
(ii) R is mid-point of CD [In 'ADC, Q is mid-point of AC and QR // AD]
So, QR = 1 AD or, QR = 1 × 4 cm ? QR = 2 cm
2 2
Also, PR = PQ + QR = 3 cm + 2 cm = 5 cm
Hence, BC is 6 cm and PR is 5 cm.
Example 2: In ∆ ABC, P and Q are the mid–points of AB and BC
Solution: respectively. R is any point on AC. Prove that PQ
bisects BR at S.
Given: P and Q are the mid–points of the sides AB and BC of the
triangle ABC respectively. R is any point on AC.
To prove: PQ bisects BR at S.
Proof
Statements Reasons
1. In ∆ABC, PQ // AC 1. PQ joins the mid–points of the sides AB and BC.
2. In ∆ABR, PS // AR 2. PS is a portion of PQ and AR is a portion of AC.
3. S is the mid–point of BR 3. P is the mid–point of AB and PS // AR.
4. BS = SR 4. From statement (3)
i.e. PQ bisects BR at S
Example 3: In the given ∆PQR, PA is the bisector of P Proved
QPR. If QO A PA and QB = BR, prove that
O R
(i) OB // PR (ii) OB = 1 (PR – PQ) Q R
2 Q AB
Solution: P
Given: QPA = APR, QO A PA and QB = BR S
O
To prove: (i) OB // PR (ii) OB = m12 e(ePtRP–RPaQt S) .
Construction: QO is produced to AB
Proof:
Statements Reasons
1. In ∆ POQ and ∆ POS
(i) In QPO = SPO (A) 1.
(i) Given
(ii) PO = PO (S) (ii) Common side
(iii) QS A PA
(iii) POQ = POS (A) (iv) A.S.A axiom
(iv) ∆POQ # ∆POS
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2. (i) QO = OS , PQ = PS 2. (i) Corresponding sides of congruent triangles
(ii) QB = BR (ii) Given
(iii) OB // SR, i.e. OB // PR In ∆QSR, OB joins the mid-points of the sides QS
3. (i) OB = 1 SR (iii) and QR
2 3. (i) Same as reason 2. (iii)
(ii) OB = 1 (PR – PS) (ii) SR = PR – PS
2
iii) From statement 2 (i)
(iii) OB = 1 (PR – PQ)
2
Proved
Example 4: In trapezium ABCD, AD // BC, M and N are mid-points A D
M N
of AB and CD respectively. Prove that B C
(i) AD // MN // BC D E
1 N
(ii) MN = 2 (AD + BC)
C
Solution:
Given: ABCD is a trapezium in which AD // BC, A
M and N are mid-point of sides AB and CD M
respectively.
To prove: (i) AD // MN // BC (ii) MN = 1 (AD + BC) B
2
Construction: AN is joined and produced to meet BC produced at E.
Statements Reasons
1. In ∆ ADN and ∆ NEC 1.
(i) NAD = CEN (A) (i) AD // BE, alternate angles
(ii) ADN = NCE (A) (ii) AB // BE, alternate angles
(iii) DN = NC (S) (iii) Given ( N is the mid-point of DC)
(iv) ∆ADN # ∆NEC (iv) By A.A.S. axiom
2. AN = NE and AD = CE
2. Corresponding sides of congruent
triangles
3. MN // BE and MN = 1 BE 3. M and N being mid-points of AB and AE
2 in 'ABE.
4. MN // BC and MN = 1 (BC + CE) 4. From (3)
2
5. AD // MN // BC and MN = 1 (AD + BC) 5. Given and from statements (2) and (4).
2
EXERCISE 11.5 M N
General section
1. a) In the adjoining figure, M and N are the mid-points of AB and
AC respectively. Write the relation between MN and BC.
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b) In the given triangle, X is the mid-point of PR and
XY // PQ. What is the relation between QY and YR?
c) In the 'ABC given alongside, AP = PC and BQ = QC. If A P
PQ = 3.2 cm and AB = x cm, find the value of x. x
BQ C
d) In the triangle XYZ, A is the mid-point of XY and 5.8 cm
AB // XZ. If XZ is 6.4 cm, what is the length of AB?
