Mathematics Part ii Solutions Solutions for Class 10 Maths Chapter 2 - Pythagoras Theorem

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Page No 38:

Question 1:

Identify, with reason, which of the following are Pythagorean triplets.
(i) (3, 5, 4)
(ii) (4, 9, 12)
(iii) (5, 12, 13)
(iv) (24, 70, 74)
(v) (10, 24, 27)
(vi) (11, 60, 61)

Answer:

(i) In the triplet (3, 5, 4),
3² = 9, 5² = 25, 4² = 16 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 5, 4) is a pythagorean triplet.

(ii) In the triplet (4, 9, 12),
4² = 16, 9² = 81, 12² = 144 and 16 + 81 = 97 ≠ 144

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (4, 9, 12) is not a pythagorean triplet.

(iii) In the triplet (5, 12, 13),
5² = 25, 12² = 144, 13² = 169 and 25 + 144 = 169

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (5, 12, 13) is a pythagorean triplet.

(iv) In the triplet (24, 70, 74),
24² = 576, 70² = 4900, 74² = 5476 and 576 + 4900 = 5476

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (24, 70, 74) is a pythagorean triplet.

(v) In the triplet (10, 24, 27),
10² = 100, 24² = 576, 27² = 729 and 100 + 576 = 676 ≠ 729

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (10, 24, 27) is not a pythagorean triplet.

(vi) In the triplet (11, 60, 61),
11² = 121, 60² = 3600, 61² = 3721 and 121 + 3600 = 3721

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (11, 60, 61) is a pythagorean triplet.

Page No 38:

Question 2:

In the given figure, ∠MNP = 90°, seg NQ ⊥seg MP, MQ = 9, QP = 4, find NQ.

Answer:

Page No 38:

Question 3:

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

Answer:

Page No 39:

Question 4:

In the given figure. Find RP and PS using the information given in âˆ†PSR.

Answer:

In âˆ†PSR,
∠S = 90^âˆ˜, ∠P = 30^âˆ˜, ∴ ∠R = 60^âˆ˜

By 30^âˆ˜− 60^âˆ˜ − 90^âˆ˜ theorem,

$RS = \frac{1}{2} \times RP \Rightarrow 6 = \frac{1}{2} \times RP \Rightarrow 6 \times 2 = RP \Rightarrow RP = 12 . . . (1) PS = \frac{\sqrt{3}}{2} \times RP = \frac{\sqrt{3}}{2} \times 12 = 6 \sqrt{3} . . . (2)$

Hence, RP = 12 and PS = 6 $\sqrt{3}$ .

Page No 39:

Question 5:

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........

$∴ ∠ BAC = ∴ AB = BC = \times AC = \times \sqrt{8} = \times 2 \sqrt{2} =$

Answer:

In âˆ†ABC,
∠B = 90^âˆ˜, AC = $\sqrt{8}$ , AB = BC, ∴ ∠A = ∠C = 45^âˆ˜

By 45^âˆ˜− 45^âˆ˜ − 90^âˆ˜ theorem,

$AB = BC = \frac{1}{\sqrt{2}} \times AC = \frac{1}{\sqrt{2}} \times \sqrt{8} = \frac{1}{\sqrt{2}} \times 2 \sqrt{2} = 2$

Hence, AB = 2 and BC = 2.

Hence, the completed activity is

AB = BC .......... $given$

$∴ ∠ BAC = 45^{°} ∴ AB = BC = \frac{1}{\sqrt{2}} \times AC = \frac{1}{\sqrt{2}} \times \sqrt{8} = \frac{1}{\sqrt{2}} \times 2 \sqrt{2} = 2$

Page No 39:

Question 6:

Find the side and perimeter of a square whose diagonal is 10 cm.

Answer:

It is given that ABCD is a square.

∴ AB = BC = CD = DA = a (say)

According to Pythagoras theorem, in âˆ†ABD

${AB}^{2} + {AD}^{2} = {BD}^{2} \Rightarrow a^{2} + a^{2} = 10^{2} \Rightarrow 2 a^{2} = 100 \Rightarrow a^{2} = 50 \Rightarrow a = \sqrt{50} \Rightarrow a = 5 \sqrt{2} cm$

Hence, the side of the square is 5 $\sqrt{2}$ cm.

Now,
Perimeter of a square = $4 \times (side)$
                                   = $4 \times a$
                                   = $4 \times 5 \sqrt{2}$
                                   = $20 \sqrt{2}$ cm

Hence, the perimeter of the square is 20 $\sqrt{2}$ cm.

