Chapter 2 – Pythagoras Theorem
Theorems to study for the Chapter
Practice set 2.1
1. Identify, with reason, which of the following are Pythagorean triplets.
(i) (3, 5, 4)
Solution:
5² = 25 …(i)
3² + 4² = 9 + 16
∴ 3² + 4² = 25 …(ii)
From (i) and (ii),
5² = 3² + 4²
∴ 3, 5, 4 is a Pythagorean triplet.
(ii) (4, 9, 12)
Solution:
12² = 144 …(i)
4² + 9² = 16 + 81
∴ 4² + 9² = 97 …(ii)
From (i) and (ii),
12² ≠ 4² + 9²
∴ 4, 9, 12 is not a Pythagorean triplet.
(iii) (5, 12, 13)
Solution:
13² = 169 …(i)
5² + 12² = 25 + 144
∴ 5² + 12² = 169 …(ii)
From (i) and (ii),
13² = 5² + 12²
∴ 5, 12, 13 is a Pythagorean triplet.
(iv) (24, 70, 74)
Solution:
74² = 5476 …(i)
24² + 70² = 576 + 4900
∴ 24² + 70² = 5476 …(ii)
From (i) and (ii),
74² = 24² + 70²
∴ 24, 70, 74 is a Pythagorean triplet.
(v) (10, 24, 27)
Solution:
27² = 729 …(i)
10² + 24² = 100 + 576
∴ 10² + 24² = 676 …(ii)
From (i) and (ii),
27² ≠ 10² + 24²
∴ 10, 24, 27 is not a Pythagorean triplet.
(vi) (11, 60, 61)
Solution:
61² = 3721 …(i)
60² + 11² = 3600 + 121
∴ 60² + 11² = 3721 …(ii)
From (i) and (ii),
61² = 60² + 11²
∴ 11, 60, 61 is a Pythagorean triplet.
2. In figure 2.17, ∠MNP = 90°, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.
Given:
In ∆MNP,
∠MNP = 90⁰
seg NQ ⊥ seg MP
MQ = 9
QP = 4
To find:
NQ.
Solution:
In ∆MNP,
∠MNP = 90⁰…[Given]
and seg NQ ⊥ hypotenuse MP …[Given]
∴ NQ² = MQ × PQ …[By the Theorem of Geometric mean]
∴ NQ² = 9 × 4
∴ NQ² = 36
∴ \(\small \sqrt {NQ²}\) = \(\small \sqrt {36}\)
∴ NQ = 6 units
Ans: NQ is 6 units.
3. In figure 2.18, ∠QPR = 90°, seg PM ⊥ seg QR and Q – M – R, PM = 10, QM = 8, find QR.
Given:
In ∆QPR,
∠QPR = 90⁰
seg PM ⊥ seg QR
Q – M – R
PM = 10
QM = 8
To find:
QR
Solution:
In ∆QPR,
∠QPR = 90⁰ …[Given]
and seg PM ⊥ hypotenuse QR …[Given]
∴ PM² = QM × RM …[By the Theorem of Geometric mean]
∴ 10² = 8 × RM
∴ RM = \(\large \frac {100}{8}\)
∴ RM = 12.5 units
Now,
QR = QM + RM …[Q – M – R]
∴ QR = 8 + 12.5
∴ QR = 20.5 units
Ans: QR is 20.5 units.
4. See figure 2.19. Find RP and PS using the information given in ∆PSR.
Given:
In ∆PSR,
∠S = 90⁰
∠P = 30⁰
SR = 6 units
To find:
RP and PS
Solution:
In ∆PSR,
∠R = 180 – (90 + 30) …[Sum of all angles of a triangle is 180⁰]
∴ ∠R = 180 – (90 + 30)
∴ ∠R = 180 – 120
∴ ∠R = 60⁰
∴ ∆PSR is 30⁰ – 60⁰ – 90⁰ triangle
By 30⁰ – 60⁰ – 90⁰ triangle theorem,
RS = \(\large \frac {1}{2}\) × PR …[side opposite to 30⁰]
∴ 6 = \(\large \frac {1}{2}\) × PR
∴ PR = 12 units
PS = \( \frac {\large \sqrt {3}}{\large 2}\) × PR …[side opposite to 60⁰]
∴ PS = \( \frac {\small \sqrt {3}}{\large 2}\) × 12
∴ PS = 6\(\small \sqrt {3}\) units
Ans: PR is 12 units and PS = 6\(\small \sqrt {3}\) units.
5. For finding AB and BC with the help of information given in figure 2.20, complete the following activity.
Solution:
AB = BC …(□)
∴ ∠BAC = □
∴ AB = BC = □ AC
∴ AB = □ \( \small\sqrt {8}\)
∴ AB = □ 2 \( \small \sqrt {2}\)
∴ AB = □
Solution:
AB = BC …(Side opposite to congruent angle)
∴ ∠BAC = 45⁰
∴ AB = BC = \(\frac{\large 1}{\large \sqrt {2}}\) × AC
∴ AB = \(\frac{\large 1}{\large \sqrt {2}}\) × \( \small \sqrt {8}\)
∴ AB = \(\frac{\large 1}{\large \sqrt {2}}\) × 2 \(\small \sqrt {2}\)
∴ AB = 2 units
6. Find the side and perimeter of a square whose diagonal is 10 cm.
Given:
□PQRS is a square
PR = 10 cm
To find:
Side PQ and perimeter of □PQRS
Solution:
PQRS is a square …[Given]
∴ PQ = QR …(i) [Sides of a square]
In ∆PQR,
∠Q = 90⁰…[Each angle of a square is 90⁰]
∴ PR² = PQ² + QR² …[By Pythagoras theorem]
∴ 10² = PQ² + PQ²…[From (i)]
∴ 2PQ² = 100
∴ PQ² = \(\large \frac {100}{2}\)
∴ PQ² = 50
∴ \(\small \sqrt {PQ²}\) = \(\small \sqrt {50}\)
∴ PQ = 5\(\small \sqrt {2}\) cm
Now,
Perimeter of □PQRS = 4 × side
∴ Perimeter of □PQRS = 4 × PQ
∴ Perimeter of □PQRS = 4 × 5\(\small \sqrt {2}\)
∴ Perimeter of □PQRS = 20\(\small \sqrt {2}\) cm
Ans: Perimeter of □PQRS is 20\(\small \sqrt {2}\) cm.