2. a) In ∆ ABC, M and N are the mid-points of BC and AC
respectively. If MN = 2.6 cm, CNM = 30°, find the length
of AB and the size of BAC.
b) In ∆ PQR, A, B and C are the mid-points of PQ, QR and P
PR respectively. Find the sizes of angles represented bO
by letters. A ef C
c 30° d 95°
Q
B a R
P
c) In 'PQR, X and Y are mid-points of PQ and PR respectively. Q XY
If P + Q = 50° and P + R = 150°, find the size of R
PXY and PYX.
A
Creative section - A
3. a) In the given 'ABC, P, Q, and R are the mid-points of sides AB, P R
QC
BC and CA respectively. If the perimeter of 'PQR is 15 cm, find
the perimeter of 'ABC. B
L
b) In the adjoining 'LMN, X, Y and Z are the middle X Z
points of sides LM, MN and NL respectively. If the Y
perimeter of 'LMN is 36 cm find the perimeter of M N
'XYZ.
Vedanta Excel in Mathematics - Book 9 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry - Triangle
A D
QR
4. a) In the given figure, AD // PR // BC and P is the mid-point of P
AB. If BC = 8 cm and QR = 3 cm, find the lengths of PQ and C
S
AD.
Z
B
R
P
b) In the given figure, PS // XZ and X is the mid-point of PQ. X
If XY = 2.5 cm and XZ = 7 cm, find the lengths of PS and Y
QR.
Creative section - B Q
5. a) In the adjoining figure, P and Q are the mid–points of the
sides AB and AC of ∆ABC respectively. X is a point on
PQ. Prove that AX = XD.
b) In the adjoining triangle XYZ, A and B are the
mid-points of the sides XY and YZ respectively. P is
any point on XZ. Prove that AB bisects PY at Q.
c) In the adjoining equilateral triangle PQR, X, Y and Z are
the middle points of the sides PQ, QR and RP respectively.
Prove that XYZ is also an equilateral triangle.
d) In the given figure, AB // DC. If E is the mid–point of
BC and F is the mid–point of AC, prove that G is the
mid-point of AD.
e) In the adjoining figure, AB // CD // PQ and AP = PC.
Prove that AB = CD.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 9
Geometry - Triangle
f) In the figure alongside, AD // PQ // BC and DQ = QC. Prove
that AD + BC = 2PQ.
A
6. a) In the figure alongside P is the mid-point of BC, QR C
b) PQ // CA and QR // BC. Prove that BC = 4QR.
P B
BA Q
A
In the given 'ABC, AX and BY are medians. Z is a R
point on BC such that YZ // AX. Prove that BC = 4CZ. Y
C
c) In the given figure, A is the mid-point of QR and B is ZX
C
the mid-point of PA. Prove that PC = 1 PQ.
3 C
(Hint: Draw AD // BC where D is a point on PQ) B
P
7. a) In the given right angled triangle ABC, right
b)
angled at B, P is the mid–point of AC. Prove that
BP = 1 AC.
2
In the given ∆ ABC, AP is the bisector of BAC. A
If BO A AP and OQ // AC, prove that BQ = QC.
O
B PQ
c) In 'ABC, AD is the bisctor of BAC. If BO A AD and O
BE = EC then prove that OE // AC. DE
d) In the given trapezium PQRS, A and B are the mid–points
of the diagonals QS and PR respectively. Prove that
(i) AB // SR (ii) AB = 1 (SR – PQ)
2
(Hint: Join Q and B and produce QB to meet SR at C then show 'PQB # BCR)
Vedanta Excel in Mathematics - Book 9 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Geometry - Similarity
12
12.1 Similar triangles - review
In the figure given alongside, each C
angle of 'ABC is respectively 30° R
equal to the corresponding angle of 30°
'PQR. Therefore, 'ABC and 'PQR 115°
P
have the same shape and they are 115° 35° B 35° Q
said to be the similar triangle. We A
write it as 'ABC ~ 'PQR. However,
'ABC and 'PQR do not have the same size. FZ
On the other hand, 'DEF and 'XYZ are not 85° 40°
the similar triangles. Because, none of the 110°
X
angles of these triangles are equal. D 50° 45° 30° Y
E
Following are the required conditions for the similarity between triangles. These are
also called the properties of similar triangles.