Page No 39:

Question 7:

In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

Answer:

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.

Here, seg GF ⊥ seg ED

$∴ {GF}^{2} = EG \times GD \Rightarrow 12^{2} = EG \times 8 \Rightarrow 144 = EG \times 8 \Rightarrow EG = \frac{144}{8} \Rightarrow EG = 18$

Hence, EG = 18.

Now,
According to Pythagoras theorem, in âˆ†DGF

${DG}^{2} + {GF}^{2} = {FD}^{2} \Rightarrow 8^{2} + 12^{2} = {FD}^{2} \Rightarrow 64 + 144 = {FD}^{2} \Rightarrow {FD}^{2} = 208 \Rightarrow FD = 4 \sqrt{13}$

In âˆ†EGF

${EG}^{2} + {GF}^{2} = {EF}^{2} \Rightarrow 18^{2} + 12^{2} = {EF}^{2} \Rightarrow 324 + 144 = {EF}^{2} \Rightarrow {EF}^{2} = 468 \Rightarrow EF = 6 \sqrt{13}$

Hence, FD = $4 \sqrt{13}$ and EF = $6 \sqrt{13}$ .

Page No 39:

Question 8:

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Answer:

According to Pythagoras theorem, in âˆ†ABC

${AB}^{2} + {BC}^{2} = {AC}^{2} \Rightarrow 35^{2} + 12^{2} = {AC}^{2} \Rightarrow 1225 + 144 = {AC}^{2} \Rightarrow {AC}^{2} = 1369 \Rightarrow AC = 37 cm$

Hence, the length of the diagonal is 37 cm.

Page No 39:

Question 9:

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ²= 4PM²– 3PR²

Answer:

According to Pythagoras theorem,

In âˆ†PRM

${PR}^{2} + {RM}^{2} = {PM}^{2} \Rightarrow {RM}^{2} = {PM}^{2} - {PR}^{2} . . . (1)$

In âˆ†PRQ

${PR}^{2} + {RQ}^{2} = {PQ}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(RM + MQ)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(RM + RM)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(2 RM)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + 4 {RM}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + 4 ({PM}^{2} - {PR}^{2}) (from (1)) \Rightarrow {PQ}^{2} = {PR}^{2} + 4 {PM}^{2} - 4 {PR}^{2} \Rightarrow {PQ}^{2} = 4 {PM}^{2} - 3 {PR}^{2}$

Hence, PQ²= 4PM²– 3PR².

Page No 39:

Question 10:

Walls of two buildings on either side of a street are parellel 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 4.2 m. Find the width of the street.

Answer:

Let the length of the ladder be 5.8 m.

According to Pythagoras theorem, in âˆ†EAB

${EA}^{2} + {AB}^{2} = {EB}^{2} \Rightarrow {(4.2)}^{2} + {AB}^{2} = {(5.8)}^{2} \Rightarrow 17.64 + {AB}^{2} = 33.64 \Rightarrow {AB}^{2} = 33.64 - 17.64 \Rightarrow {AB}^{2} = 16 \Rightarrow AB = 4 m . . . (1)$

In âˆ†DCB

${DC}^{2} + {CB}^{2} = {DB}^{2} \Rightarrow {(4)}^{2} + {CB}^{2} = {(5.8)}^{2} \Rightarrow 16 + {CB}^{2} = 33.64 \Rightarrow {CB}^{2} = 33.64 - 16 \Rightarrow {CB}^{2} = 17.64 \Rightarrow CB = 4.2 m . . . (2)$

From (1) and (2), we get

$AB + BC = (4 + 4.2) m = 8.2 m$

Hence, the width of the street is 8.2 m.

Page No 43:

Question 1:

In âˆ†PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.

Answer:

In âˆ†PQR, point S is the midpoint of side QR.

$QS = SR = \frac{1}{2} QR$

${PQ}^{2} + {PR}^{2} = 2 {PS}^{2} + 2 {QS}^{2} (by Apollonius theorem) \Rightarrow 11^{2} + 17^{2} = 2 {(13)}^{2} + 2 {QS}^{2} \Rightarrow 121 + 289 = 2 (169) + 2 {QS}^{2} \Rightarrow 410 = 338 + 2 {QS}^{2} \Rightarrow 2 {QS}^{2} = 410 - 338 \Rightarrow 2 {QS}^{2} = 72 \Rightarrow {QS}^{2} = 36 \Rightarrow QS = 6 ∴ QR = 2 \times QS = 2 \times 6 = 12$

Hence, QR = 12.