7. In figure 2.21, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find
(1) EG
(2) FD and
(3) EF
Given:
In ∆DFE,
∠DFE = 90°
FG ⊥ ED
GD = 8
FG = 12
To find:
EG, FD and EF
Solution:
(i) For EG:
In ∆DFE,
∠DFE = 90⁰…[Given]
and seg FG ⊥ hypotenuse DE …[Given]
∴ FG² = DG × EG …[By the Theorem of Geometric mean]
∴ 12² = 8 × EG
∴ EG = \(\large \frac {144}{8}\)
∴ EG = 18 units …(i)
(ii) For FD:
In ∆FGD,
∠FGD = 90⁰ …[Given]
∴ FD² = FG² + GD² …[By Pythagoras theorem]
∴ FD² = 12² + 8²
∴ FD² = 144 + 64
∴ FD² = 208
∴ \(\small \sqrt {FD²}\) = \(\small \sqrt {208}\) …[Taking square roots]
∴ FD = 4\(\small \sqrt {13}\) units
(iii) For EF:
In ∆FGE
m∠FGE = 90⁰ …[Given]
∴ EF² = EG² + FG² …[By Pythagoras theorem]
∴ EF² = 18² + 12² …[From (i)]
∴ EF² = 324 + 144
∴ EF² = 468
∴ \(\small \sqrt {EF²}\) = \(\small \sqrt {468}\) …[Taking square roots]
∴ EF = 6\(\small \sqrt {13}\) units
Ans: EG is 18 units, FD is 4\(\small \sqrt {13}\) units, and EF is 6\(\small \sqrt {13}\).
8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
Given:
In □ABCD,
AB = 35 cm
BC = 12 cm
To find:
diagonal AC
Solution:
In ∆ABC,
∠ABC = 90⁰…[Each angle of a rectangle is 90⁰]
∴ AC² = AB² + BC² …[By Pythagoras theorem]
∴ AC² = 35² + 12²
∴ AC² = 1225 + 144
∴ AC² = 1369
∴ \(\small \sqrt {AC²}\) = \(\small \sqrt {1369}\) …[Taking square roots]
∴ AC = 37 cm
Ans: Length of the diagonal of the rectangle is 37 cm.
9. In the figure 2.22, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ² = 4PM² – 3PR²
Given:
In ∆PRQ,
M is the midpoint of QR
∠PRQ = 90°
To prove:
PQ² = 4PM² – 3PR²
Solution:
In ∆PRQ,
∠PRQ = 90⁰ …[Given]
∴ PQ² = PR² + QR² …(i) [By Pythagoras theorem]
But, QR = 2RM …(ii) [M is the midpoint of seg QR]
∴ PQ² = PR² + (2 RM)² …[From (i) and (ii)]
∴ PQ² = PR² + 4 RM² …(iii)
In ∆PRM,
∠PRM = 90⁰
∴ PM² = PR² + RM² …[By Pythagoras theorem]
∴ RM² = PM² – PR² …(iv)
∴ PQ² = PR² + 4 (PM² – PR²) …[From (iii) and (iv)]
∴ PQ² = PR² + 4 PM² – 4 PR²
∴ PQ² = 4 PM² – 3 PR²
Hence proved.
10. Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height of 4.2 m. Find the width of the street
Given:
Let,
RD represents the width of the street.
BD represents the first building.
AR represents the second building
CA and CB are two different positions of the same ladder from point C.
∴ AR = 4.2 m
BD = 4 m
AC = BC = 5.8 m
To find:
RD
Solution:
In ∆ARC
∠R = 90⁰ …[Given]
∴ AC² = AR² + CR² …[By Pythagoras theorem]
∴ (5.8)² = (4.2)² + CR²
∴ CR² = 33.64 – 17.64
∴ CR² = 16
∴ \(\small \sqrt {CR²}\) = \(\small \sqrt {16}\) …[Taking square roots]
∴ CR = 4 m
In ∆BDC,
∠D = 90⁰ …[Given]
∴ BC² = CD² + BD² …[By Pythagoras theorem]
∴ 5.8² = CD² + 4²
∴ CD² = 33.64 – 16
∴ CD² = 17.64
∴ \(\small \sqrt {CD²}\) = \(\small \sqrt {17.64}\) …[Taking square roots]
∴ CD = 4.2 m
Now,
RD = RC + CD …[R – C – D]
∴ RD = 4 + 4.2
∴ RD = 8.2 m
Ans: Width of the street is 8.2 m.
Practice set 2.2
1. In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.
Given:
In ∆PQR,
QS = SR
PQ = 11
PR = 17
PS =13
To find:
QR
Solution:
In ∆PQR,
PS is the median …[∵ S is the midpoint of QR]
∴ PQ² + PR² = 2PS² + 2QS² …[By Apollonius theorem]
∴ 11² + 17² = 2 × 13² + 2 × QS²
∴ 121 + 289 = 2 (169 + QS²)
∴ \(\large \frac {410}{2}\) = 169 + QS²
∴ 205 = 169 + QS²
∴ QS² = 205 – 169
∴ QS² = 36
∴ \(\small \sqrt{QS}\)² = \(\small \sqrt{36}\) …[Taking square roots]
∴ QS = 6 units
Now,
QR = 2 QS …[∵ S is the midpoint of QR]
∴ QR = 2 × 6
∴ QR = 12 units
Ans: QR is 12 units.