(i) When all angles of one triangle are respectively equal to the corresponding angles of
another triangle, the triangles are said to be similar.
For example A P
In ∆s ABC and PQR, B
CQ R
A = P
B = Q
C = R
? ∆ BC a ∆PQR
‘a' is the symbol of similarity between the similar triangles.
Note: Of course, if two angles of a triangle are respectively equal to two corresponding
angles of another triangle, the remaining angles must be equal. So the triangles are said
to be similar.
(ii) When the ratios of the corresponding sides of triangles are equal, i.e. the corresponding
sides of triangles are proportional, the triangles are said to be similar.
For example K
In ∆s KLM and XYZ, X
KL = LM = MK
XY YZ ZX
? ∆ KLM a ∆ XYZ L MY Z
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 9
Geometry - Similarity
Here,
a) KL is the opposite side of M and M = Z; the opposite side of Z is XY. So, KL
and XY are corresponding sides.
b) LM is the opposite side of K and K = X; the opposite side of X is YZ. So, LM
and YZ are corresponding sides.
c) MK is the opposite side of L and L = Y; the opposite side of Y is ZX. So, MK
and ZX are the corresponding sides.
Theorem 9
Two equiangular triangles are similar.
Experimental verification
Step 1: Three pairs of triangles ABC and PQR with two equal corresponding angles are
drawn.
Here, in ∆ABC and ∆PQR, A = P and B = Q.
Step 2: The remaining angles and all the corresponding sides of each pair of triangles are
measured. The ratios of corresponding sides are found.
Fig. C R Result ∆ABC ∆PQR Ratio of corres- Result
AB BC CA PQ QR RP ponding sides
AB BC CA
PQ QR RP
(i) C = R AB = BC =CRAP
PQ QR
(ii) C = R AB = BC =CRAP
PQ QR
(iii) C = R AB = BC =CRAP
PQ QR
Conclusion: As the corresponding sides of equiangular triangles are proportional, each pair
of triangles are similar.
Vedanta Excel in Mathematics - Book 9 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry - Similarity
Theorem 10
If any two corresponding sides of triangles are proportional and the angles included by
them are equal, the triangles are similar.
Experimental verification
Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair
AB = BC and B = Q.
PQ QR
Step 2: The remaining angles and the remaining sides of each pair of triangles are measured.
Then the ratios of the remaining corresponding sides are found.
Fig Corresponding angles Corresponding sides AC Results
A P Result C R AC PR PR
A = P Result
(i) C = R AB = BC = AC
PQ QR PR
(ii) A = P C = R AB = BC = AC
PQ QR PR
(iii) A = P C = R AB = BC = AC
PQ QR PR
Conclusion: If two corresponding sides of triangles are proportional and the angle included
by them are equal, the triangles are similar.
Theorem 11
If two triangles have their corresponding sides proportional, the triangles are similar.
Experimental verification
Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair
AB = BC = AC .
PQ QR PR
Step 2: The corresponding angles of each pair of triangles are measured.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 9
Geometry - Similarity
Fig. A P Result B Q Result C R Result
(i) A = P B = Q C = R
(ii) A = P B = Q C = R
(iii) A = P B = Q C = R
Conclusion: If two triangles have their corresponding sides proportional, the triangles are
similar.
12.2 Similar polygons
In the adjoining figure, ABCDE and PQRST are two polygons. These are pentagons.
These polygons can be similar under the following conditions.
(i) If two or more polygons are equiangular, they
are similar. In the figure, A = P,
B = Q, C = R, D = S, and
E = T
? Pentagon ABCDE a pentagon PQRST
(ii) If the corresponding sides of two polygons are
proportional, they are similar.
In the figure, AB = BC = CD = DE = EA
PQ QR RS ST TP
? pentagon ABCDE a pentagon PQRST
(iii) If the corresponding diagonals of the polygons
are proportional to their corresponding sides,
they are similar.
In the figure, AC = BD = AB
PR QS PQ
? quad. ABCD a quad. PQRS
(iv) If the corresponding diagonals divide the
polygons into the same number of similar
triangles, the polygons are similar.