Page No 43:

Question 2:

In âˆ†ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB

Answer:

In âˆ†ACB, point D is the midpoint of side AB.

$AD = BD = \frac{1}{2} AB = 5$

${CA}^{2} + {CB}^{2} = 2 {DC}^{2} + 2 {AD}^{2} (by Apollonius theorem) \Rightarrow 7^{2} + 9^{2} = 2 {DC}^{2} + 2 (5^{2}) \Rightarrow 49 + 81 = 2 {DC}^{2} + 2 (25) \Rightarrow 130 = 2 {DC}^{2} + 50 \Rightarrow 2 {DC}^{2} = 130 - 50 \Rightarrow 2 {DC}^{2} = 80 \Rightarrow {DC}^{2} = 40 \Rightarrow DC = 2 \sqrt{10}$

Hence, the length of the median drawn from point C to side AB is $2 \sqrt{10}$ .

Page No 43:

Question 3:

In the given figure seg PS is the median of âˆ†PQR and PT ⊥ QR. Prove that,

(1) ${PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}$

(2) ${PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}$

Answer:

According to Pythagoras theorem, in âˆ†PTQ

${PQ}^{2} = {PT}^{2} + {QT}^{2} . . . (1)$

In âˆ†PTS

${PS}^{2} = {PT}^{2} + {TS}^{2} . . . (2)$

In âˆ†PTR

${PR}^{2} = {PT}^{2} + {RT}^{2} . . . (3)$

In âˆ†PQR, point S is the midpoint of side QR.

$QS = SR = \frac{1}{2} QR . . . (4)$

${PQ}^{2} + {PR}^{2} = 2 {PS}^{2} + 2 {QS}^{2} (by Apollonius theorem) \Rightarrow {PR}^{2} = 2 {PS}^{2} + 2 {QS}^{2} - {PQ}^{2} \Rightarrow {PR}^{2} = {PS}^{2} + {PS}^{2} + {QS}^{2} + {QS}^{2} - {PQ}^{2} \Rightarrow {PR}^{2} = {PS}^{2} + ({PT}^{2} + {TS}^{2}) + {QS}^{2} + {(\frac{QR}{2})}^{2} - ({PT}^{2} + {QT}^{2}) (From (1), (2) and (4)) \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {PT}^{2} + {TS}^{2} + {QS}^{2} - {PT}^{2} - {QT}^{2} \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} + ({TS}^{2} - {QT}^{2}) \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} + (TS + QT) (TS - QT) \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} + (QS) (TS - QT) \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} + QS \times TS - QS \times QT \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + QS \times TS + {QS}^{2} - QS \times QT \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + QS \times TS + QS (QS - QT) \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + QS \times TS + QS \times TS \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + 2 QS \times TS \Rightarrow {PR}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + QR \times TS$

Hence, ${PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}$ .

Now,

${PQ}^{2} + {PR}^{2} = 2 {PS}^{2} + 2 {QS}^{2} (by Apollonius theorem) \Rightarrow {PQ}^{2} = 2 {PS}^{2} + 2 {QS}^{2} - {PR}^{2} \Rightarrow {PQ}^{2} = {PS}^{2} + {PS}^{2} + {QS}^{2} + {QS}^{2} - {PR}^{2} \Rightarrow {PQ}^{2} = {PS}^{2} + ({PT}^{2} + {TS}^{2}) + {QS}^{2} + {(\frac{QR}{2})}^{2} - ({PT}^{2} + {RT}^{2}) (From (2), (3) and (4)) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {PT}^{2} + {TS}^{2} + {QS}^{2} - {PT}^{2} - {RT}^{2} \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} - ({RT}^{2} - {TS}^{2}) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} - (TS + RT) (RT - TS) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} - (TS + RT) (RS) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} - (TS + RT) (QS) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} + {QS}^{2} - QS \times TS - QS \times RT \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - QS \times TS + {QS}^{2} - QS \times RT \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - QS \times TS - QS (RT - QS) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - QS \times TS - QS (RT - SR) \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - QS \times TS - QS \times TS \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - 2 QS \times TS \Rightarrow {PQ}^{2} = {PS}^{2} + {(\frac{QR}{2})}^{2} - QR \times TS$

Hence, ${PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}$ .