2. In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB.
Given:
In ∆ABC,
seg CM is a median
AB = 10
AC = 7
BC = 9
To find:
CM
Solution:
In ∆ABC,
BM = \(\large \frac {1}{2}\) × AB …[∵ M is the midpoint of AB]
∴ BM = \(\large \frac {1}{2}\) × 10
∴ BM = 5 units …(i)
CM is the median …[Given]
∴ AC² + BC² = 2CM² + 2BM² …[By Apollonius theorem]
∴ 7² + 9² = 2CM² + 2 × 5²
∴ 49 + 81 = 2(CM² + 25)
∴ \(\large \frac {130}{2}\) = CM² + 25
∴ 65 = CM² + 25
∴ CM² = 65 – 25
∴ CM² = 40
∴ \(\small \sqrt{CM}\)² = \(\small \sqrt{40}\) …[Taking square roots]
∴ CM = 2\(\small \sqrt{10}\) units
Ans: CM is 2\(\small \sqrt{10}\) units
3. In the figure 2.28 seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,
(i) PR² = PS² + QR × ST + \(\large (\frac {QR}{2})\)²
(ii) PQ² = PS² – QR × ST + \(\large (\frac {QR}{2})\)²
Given:
In ∆PQR,
seg PS is the median
PT ⊥ QR
To prove:
(i) PR² = PS² + QR × ST + \(\large (\frac {QR}{2})\)²
(ii) PQ² = PS² – QR × ST + \(\large (\frac {QR}{2})\)²
Proof:
In ∆PTS,
QS = RS = \(\large \frac {QR}{2}\) …(i) [∵ S is the midpoint of seg QR]
∠PTS = 90⁰ …[Given]
∴ PS² = PT² + ST²…[By Pythagoras theorem]
∴ PT² = PS² – ST² …(ii)
(i) In ∆PTR,
∠PTR = 90⁰ …[Given]
∴ PR² = PT² + TR² …[By Pythagoras theorem]
∴ PR² = PS² – ST² + (RS + ST)² …[From (ii), R – S – T]
∴ PR² = PS² – ST² + RS² + 2 × RS × ST + ST² …[∵ (a + b)² = a² + 2ab + b²]
∴ PR² = PS² + 2RS × ST + RS²
∴ PR² = PS² + 2 × \(\large \frac {QR}{2}\) × ST + \(\large (\frac {QR}{2})\)² …[From (i)]
∴ PR = PS + QR × ST \(\large (\frac {QR}{2})\)²
Hence proved.
(ii) In ∆PTQ,
∠PTQ = 90° …[Given]
∴ PQ² = PT² + QT² …[By Pythagoras theorem]
∴ PQ² = PS² – ST² + (QS – ST)²…[From (ii), Q – T – S]
∴ PQ² = PS² – ST² + QS² – ² QS × ST + ST² …[∵ (a + b)² = a² + 2ab + b²]
∴ PQ² = PS² – 2QS × ST + QS²
∴ PQ² = PS² – 2 × \(\large \frac {QR}{2}\) × ST + \(\large (\frac {QR}{2})\)² …[From (i)]
∴ PQ² = PS² – QR × ST + \(\large (\frac {QR}{2})\)²
Hence proved.
4. In ∆ABC, point M is the midpoint of side BC. If AB² + AC² = 290 cm², AM = 8 cm, find BC.
Given:
In ∆ABC,
BM = MC
AB² + AC² = 290 cm²
AM = 8 cm
To find:
BC
Solution:
In ∆ABC,
seg AM is median …[Given]
∴ AB² + AC² = 2AM² + 2BM² …[By Apollonius theorem]
∴ 290 = 2 × 8² + 2 × BM²
∴ 290 = 2 (8² + BM²)
∴ \(\large \frac {290}{2}\) = 64 + BM²
∴ 145 = 64 + BM²
∴ 145 – 64 = BM²
∴ BM² = 81
∴ \(\small \sqrt{BM}\)² = \(\small \sqrt{81}\) …[Taking square roots]
∴ BM = 9 units
Now,
BC = 2 BM …[∵ M is the midpoint of seg BC]
∴ BC = 2 × 9
∴ BC = 18 cm
Ans: BC is 18 cm.
5. In figure 2.30, point T is in the interior of rectangle PQRS, Prove that, TS² + TQ² = TP² + TR². (As shown in the figure, draw seg AB || side SR and A – T – B)
Given:
□PQRS is a rectangle
T is in the interior of □PQRS
To prove:
TS² + TQ² = TP² + TR²
Construction:
Draw a line parallel to side SR, through point T, intersecting sides PS and QR at point A and B such that (A – T – B).
Proof:
□PQRS is a rectangle …[Given]
∴ ∠SPQ = ∠PSR = ∠SRQ = ∠PQR = 90° …(i) [Each angle of a rectangle is 90⁰]
∴ seg PQ || seg SR …(ii) [Opposite sides of rectangle are parallel]
But,
seg AB || seg SR …(iii) [Construction]
∴ seg PQ || seg SR || seg AB …(iv) [From (i), (ii) & (iii)]
seg PQ || seg AB and PS is the transversal …[From (iv)]
∴ ∠QPS = ∠BAS = 90° …(v) [Corresponding angles test]
And,
seg PQ || seg AB and QR is the transversal …[From (iv)]
∴ ∠PQB = ∠ABR = 90° …(vi) [Corresponding angles test]
Now,
∠BAS = 90° …[From (v)]
∴ ∠PAB = 90° …(vii) [Linear angles test]
Also,
∠ABR = 90° …[From (vi)]
∴ ∠ABQ = 90° …(viii) [Linear angles test]
In ∆PAT,
∠PAT = 90° …[From (vii), A – T – B]
∴ TP² = PA² + AT² …(ix) [By Pythagoras theorem]
In ∆SAT,
∠SAT = 90° …[From (v), A – T – B]
∴ TS² = AS² + AT² …(x) [By Pythagoras theorem]
In ∆QBT,
∠QBT = 90° …[From (viii), A – T – B]
∴ TQ² = QB² + BT² …(xi) [By Pythagoras theorem]
In ∆RBT,
∠RBT = 90° …[From (vi), A – T – B]
∴ TR² = BR² + BT² …(xii) [By Pythagoras theorem]
In □ABRS,
seg AB || seg SR …[From (iii)]
seg AS || seg BR …[Opposite sides of rectangle are parallel and P – A – S, Q – B – R]
∴ □ABRS is parallelogram …[By definition]
∴ AS = BR …(xiii) [Opposite sides of parallelogram are equal]
And,
In □ ABQP,
seg AB || seg PQ …[From (iv)]
seg AP || seg BQ …[Opposite sides of rectangle are parallel and P – A – S, Q – B – R]
∴ □ABPQ is parallelogram …[By definition]
∴ AP = QB …(xiv) [Opposite sides of parallelogram are equal]
Adding (x) and (xi), we get,
TS² + TQ² = AS² + AT² + QB² + BT²
∴ TS² + TQ² = BR² + AT² + AP² + BT²
∴ TS² + TQ² = BR² + BT² + AP² + AT²
∴ TS² + TQ² = TR² + TP² …[From (ix) and (xii)]
Hence proved.