In the figure,
∆ABC a ∆PQR, ∆ACD a ∆PRS and
∆ADE a ∆PST
? pentagon ABCDE a pentagon PQRST
(v) If the straight lines that join the corresponding vertices of
polygons are concurrent, the polygons are similar.
In the figure,
The straight lines joining the corresponding vertices of quad.
ABCD and quad. PQRS are passing through the same point O.
So, they are concurrent.
? quad. ABCD a quad. PQRS.
Vedanta Excel in Mathematics - Book 9 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry - Similarity
Worked-out examples
Example 1: In the adjoining figure, show that A y cm P 9 cm C
Solution: ∆ABC a ∆PQC. Then, find the lengths of 10 cm
sides marked with x and y. 6 cm Q x cm
1. In ∆ ABC and ∆ PQC, 2 cm
(i) ACB = PCQ [Common angle] B
(ii) ABC = PQC [AB // PQ and corresponding angles]
(iii) BAC = QPC [Remaining angles of triangles]
(iv) ∆ABC a ∆PQC [A.A.A. axiom]
2. AB = BC = AC [Corresponding sides of similar triangles]
PQ QC PC
or, 10 = 2 + x = 9+y
6 x 9
5 2 + x 5 9 + y
or, 3 = x Also, 3 = 9
or, 5x = 6 + 3x or, 27 + 3y= 45
or, 2x = 6 or, 3y = 18
? x = 3 ? y =6
Hence, x = 3 cm and y = 6 cm.
Example 2: In the adjoining figure, AC // DE. Prove that
AB.DE = AC.BE. Also, find the value of x.
Solution:
1. In ∆ ABC and ∆BDE,
(i) ACB = BDE [AC // DE and alternate angles]
(ii) ABC = DBE [Vertically opposite angles]
(iii) BAC = BED [Remaining angles of the
triangles]
(iv) ? ∆ABC a ∆BDE [A.A.A. axiom]
2. AB = AC = BC [Corresponding sides of similar triangle are proportional]
BE DE BD
or, AB.DE = AC.BE proved.
Again, AB = BC
BE BD
or, 5.6 = x P
3.6 4.5 R
or, x = 7 cm. 4 cm
5 cm
Example 3: In the given figure, show that 'PQR ~ 'XQR. If X
PX = 5 cm and QX = 4 cm, find the length of QR.
Q
Solution:
1. In ∆ PQR and ∆ XQR,
(i) PQR = XQR [Common angle]
(ii) QPR = XRQ [Given]
(iii) PRQ = QXR [Remaining angles of the triangles]
(iv) ? ∆PQR a ∆XQR [A.A.A. axiom]
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199 Vedanta Excel in Mathematics - Book 9
Geometry - Similarity
2. PQ = QR [Corresponding sides of similar triangles]
QR XQ
or, 9 cm = QR
QR 4 cm
or, QR2 = 36 cm2 ? QR = 6 cm
Hence, the length of QR is 6 cm. A M D
O
Example 4: In the figure alongside, ABCD is a
parallelogram, in which M is the mid-point
of AD, prove that: OB = 2OD.
Solution: B C
Given: In parallelogram ABCD, M is the mid-point of AD.
To prove: OB = 2OD
Proof:
Statements Reasons
1.
1. In 'MOD and ∆ BOC (i) AD // BC and alternate angles
(ii) Same as (i)
(i) OMD = OCB (A) (iii) Vertically opposite angles.
(iv) By A.A.A axiom
(ii) ODM = OBC (A)
(iii) MOD = BOC (A)
(iv) ∆MOD a ∆BOC
2. OM = OD = MD 2. Corresponding sides of similar triangles
OC OB BC
3. BC = AD = 2MD 3. Opposite sides of parallelogram and
given.
4. OD = MD 4. From statements (3) and (4)
OB 2MD
? OB = 2OD
Proved
EXERCISE 12.1 P A
B Q
General section C
1. a) In the given figure, ∆ABC ~ ∆APQ. If AB = 15 cm,
AP = 10 cm and BC = 12 cm, find the length of PQ.
b) In the adjoining figure, PQ // XZ, PX = 6 cm, XY = 15 cm X
P
and YQ = 3 cm, find the length of QZ. Y
QZ
Vedanta Excel in Mathematics - Book 9 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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