Page No 43:

Question 4:

In âˆ†ABC, point M is the midpoint of side BC.
If, AB²+ AC²= 290 cm², AM = 8 cm, find BC.

Answer:

In âˆ†ABC, point M is the midpoint of side BC.

$BM = MC = \frac{1}{2} BC$

${AB}^{2} + {AC}^{2} = 2 {AM}^{2} + 2 {BM}^{2} (by Apollonius theorem) \Rightarrow 290 = 2 {(8)}^{2} + 2 {BM}^{2} \Rightarrow 290 = 2 (64) + 2 {BM}^{2} \Rightarrow 290 = 128 + 2 {BM}^{2} \Rightarrow 2 {BM}^{2} = 290 - 128 \Rightarrow 2 {BM}^{2} = 162 \Rightarrow {BM}^{2} = 81 \Rightarrow BM = 9 ∴ BC = 2 \times BM = 2 \times 9 = 18 cm$

Hence, BC = 18 cm.

Page No 43:

Question 5:

In the given figure, 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)

Answer:

According to Pythagoras theorem, in âˆ†PAT

${PT}^{2} = {PA}^{2} + {AT}^{2} . . . (1)$

In âˆ†ATS

${TS}^{2} = {AT}^{2} + {AS}^{2} . . . (2)$

In âˆ†QBT

${QT}^{2} = {QB}^{2} + {BT}^{2} . . . (3)$

In âˆ†BRT

${TR}^{2} = {BT}^{2} + {BR}^{2} . . . (4)$

Now,
${TS}^{2} + {TQ}^{2} = ({AS}^{2} + {AT}^{2}) + ({QB}^{2} + {BT}^{2}) (From (2) and (3)) \Rightarrow {TS}^{2} + {TQ}^{2} = ({BR}^{2} + {AT}^{2}) + ({PA}^{2} + {BT}^{2}) (∵ AS = BR and PA = QB) \Rightarrow {TS}^{2} + {TQ}^{2} = ({BR}^{2} + {BT}^{2}) + ({PA}^{2} + {AT}^{2}) \Rightarrow {TS}^{2} + {TQ}^{2} = ({TR}^{2}) + ({PT}^{2}) (From (1) and (4)) \Rightarrow {TS}^{2} + {TQ}^{2} = {TR}^{2} + {PT}^{2}$

Hence, TS²+ TQ²= TP²+ TR².

Page No 43:

Question 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)

(2) In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?

(A) 15 (B) 13 (C) 5 (D) 12

(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

(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

(5) Find 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

(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}

(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

(8) In âˆ†ABC, AB =

6 \sqrt{3}

cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.

(A) 30° (B) 60° (C) 90° (D) 45°

Answer:

(1) (A) In the triplet (1, 5, 10),
1² = 1, 5² = 25, 10² = 100 and 1 + 25 = 26 ≠ 100

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (1, 5, 10) is not a pythagorean triplet.

(B) In the triplet (3, 4, 5),
3² = 9, 4² = 16, 5² = 25 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 4, 5) is a pythagorean triplet.

(C) In the triplet (2, 2, 2),
2² = 4, 2² = 4, 2² = 4 and 4 + 4 = 8 ≠ 4

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (2, 2, 2) is not a pythagorean triplet.

(D) In the triplet (5, 5, 2),
2² = 4, 5² = 25, 5² = 25 and 4 + 25 = 29 ≠ 25

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (5, 5, 2) is not a pythagorean triplet.

Hence, the correct option is (B).

(2) According to the pythagoras theorem,
Sum of the squares of the sides making right angle is equal to the square of the third side.

∴ 169 = square of the hypotenuse
⇒ Length of the hypotenuse = $\sqrt{169}$
                                              = 13

Hence, the correct option is (B).

(3) (A) In the triplet 15/08/17,
15² = 225, 8² = 64, 17² = 289 and 225 + 64 = 289

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 15/08/17 is a pythagorean triplet.

(B) In the triplet 16/08/16,
16² = 256, 8² = 64, 16² = 256 and 256 + 64 = 320 ≠ 256

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 16/08/16 is not a pythagorean triplet.

(C) In the triplet 3/5/17,
3² = 9, 5² = 25, 17² = 289 and 9 + 25 = 34 ≠ 289

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 3/5/17 is not a pythagorean triplet.

(D) In the triplet 4/9/15,
4² = 16, 9² = 81, 15² = 225 and 16 + 81 = 97 ≠ 225

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 4/9/15 is not a pythagorean triplet.