Problem Set 2
1. Some questions and their alternative answers are given. Select the correct alternative.
(1) Out of the following, which is the Pythagorean triplet?
(A) (1, 5, 10)
(B) (3, 4, 5)
(C) (2, 2, 2)
(D) (5, 5, 2)
Ans: Option (B) : (3, 4, 5)
Solution:
5² = 25 …(i)
3² + 4² = 9 + 16
∴ 3² + 4² = 25 …(ii)
From (i) and (ii),
5² = 3² + 4²
∴ 3, 5, 4 is a Pythagorean triplet.
(2) In a right angled triangle, if the sum of the squares of the sides making the right angle is 169 then what is the length of the hypotenuse?
(A) 15
(B) 13
(C) 5
(D) 12
Ans: Option (B) : 13
Solution:
(side 1)² + (side 2)² = 169
In a right angled triangle,
By Pythagoras Theorem,
(Hypotenuse)² = (side 1)² + (side 2)²
(Hypotenuse)² = 169
Hypotenuse = 13 …[Taking Square Roots]
(3) Out of the dates given below, which date constitutes a Pythagorean triplet ?
(A) 15/08/17
(B) 16/08/16
(C) 3/5/17
(D) 4/9/15
Ans: Option (A) : 15/08/17
Solution:
17² = 289 …(i)
8² + 15² = 64 + 225
∴ 8² + 15² = 289 …(ii)
From (i) and (ii),
17² = 8² + 17²
∴ 8, 15, 17 is a Pythagorean triplet.
(4) If a, b, c are sides of a triangle and a² + b² = c², name the type of triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Ans: Option (C) : Right angled triangle
Explanation:
In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right-angled triangle.
(5) Find the perimeter of a square if its diagonal is 10\(\sqrt{2}\) cm.
(A)10 cm
(B) 40\(\sqrt{2}\) cm
(C) 20 cm
(D) 40 cm
Ans: Option (D) : 40 cm
Solution:
Diagonal = 10\(\sqrt{2}\) cm
We can take the diagonal as the hypotenuse and the two sides of the square as remaining sides of the right angled triangle.
By Pythagoras Theorem, we get,
(Hypotenuse)² = (side)² + (side)²
∴ (10\(\sqrt{2}\))² = 2 (side)²
∴ 200 = 2 (side)²
∴ (side)² = \(\large \frac {200}{2}\)
∴ (side)² = 100
∴ side = 10 …[Taking square roots]
Perimeter of the square = 4 × side
∴ Perimeter of the square = 4 × 10
∴ Perimeter of the square = 40 cm
(6) Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm
(B) 4 cm
(C) 6 cm
(D) 2 \(\sqrt{6}\) cm
Ans: Option (C) : 6 cm
Solution:
We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.
∴ (Altitude)² = (Part 1)² × (Part 2)²
∴ (Altitude)² = 4 × 9
∴ (Altitude)² = 36
∴ Altitude = 6 cm
(7) Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Ans: Option (B) : 30 cm
Solution:
In a right angled triangle,
By Pythagoras Theorem,
(Hypotenuse)² = (Height)² + (base)²
∴ (Hypotenuse)² = 24² + 18²
∴ (Hypotenuse)² = 324 + 576
∴ (Hypotenuse)² = 900
∴ Hypotenuse = 30 …[Taking Square Roots]
(8) In ∆ABC, AB = 6\(\sqrt{3}\) cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Ans: Option (A) : 30°
Solution:
In ∆ABC,
AB = 6\(\sqrt{3}\) cm
AC = 12 cm
BC = 6 cm
AB² = (6\(\sqrt{3}\))² = 108
AC² = (12)² = 144
BC² = (6)² = 36
∴ 108 + 36 = 144
i.e. AB² + BC² = AC²
In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.
In a right angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is 30°. Here, BC is half of AC.
Thus, measure of ∠A is 30⁰
2. Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a.
Given:
∆ABC is an equilateral triangle
AB = 2A
To find:
AM
Construction:
Seg AM ⊥ side BC, B – M – C
Solution:
In ∆AMB,
∠AMB = 90⁰ …[Given]
∠B = 60⁰ …[Each angle of an equilateral triangle is 60⁰]
∠BAM = 30⁰ …[Sum of all angles of a triangle is 180⁰]
∴ ∆AMB is 30⁰ – 60⁰ – 90⁰ triangle
By 30⁰ – 60⁰ – 90⁰ triangle theorem,
AM = \(\large \frac {\sqrt{3}}{2}\) × AB …[Side opposite to 60⁰]
AM = \(\large \frac {\sqrt{3}}{2}\) × AB
∴ AM = \(\large \frac {\sqrt{3}}{2}\) × 2a
∴ AM = \(\sqrt{3}\) a
Ans: The value of AM is \(\sqrt{3}\) a.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give a reason.
Given:
Sides of a triangle are 7 cm, 24 cm, 25 cm
To find:
Do they form a right angled triangle
Solution:
25² = 625 …(i)
7² + 24² = 49 + 576
7² + 24² = 625 …(ii)
∴ 25² = 7² + 24² ...[From (i) and (ii)]
∴ The given measurements form a right-angled triangle …[By converse of Pythagoras theorem]
Ans: The given measurements form a right-angled triangle.
(3) Find the length of a diagonal of a rectangle having sides 11 cm and 60cm.
Given:
In □ABCD,
AB = 60 cm
BC = 11 cm.
To find:
AC
Solution:
In ∆ABC,
∠ABC = 90⁰ …[Each angle of a rectangle is 90⁰]
∴ AC² = AB² + BC² …[By Pythagoras theorem]
AC² = (60)² + (11)²
AC² = 3600 + 121
AC² = 3721
∴ AC = 61 cm …[Taking square roots]
Ans: The length of the diagonal of the rectangle is 61 cm.