Hence, the correct option is (A).

(4) 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.

Hence, the correct option is (C).

(5)

It is given that ABCD is a square.

∴ AB = BC = CD = DA = x (say)

According to Pythagoras theorem, in âˆ†ABD

${AB}^{2} + {AD}^{2} = {BD}^{2} \Rightarrow x^{2} + x^{2} = {(10 \sqrt{2})}^{2} \Rightarrow 2 x^{2} = 200 \Rightarrow x^{2} = 100 \Rightarrow x = \sqrt{100} \Rightarrow x = 10 cm$

Hence, the side of the square is 10 cm.

Now,
Perimeter of a square = $4 \times (side)$
                                   = $4 \times x$
                                   = $4 \times 10$
                                   = $40$ cm

Hence, the correct option is (D).

(6)

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.

$∴ {AD}^{2} = CD \times DB = 4 \times 9 = 36 \Rightarrow AD = 6 cm$

Hence, the correct option is (C).

(7) According to Pythagoras theorem,

${(Hypotenuse)}^{2} = {(Base)}^{2} + {(Height)}^{2} = {(18)}^{2} + {(24)}^{2} = 324 + 576 = 900 ∴ Hypotenuse = 30$

Hence, the correct option is (B).

(8) In âˆ†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

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 30°

Hence, the correct option is (A).

Page No 44:

Question 2:

Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle ? Give reason.
(3) Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.
(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.
(6) In âˆ†PQR; PQ = $\sqrt{8}$ , QR = $\sqrt{5}$ , PR = $\sqrt{3}$ . Is âˆ†PQR a right angled triangle ? If yes, which angle is of 90°?

Answer:

(1)

Since, ABC is an equilateral triangle, AD is the perpendicular bisector of BC.

Now, According to Pythagoras theorem,
In âˆ†ABD

${AB}^{2} = {AD}^{2} + {BD}^{2} \Rightarrow {(2 a)}^{2} = {AD}^{2} + a^{2} \Rightarrow 4 a^{2} - a^{2} = {AD}^{2} \Rightarrow {AD}^{2} = 3 a^{2} \Rightarrow AD = \sqrt{3} a$

Hence, the height of an equilateral triangle is $\sqrt{3} a$ .

(2) In the triplet (7, 24, 25),
7² = 49, 24² = 576, 25² = 625 and 49 + 576 = 625

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ Sides 7 cm, 24 cm, 25 cm form a right angled triangle.

(3)

According to Pythagoras theorem,

In âˆ†ABC

${AB}^{2} + {BC}^{2} = {AC}^{2} \Rightarrow {(60)}^{2} + {(11)}^{2} = {AC}^{2} \Rightarrow 3600 + 121 = {AC}^{2} \Rightarrow {AC}^{2} = 3721 \Rightarrow AC = 61 cm$

Hence, the length of a diagonal of the rectangle is 61 cm.

(4) In a right angled triangle,

According to Pythagoras theorem.

${(Hypotenuse)}^{2} = {(Base)}^{2} + {(Height)}^{2} = {(9)}^{2} + {(12)}^{2} = 81 + 144 = 225 ∴ Hypotenuse = 15 cm$

Hence, the length of the hypotenuse is 15 cm.

(5) It is given that, a side of an isosceles right angled triangle is x.

Then, the other side of the triangle is also x.

According to Pythagoras theorem.

${(Hypotenuse)}^{2} = x^{2} + x^{2} = 2 x^{2} ∴ Hypotenuse = \sqrt{2} x$

Hence, its hypotenuse is $\sqrt{2} x$ .

(6) In âˆ†PQR, PQ = $\sqrt{8}$ , QR = $\sqrt{5}$ , PR = $\sqrt{3}$ .
( $\sqrt{8}$ )² = 8, ( $\sqrt{5}$ )² = 5, ( $\sqrt{3}$ )² = 3 and 3 + 5 = 8

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ âˆ†PQR form a right angled triangle, where angle R is of 90°.

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Question 3:

In âˆ†RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.

Answer:

In âˆ†RST,
∠S = 90^âˆ˜, ∠T = 30^âˆ˜, ∴ ∠R = 60^âˆ˜

By 30^âˆ˜− 60^âˆ˜ − 90^âˆ˜ theorem,

$RS = \frac{1}{2} \times RT \Rightarrow RS = \frac{1}{2} \times 12 \Rightarrow RS = 6 cm . . . (1) ST = \frac{\sqrt{3}}{2} \times RT \Rightarrow ST = \frac{\sqrt{3}}{2} \times 12 \Rightarrow ST = 6 \sqrt{3} cm . . . (2)$

Hence, RS = 6 cm and ST = 6 $\sqrt{3}$ cm.