(4) Find the length of the hypotenuse of a right angled triangle if the remaining sides are 9 cm and 12 cm.
Given:
In ∆ABC,
∠ABC = 90⁰
AB = 9 cm
BC = 12 cm
To find:
AC
Solution:
In ∆ABC,
∠ABC = 90⁰ …[Given]
∴ AC² = AB² + BC² …[By Pythagoras theorem]
∴ AC² = (9)² + (12)²
∴ AC² = 81 + 144
∴ AC² = 225
∴ AC = 15 cm …[Taking square roots]
Ans: Length of the hypotenuse is 15 cm.
(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.
Given:
PQ = QR = x
To find:
PR
Solution:
In ∆PQR
∠PQR = 90⁰ …[Given]
∴ PR² = PQ² + QR² …[By Pythagoras theorem]
∴ PR² = x² + x²
∴ PR² = 2x²
∴ PR = 2x …[Taking square roots]
Ans: The length of the hypotenuse is \(\sqrt{2}\) x units.
(6) In ∆PQR; PQ = \(\sqrt{8}\), QR = \(\sqrt{5}\), PR = \(\sqrt{3}\). Is ∆PQR a right angled triangle? If yes, which angle is 90°?
Given:
In ∆PQR,
PQ = \(\sqrt{8}\)
QR = \(\sqrt{5}\)
PR = \(\sqrt{3}\)
To find:
Is ∆PQR a right angled triangle
Solution:
PQ² = \((\sqrt{8})\)² = 8 …(i)
PR² + QR² = \((\sqrt{3})\)² + \((\sqrt{5})\)²
∴ PR² + QR² = 3 + 5
∴ PR² + QR² = 8 …(ii)
∴ PQ² = PR² + QR² …[From (i) and (ii)]
∴ ∆PQR is a right angled triangle.
∴ ∠R = 90⁰…[Converse of Pythagoras theorem]
Ans: ∆PQR is a right angled triangle and ∠R = 90⁰.
3. In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.
Given:
In ∆RST
∠S = 90°
∠T = 30°
RT = 12 cm
To find:
RS and ST
Solution:
In ∆PSR,
∠S = 90° …[Given]
∠T = 30⁰ …[Given]
∴ ∠R = 60⁰ …[Sum of all angles of a triangle is 180⁰]
∴ ∆PSR is 30⁰ – 60⁰ – 90⁰ triangle
By 30⁰ – 60⁰ – 90⁰ triangle theorem,
RS = \(\large \frac {1}{2}\) × RT …[Side opposite to 30⁰]
∴ RS = \(\large \frac {1}{2}\) × 12
∴ RS = 6 cm
ST = \(\large \frac {\sqrt{3}}{2}\) × RT …[Side opposite to 60⁰]
ST = \(\large \frac {\sqrt{3}}{2}\) × 12
∴ ST = 6 \(\sqrt{3}\) cm
Ans: The value of RS = 6 cm and ST = 6 \(\sqrt{3}\) cm.
4. Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.
Given:
□ABCD is a rectangle
AB = 16 cm
A(□\, ABCD) = 192 sq. cm.
To find:
AC
Solution:
□ ABCD is a rectangle
A(□\, ABCD) = length × breadth
∴ 192 = AB × BC
∴ 192 = 16 × BC
∴ \(\large \frac {192}{16}\) = BC
∴ BC = 12 cm
In ∆ABC,
∠ABC = 90⁰ …[Each angle of a rectangle is 90⁰]
∴ AC² = AB² + BC² …[By Pythagoras theorem]
AC² = (16)² + (12)²
∴ AC² = 256 + 144
∴ AC² = 400
∴ AC = 20 cm …[Taking square roots]
Ans: Length of the diagonal is 20 cm.
5. Find the length of the side and perimeter of an equilateral triangle whose height is \(\sqrt{3}\) cm.
Given:
∆ABC is an equilateral triangle
seg AM ⊥ side BC, B – M – C
AM = 3 cm
To find:
AB
Perimeter of ∆ABC
Solution:
In ∆AMB,
∠AMB = 90⁰ …[Given]
∠B = 60⁰ …[Each angle of an equilateral triangle is 60⁰]
∴ ∠BAM = 30⁰ …[Sum of all angles of a triangle is 180⁰]
∴ ∆AMB is 30⁰ – 60⁰ – 90⁰ triangle
By 30⁰ – 60⁰ – 90⁰ triangle theorem,
AM = \(\large \frac {\sqrt{3}}{2}\) × AB …[Side opposite to 60⁰]
∴ AB = \(\large \frac {2}{\sqrt{3}}\) × AM
∴ AB = \(\sqrt{3}\) × \(\large \frac {2}{\sqrt{3}}\)
∴ AB = 2 cm
Perimeter of ∆ABC = 3 × AB
Perimeter of ∆ABC = 3 × 2
Perimeter of ∆ABC = 6 cm
Ans: The value of AB is 2 cm and perimeter of ∆ABC is 6 cm.
6. In ∆ABC seg AP is a median. If BC = 18, AB² + AC² = 260 Find AP.
Given:
In ∆ABC
Seg AP is a median
BC = 18
AB² + AC² = 260
To find:
AP
Solution:
In ∆ABC,
Seg AP is the median on side BC
∴ BP = \(\large \frac {1}{2}\) × BC
∴ BP = \(\large \frac {1}{2}\) × 18
∴ BP = 9 units …(i)
In ∆ABC, seg AP is median …[Given]
∴ AB² + AC² = 2BP² + 2AP² …[By Apollonius theorem]
∴ 260 = 2(9² + AP²) …[From (i)]
∴ \(\large \frac {260}{2}\) = 81 + AP²
∴ AP² = 130 – 81
∴ AP² = 49
∴ AP = 7 units …[Taking square roots]
Ans: The value of AP is 7 units.
7. ∆ABC is an equilateral triangle. Point P is on base BC such that PC = \(\large \frac {1}{3}\) BC, if AB = 6 cm find AP.
Given:
∆ABC is an equilateral triangle.
AB = 6 cm
PC = \(\large \frac {1}{3}\) BC
To find:
AP
Construction:
Draw seg AM ⊥ BC, C – M – B
Solution:
AB = BC = AC …[Sides of an equilateral triangle are equal]
∵ AB = 6 cm
∴ BC = AC = 6 cm …(i)
PC = \(\large \frac {1}{3}\) BC …[Given]
∴ PC = \(\large \frac {1}{3}\) × 6
∴ PC = 2 cm …(ii)
In ∆AMC,
∠C = 60⁰…[Each angle of an equilateral triangle is 60⁰]
∠AMC = 90⁰ …[Construction]
∴ ∠CAM = 30⁰ …[Sum of all angles of a triangle is 180⁰]
∴ ∆AMC is 30⁰ – 60⁰ – 90⁰ triangle
By 30⁰ – 60⁰ – 90⁰ triangle theorem,
∴ CM = \(\large \frac {1}{2}\) AC …[Side opposite to 30⁰]
∴ CM = \(\large \frac {1}{2}\) × 6 cm
∴ CM = 3 cm …(iii)
AM = \(\large \frac {\sqrt{3}}{2}\) × AC …[Side opposite to 60⁰]
∴ AM = \(\large \frac {\sqrt{3}}{2}\) × 6
∴ AM = 3\(\sqrt{3}\) …(iv)
PM = CM – CP …[C – P – M]
∴ PM = 3 – 2
∴ PM = 1 cm
In ∆AMP,
∠AMP = 90⁰ …[Construction]
AP² = AM² + PM² …[By Pythagoras theorem]
AP² = \(\large (\)3\(\sqrt{3} \large )\)² + 1
AP² = 27 + 1
∴ AP² = 28
∴ AP = 2\(\sqrt{7}\) cm …[Taking square roots]
Ans: The value of AP is 2\(\sqrt{7}\) cm.
8. From the information given in the figure 2.31, prove that PM = PN = \(\sqrt{3}\) × a
Given:
PQ = PR = MQ = QR = RN = a
To prove:
PM = PN = \(\sqrt{3}\) × a
Proof:
Point Q is the midpoint of seg MR …(i)
In ∆PMR,
Seg PQ is a median …[From (i) and by Definition]
∴ PM² + PR² = 2PQ² + 2QM² …[By Apollonius theorem]
∴ PM² + a² = 2a² + 2a²
∴ PM² + a² = 4a²
∴ PM² = 4a² – a²
∴ PM² = 3a²
∴ PM = \(\sqrt{3}\)a …[Taking square roots]
Point R is the midpoint of seg QN..(ii)
In ∆PQN,
Seg PR is a median …[From (ii) and by Definition]
∴ PQ² + PN² = 2PR² + 2RN² …[By Apollonius theorem]
∴ a² + PN² = 2a² + 2a²
∴ a² + PN² = 4a²
∴ PN² = 4a² – a²
∴ PN² = 3a²
∴ PN = \(\sqrt{3}\)a …[Taking square roots]
∴ PM = PN = \(\sqrt{3}\)a
Hence Proved.
9. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Given:
ABCD is a parallelogram
Diagonals AC and BD intersect each other at point M.
To prove:
AC² + BD² = AB² + BC² + CD² + AD²
Proof:
□ ABCD is a parallelogram …[Given]
AM = CM = \(\large \frac {1}{2}\) AC …(i)
BM = DM = \(\large \frac {1}{2}\) BD …(ii) [Diagonals of a parallelogram bisect each other]
In ∆ABC
Seg BM is a median …[From (i)]
∴ AB² + BC² = 2BM² + 2AM² …(iii) [By Apollonius theorem]
In ∆ADC,
Seg DM is a median …[From (i)]
∴ CD² + AD² = 2DM² + 2AM² …(iv) [By Apollonius theorem]
Adding (iii) and (iv)
∴ AB² + BC² + CD² + AD² = 2BM² + 2AM² + 2DM² + 2AM²
∴ AB² + BC² + CD² + AD² = 2BM² + 2DM² + 4AM²
∴ AB² + BC² + CD² + AD² = 2BM² + 2BM² + 4AM² …[From (ii)]
∴ AB² + BC² + CD² + AD² = 4BM² + 4AM²
∴ AB² + BC² + CD² + AD² = 4 [BM² + AM²]
∴ AB² + BC² + CD² + AD² = 4 \(\large [(\frac {1}{2}\) BD \(\large )\)² + \(\large (\frac {1}{2}\) AC \(\large )]\)² …[From (i) and (ii)]
∴ AB² + BC² + CD² + AD² = 4 \(\large [\frac {1}{4}\) BD² + \(\large \frac {1}{4}\) AC² \(\large ]\)
∴ AB² + BC² + CD² + AD² = 4 × \(\large \frac {1}{4}\) [BD² + AC²]
∴ AB² + BC² + CD² + AD² = BD² + AC²
∴ AC² + BD² = AB² + BC² + CD² + AD²
Hence Proved.
10. Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours, the distance between them was 15\(\sqrt{2}\) km. Find their speed per hour.
Given:
The distance between them was 15\(\sqrt{2}\)km
To find:
Their speed per hour
Solution:
Let B represents the starting point of journey.
BA is the distance travelled by Prasad in the North direction.
BC is the distance travelled by Pranali in the East direction.
AC is the distance between Pranali and Prasad after two hours.
Let the speed of each one be x km/hr.
∴ Distance travelled by each one hour is 2x km.
i.e. AB = BC = 2x km
In ∆ABC,
∠B = 90⁰ …[Line joining adjacent direction are ⊥ to each other]
∴ AB² + BC² = AC² …[By Pythagoras theorem]
∴ (2x)² + (2x)² = 15 \((\sqrt{2})\)²
∴ 4x² + 4x² = 225 × 2
∴ 8x² = 225 × 2
∴ x² = \(\large \frac {225 \,×\, 2}{8}\)
∴ x² = \(\large \frac {225}{4}\)
∴ x = \(\large \frac {15}{2}\) …[Taking square roots]
∴ x = 7.5 km/hr
Ans: Their speed is 7.5 km/hr.