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Question 4:

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Answer:

It is given that, area of rectangle is 192 sq.cm.

$Area = Length \times Breadth \Rightarrow 192 = 16 \times BC \Rightarrow BC = \frac{192}{16} \Rightarrow BC = 12 cm . . . (1)$

According to Pythagoras theorem,

In âˆ†ABC

${AB}^{2} + {BC}^{2} = {AC}^{2} \Rightarrow {(16)}^{2} + {(12)}^{2} = {AC}^{2} \Rightarrow 256 + 144 = {AC}^{2} \Rightarrow {AC}^{2} = 400 \Rightarrow AC = 20 cm$

Hence, the length of a diagonal of the rectangle is 20 cm.

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Question 5:

Find the length of the side and perimeter of an equilateral triangle whose height is $\sqrt{3}$ cm.

Answer:

Since, ABC is an equilateral triangle, CD is the perpendicular bisector of AB.

Now, According to Pythagoras theorem,
In âˆ†ACD

${AC}^{2} = {AD}^{2} + {CD}^{2} \Rightarrow {(2 a)}^{2} = a^{2} + {(\sqrt{3})}^{2} \Rightarrow 4 a^{2} - a^{2} = 3 \Rightarrow 3 a^{2} = 3 \Rightarrow a^{2} = 1 \Rightarrow a = 1 cm$

Hence, the length of the side of an equilateral triangle is 2 cm.

Now,
Perimeter of the triangle = (2 + 2 + 2) cm
= 6 cm

Hence, perimeter of an equilateral triangle is 6 cm.

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Question 6:

In âˆ†ABC seg AP is a median. If BC = 18, AB²+ AC²= 260 Find AP.

Answer:

In âˆ†ABC, point P is the midpoint of side BC.

$BP = PC = \frac{1}{2} BC = 9$

${AB}^{2} + {AC}^{2} = 2 {AP}^{2} + 2 {BP}^{2} (by Apollonius theorem) \Rightarrow 260 = 2 {AP}^{2} + 2 (9^{2}) \Rightarrow 260 = 2 {AP}^{2} + 2 (81) \Rightarrow 260 = 2 {AP}^{2} + 162 \Rightarrow 2 {AP}^{2} = 260 - 162 \Rightarrow 2 {AP}^{2} = 98 \Rightarrow {AP}^{2} = 49 \Rightarrow AP = 7$

Hence, AP = 7.

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Question 7:

âˆ†ABC is an equilateral triangle. Point P is on base BC such that PC = $\frac{1}{3}$ BC, if AB = 6 cm find AP.

Answer:

âˆ†ABC is an equilateral triangle.

It is given that,
$PC = \frac{1}{3} BC \Rightarrow PC = \frac{1}{3} \times 6 \Rightarrow PC = 2 cm \Rightarrow BP = 4 cm$

Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
∴ OC = 3 cm
⇒ OP = OC − PC
= 3 − 2
= 1 ...(1)

Now, According to Pythagoras theorem,
In âˆ†AOB,

${AB}^{2} = {AO}^{2} + {OB}^{2} \Rightarrow {(6)}^{2} = {AO}^{2} + {(3)}^{2} \Rightarrow 36 - 9 = {AO}^{2} \Rightarrow {AO}^{2} = 27 \Rightarrow AO = 3 \sqrt{3} cm . . . (2)$

In âˆ†AOP,

${AP}^{2} = {AO}^{2} + {OP}^{2} \Rightarrow {AP}^{2} = {(3 \sqrt{3})}^{2} + {(1)}^{2} (From (1) and (2)) \Rightarrow {AP}^{2} = 27 + 1 \Rightarrow {AP}^{2} = 28 \Rightarrow AP = 2 \sqrt{7} cm$

Hence, AP = 2 $\sqrt{7}$ cm.