11. In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that : 4(BL² + CM²) = 5 BC²
Given:
In ∆ABC
∠BAC = 90°
seg BL and seg CM are medians of ∆ABC
To prove:
4(BL² + CM²) = 5 BC²
Proof:
In ∆BAC,
∠BAC = 90⁰ …[Given]
∴ BC² = AB² + AC² …(i) [By Pythagoras theorem]
In ∆BAL,
∠BAC = 90⁰ …[Given]
∴ BL² = AB² + AL² …(ii) [By Pythagoras theorem]
In ∆CAM,
∠CAM = 90⁰ …[Given]
∴ CM² = AC² + AM² …(iii) [By Pythagoras theorem]
Adding (ii) and (iii),
BL² + CM² = AB² + AL² + AC² + AM²
∴ BL² + CM² = AB² + AC² + AL² + AM²
∴ BL² + CM² = BC² + AL² + AM² …[From (i)]
∴ BL² + CM² = BC² + \(\large (\frac {1}{2}\) AC \(\large )\)² + \(\large (\frac {1}{2}\) AB \(\large )\)² …[∵ L and M are the midpoint of sides AC and AB respectively]
∴ BL² + CM² = BC² + \(\large (\frac {AC²}{4}\) + \(\large (\frac {AB²}{4}\)
∴ 4(BL² + CM²) = 4BC² + AC² + AB² …[Multiplying throughout by 4]
∴ 4(BL² + CM²) = 4BC² + BC² …[From (i)]
∴ 4(BL² + CM²) = 5BC²
Hence proved.
12. Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is14 cm. Find the length of the other diagonal.
Given:
□ ABCD is a parallelogram
AB² + BC² = 130 cm²
AC = 14 cm
To find:
BD
Solution:
□ ABCD is a parallelogram …[Given]
∴ BM = \(\large \frac {1}{2}\) BD …(i)
∴ AM = \(\large \frac {1}{2}\) AC …(ii) …[Diagonals of a parallelogram bisect each other]
∴ AM = \(\large \frac {1}{2}\) × 14
∴ AM = 7 cm
In ∆ABC,
seg BM is the median ...[From (ii) and Definition]
∴ AB² + BC² = 2AM² + 2BM² …[By Apollonius theorem]
∴ 130 = 2 (7² + BM²)
∴ \(\large \frac {130}{2}\) = 49 + BM²
∴ 65 = 49 + BM²
∴ 65 – 49 = BM²
∴ BM² = 16
∴ BM = 4 cm …[Taking square roots]
∴ \(\large \frac {1}{2}\) BD = 4 cm …[From (i)]
∴ BD = 8 cm
Ans: The value of BD is 8 cm.
13. In ∆ABC, seg AD ⊥ seg BC, DB = 3CD. Prove that : 2AB² = 2AC² + BC²
Given:
In ∆ABC
seg AD ⊥ seg BC
DB = 3CD
To prove:
2AB² = 2AC² + BC²
Proof:
In ∆ADB,
∠ADB = 90⁰ …[Given]
∴ AB² = AD² + DB² …[By Pythagoras theorem]
∴ AB² = AD² + (3CD)² [Given]
∴ AB² = AD² + 9CD² …(i)
In ∆ADC,
∠ADC = 90⁰ ...[Given]
∴ AC² = AD² + CD² …[By Pythagoras theorem]
∴ AD² = AC² – CD² …(ii)
AB² = AC² – CD² + 9CD² …[From (i) and (ii)]
∴ AB² = AC² + 8CD² …(iii)
But,
BC = CD + DB …[C – D – B]
∴ BC = CD + 3CD …[Given]
∴ BC = 4CD
∴ CD = \(\large \frac {BC}{4}\) …(iv)
∴ AB² = AC² + 8 \(\large (\frac {BC²}{4})\)² …[From (iii) and (iv)]
∴ AB² = AC² + 8 × \(\large \frac {BC²}{16}\)²
∴ AB² = AC² + \(\large \frac {BC²}{2}\) …[From (iii) and (iv)]
∴ 2AB² = 2AC² + BC² …[Multiplying throughout by 2]
Hence proved.
14. In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.
Given:
∆ABC is an isosceles triangle
AB = AC = 13 cm
BC = 10 cm
seg AD is the median
G is the centroid of ∆ABC
To find:
AG
Solution:
BD = \(\large \frac {1}{2}\) BC …[Median bisects opposite side]
∴ BD = \(\large \frac {1}{2}\) × 10
∴ BD = 5 cm …(i)
In ∆ABC,
seg AD is a median …[By Definition]
∴ AB² + AC2² = 2AD² + 2BD² …[By Apollonius theorem]
∴ 13² + 13² = 2 (AD² + 5²) …[From (i) and Given]
∴ 169 + 169 = 2 (AD² + 25)
∴ 338 = 2 (AD² + 25)
∴ \(\large \frac {338}{2}\) = AD² + 25
∴ 169 – 25 = AD²
∴ 144 = AD²
∴ AD = 12 cm …[Taking square roots]
AG = \(\large \frac {2}{3}\) AD …[Centroid divides each median in the ratio 3 : 1]
∴ AG = \(\large \frac {2}{3}\) × 12
∴ AG = 8 cm
Ans: The value of AG is 8 cm.
15. In a trapezium ABCD, seg AB || seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(□ABCD)
Given:
In trapezium ABCD,
seg AB || seg DC
seg BD ⊥ seg AD
seg AC ⊥ seg BC
AD = 15
BC = 15
AB = 25
To find:
A(□ABCD)
Construction:
Draw seg CM ⊥ side AB, (A – M – B)
Draw seg DN ⊥ side AB, (A – N – B)
Solution:
In ∆ABC,
∠ACB = 90⁰ …[Given]
∴ AC² + BC² = AB² …[By Pythagoras theorem]
∴ AC² + (15)² = (25)²
∴ AC² = (25)² – (15)²
∴ AC² = 625 – 225
∴ AC² = 400
∴ AC = 20 units …[Taking square roots]
A (∆ABC) = \(\large \frac {1}{2}\) × AB × CM …(i)
Also, A (∆ABC) = \(\large \frac {1}{2}\) × AC × BC …(ii)
∴ \(\large \frac {1}{2}\) × AB × CM = \(\large \frac {1}{2}\) × AC × BC
∴ 25 × CM = 20 × 15
∴ CM = \(\large \frac {20\,×\,15}{25}\)
∴ CM = 12 units …(iii)
In ∆BMC,
∠BMC = 90⁰ …[Construction]
∴ BC² = CM² + BM² …[By Pythagoras theorem]
∴ 15² = 12² + BM²
∴ BM² = 15² – 12²
∴ BM² = 225 – 144
∴ BM² = 81
∴ BM = 9 units …(iv) …[Taking square roots]
∴ CM = DN …(v) [Perpendicular distance between the same two parallel lines are equal]
In ∆BMC and ∆AND,
∠BMC ≅ ∠AND …[Each angle is 90⁰]
BC ≅ AD …[Given]
seg CM ≅ seg DN …[From (v)]
∴ ∆BMC ≅ ∆AND …[Hypotenuse side test]
∴ seg BM ≅ seg AN …(vi) …[c.s.s.t.]