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Question 8:

From the information given in the figure, prove that PM = PN = $\sqrt{3}$ × a

Answer:

Since, âˆ†PQR is an equilateral triangle, PS is the perpendicular bisector of QR.
∴ QS = SR = $\frac{a}{2}$ ...(1)

Now, According to Pythagoras theorem,
In âˆ†PQS,

${PQ}^{2} = {QS}^{2} + {PS}^{2} \Rightarrow a^{2} = {(\frac{a}{2})}^{2} + {PS}^{2} \Rightarrow {PS}^{2} = a^{2} - \frac{a^{2}}{4} \Rightarrow {PS}^{2} = \frac{4 a^{2} - a^{2}}{4} \Rightarrow {PS}^{2} = \frac{3 a^{2}}{4} \Rightarrow PS = \frac{\sqrt{3} a}{2} . . . (2)$

In âˆ†PMS,

${PM}^{2} = {MS}^{2} + {PS}^{2} \Rightarrow {PM}^{2} = {(a + \frac{a}{2})}^{2} + {(\frac{\sqrt{3}}{2} a)}^{2} \Rightarrow {PM}^{2} = {(\frac{3 a}{2})}^{2} + {(\frac{\sqrt{3}}{2} a)}^{2} \Rightarrow {PM}^{2} = \frac{9 a^{2}}{4} + \frac{3 a^{2}}{4} \Rightarrow {PM}^{2} = \frac{12 a^{2}}{4} \Rightarrow {PM}^{2} = 3 a^{2} \Rightarrow PM = \sqrt{3} a . . . (3)$

In âˆ†PNS,

${PN}^{2} = {NS}^{2} + {PS}^{2} \Rightarrow {PN}^{2} = {(a + \frac{a}{2})}^{2} + {(\frac{\sqrt{3}}{2} a)}^{2} \Rightarrow {PN}^{2} = {(\frac{3 a}{2})}^{2} + {(\frac{\sqrt{3}}{2} a)}^{2} \Rightarrow {PN}^{2} = \frac{9 a^{2}}{4} + \frac{3 a^{2}}{4} \Rightarrow {PN}^{2} = \frac{12 a^{2}}{4} \Rightarrow {PN}^{2} = 3 a^{2} \Rightarrow PN = \sqrt{3} a . . . (4)$

From (3) and (4), we get
PM = PN = $\sqrt{3}$ × a

Hence, PM = PN = $\sqrt{3}$ × a.

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Question 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.

Answer:

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In âˆ†ABD, point O is the midpoint of side BD.

$BO = OD = \frac{1}{2} BD$

${AB}^{2} + {AD}^{2} = 2 {AO}^{2} + 2 {BO}^{2} (by Apollonius theorem) . . . (1)$

In âˆ†CBD, point O is the midpoint of side BD.

$BO = OD = \frac{1}{2} BD$

${CB}^{2} + {CD}^{2} = 2 {CO}^{2} + 2 {BO}^{2} (by Apollonius theorem) . . . (2)$

Adding (1) and (2), we get

${AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = 2 {AO}^{2} + 2 {BO}^{2} + 2 {CO}^{2} + 2 {BO}^{2} \Rightarrow {AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = 2 {AO}^{2} + 4 {BO}^{2} + 2 {AO}^{2} (∵ OC = OA) \Rightarrow {AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = 4 {AO}^{2} + 4 {BO}^{2} \Rightarrow {AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = {(2 AO)}^{2} + {(2 BO)}^{2} \Rightarrow {AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = {(AC)}^{2} + {(BD)}^{2} \Rightarrow {AB}^{2} + {AD}^{2} + {CB}^{2} + {CD}^{2} = {AC}^{2} + {BD}^{2}$

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

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Question 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 distance between them was $15 \sqrt{2}$ km. Find their speed per hour.

Answer:

It is given that, Pranali and Prasad have same speed.
Thus, they cover same distance in 2 hours.
i.e. OA = OB

Let the speed be x km per hour.

According to Pythagoras theorem,
In âˆ†AOB

${AB}^{2} = {AO}^{2} + {OB}^{2} \Rightarrow {(15 \sqrt{2})}^{2} = {AO}^{2} + {OA}^{2} \Rightarrow 450 = 2 {AO}^{2} \Rightarrow {AO}^{2} = \frac{450}{2} \Rightarrow {AO}^{2} = 225 \Rightarrow AO = 15 km \Rightarrow BO = 15 km$

$Speed = \frac{Distance}{Time} = \frac{15}{2} = 7.5 km per hour$

Hence, their speed is 7.5 km per hour.