∴ AN = 9 units …(vii) ...[From (iv) and (vi)]
AB = AN + MN + BM …[A – N – M – B]
∴ 25 = 9 + MN + 9
∴ MN = 25 – 18
∴ MN= 7 units ….(viii)
In □CMND,
seg MN || seg CD …[Given, A – N – M – B]
seg CM || seg DN …[Perpendiculars drawn to the same line are parallel]
∴ □ CMND is a parallelogram …[By Definition]
∴ CD = MN …[Opposite sides of parallelogram are equal]
∴ CD = 7 units …[From (viii)]
∴ A (trapezium ABCD) = \(\large \frac {1}{2}\) (AB + CD) × CM
∴ A (trapezium ABCD) = \(\large \frac {1}{2}\) (25 + 7) × 12
∴ A (trapezium ABCD) = \(\large \frac {1}{2}\) (32) × 12
∴ A (trapezium ABCD) = 192 sq. units.
Ans: A (trapezium ABCD) = 192 square units.
16. In the figure 2.35, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = \(\large \frac {1}{3}\) QR. Prove that : 9 PS² = 7 PQ²
Given:
∆PQR is an equilateral triangle
QS = \(\large \frac {1}{3}\) QR.
To prove:
9 PS² = 7 PQ²
Construction:
Draw seg PT ⊥ side QR, (Q – S – T – R)
Proof:
∆PQR is an equilateral triangle …[Given]
PQ = QR = PR …(i) [Sides of an equilateral triangle]
In ∆PTS,
∠PTS = 90⁰ …[Construction]
∴ PS² = PT² + ST² …(ii) [By Pythagoras theorem]
In ∆PTQ,
∠PTQ = 90⁰ …[Construction]
∠PQT = 60⁰ …[All angles of an equilateral triangle are 60⁰]
∠QPT = 30⁰ …[Remaining angle]
∴ ∆PTQ is a 30° – 60° – 90° triangle
By 30° – 60° – 90° triangle theorem,
∴ PT = \(\large (\frac {\sqrt{3}}{2}\)PQ …(iii) [Side opposite to 60°]
∴ QT = \(\large \frac {1}{2}\)PQ …(iv) [Side opposite to 30⁰]
ST = QT – QS …[Q – S – T]
∴ ST = \(\large \frac {1}{2}\)PQ – \(\large \frac {1}{3}\)QR …[From (iv) and Given]
∴ ST = \(\large \frac {1}{2}\)PQ – \(\large \frac {1}{3}\)PQ ...[From (i)]
∴ ST = \(\large (\frac {3PQ\,-\,2PQ}{6}\)
∴ ST = \(\large \frac {1}{6}\)PQ …(v)
∴ PS² = \(\large (\frac {\sqrt{3}}{2}\)PQ\(\large)\)² +\(\large (\frac {1}{6}\)PQ\(\large)\)² …[From (ii) (iii) and (v)]
∴ PS² = \(\large \frac {3PQ²}{4}\) + \(\large \frac {PQ²}{36}\)
∴ PS² = \(\large \frac {27PQ²}{36}\) + \(\large \frac {PQ²}{36}\) …[Equalising the denominators]
∴ PS² = \(\large \frac {27PQ²\,+\,PQ²}{36}\) ∴ PS² = \(\large \frac {28PQ²}{36}\)
∴ PS² = \(\large \frac {7}{9}\) PQ2
∴ 9 PS² = 7 PQ²
Hence Proved.
17. Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.
Given:
Seg PM is a median of ∆PQR
PQ = 40
PR = 42
PM = 29
To find:
QR
Solution:
In ∆PQR, seg PM is the median …[Given]
∴ PQ² + PR² = 2PM² + 2QM² …[By Appollonius theorem]
∴ 40² + 42² = 2 (29)² + 2(QM)²
∴ (40)² + (42)² = 2 (29² + QM²)
∴ 1600 + 1764 = 2 (841 + QM²)
∴ 3364 = 2 (841 + QM²)
∴ \(\large \frac {3364}{2}\) = 841 + QM²
∴ 1682 = 841 + QM²
∴ 1682 – 841 = QM²
∴ QM² = 841
∴ QM = 29 …[Taking square root]
QR = 2QM …[M is midpoint of seg QR]
∴ QR = 2 × 29
∴ QR = 58 units
Ans: The value of QR is 58 units.
18. Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM.
Given:
Seg AM is a median of ∆ABC.
AB = 22
AC = 34
BC = 24
To find:
AM
Solution:
In ∆ABC,
BM = \(\large \frac {1}{2}\)BC …[M is the midpoint of BC]
∴ BM = \(\large \frac {1}{2}\) × 24
∴ BM = 12 units
In ∆ABC, seg AM is the median …[Given]
∴ AB² + AC² = 2AM² + 2BM² …[By Apollonius theorem]
∴ 222 + 342 = 2 (AM² + BM²)
∴ 484 + 1156 = 2 (AM² + 12²)
∴ 1640 = 2(AM² + 144)
∴ \(\large \frac {1640}{2}\) = AM² + 144
∴ 820 = AM² + 144
∴ 820 – 144 = AM²
∴ AM² = 676
∴ AM = 26 units …[Taking square root]
Ans: The value of AM is 26 units.