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Question 11:

In âˆ†ABC, ∠BAC = 90°, seg BL and seg CM are medians of âˆ†ABC. Then prove that:
4(BL²+ CM²) = 5 BC²

Answer:

According to Pythagoras theorem,
In âˆ†ABC

${AB}^{2} + {AC}^{2} = {CB}^{2} . . . (1)$

In âˆ†ABC, point M is the midpoint of side BD.

$BM = MA = \frac{1}{2} AB$

${AC}^{2} + {BC}^{2} = 2 {CM}^{2} + 2 {AM}^{2} (by Apollonius theorem) . . . (2)$

In âˆ†CBA, point L is the midpoint of side AC.

$CL = LA = \frac{1}{2} AC$

${CB}^{2} + {AB}^{2} = 2 {BL}^{2} + 2 {AL}^{2} (by Apollonius theorem) . . . (3)$

Adding (2) and (3), we get

${AC}^{2} + {BC}^{2} + {CB}^{2} + {AB}^{2} = 2 {CM}^{2} + 2 {AM}^{2} + 2 {BL}^{2} + 2 {AL}^{2} \Rightarrow 2 {CM}^{2} + 2 {BL}^{2} = {AC}^{2} + {BC}^{2} + {CB}^{2} + {AB}^{2} - 2 {AM}^{2} - 2 {AL}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = {AC}^{2} + 2 {BC}^{2} + {AB}^{2} - 2 {AM}^{2} - 2 {AL}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = ({AC}^{2} + {AB}^{2}) + 2 {BC}^{2} - 2 {(\frac{1}{2} AB)}^{2} - 2 {(\frac{1}{2} AC)}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = {BC}^{2} + 2 {BC}^{2} - \frac{1}{2} {AB}^{2} - \frac{1}{2} {AC}^{2} (From (1)) \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = 3 {BC}^{2} - \frac{1}{2} ({AB}^{2} + {AC}^{2}) \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = 3 {BC}^{2} - \frac{1}{2} ({BC}^{2}) \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 2 [3 {BC}^{2} - \frac{1}{2} ({BC}^{2})] \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 6 {BC}^{2} - {BC}^{2} \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 5 {BC}^{2}$

Hence, 4(BL²+ CM²) = 5 BC².

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Question 12:

Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

Answer:

It is given that,
AB² + AD² = 130 sq. cm
BD = 14 cm

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In âˆ†ABD, point O is the midpoint of side BD.

$BO = OD = \frac{1}{2} BD = 7 cm$

${AB}^{2} + {AD}^{2} = 2 {AO}^{2} + 2 {BO}^{2} (by Apollonius theorem) \Rightarrow 130 = 2 {AO}^{2} + 2 {(7)}^{2} \Rightarrow 130 = 2 {AO}^{2} + 2 \times 49 \Rightarrow 130 = 2 {AO}^{2} + 98 \Rightarrow 2 {AO}^{2} = 130 - 98 \Rightarrow 2 {AO}^{2} = 32 \Rightarrow {AO}^{2} = 16 \Rightarrow AO = 4 cm$

Sinec, point O is the midpoint of side AC.

$∴ AC = 2 AO = 8 cm$

Hence, the length of the other diagonal is 8 cm.

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Question 13:

In âˆ†ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :
2AB²= 2AC²+ BC²

Answer:

It is given that,
DB = 3 CD

∴ BC = 4 CD ....(1)

According to Pythagoras theorem,
In âˆ†ABD

${AB}^{2} = {AD}^{2} + {DB}^{2} \Rightarrow {AD}^{2} = {AB}^{2} - {DB}^{2} . . . (2)$

In âˆ†ACD

${AC}^{2} = {AD}^{2} + {CD}^{2} \Rightarrow {AC}^{2} = ({AB}^{2} - {DB}^{2}) + {CD}^{2} (From (2)) \Rightarrow {AC}^{2} = {AB}^{2} - {(3 CD)}^{2} + {CD}^{2} (Given) \Rightarrow {AC}^{2} = {AB}^{2} - 9 {CD}^{2} + {CD}^{2} \Rightarrow {AC}^{2} = {AB}^{2} - 8 {CD}^{2} \Rightarrow {AB}^{2} = {AC}^{2} + 8 {CD}^{2} \Rightarrow {AB}^{2} = {AC}^{2} + 8 {(\frac{BC}{4})}^{2} (From (1)) \Rightarrow {AB}^{2} = {AC}^{2} + \frac{{BC}^{2}}{2} \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + {BC}^{2}$

Hence, 2AB²= 2AC²+ BC².

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Question 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.

Answer:

$Area of the triangle = \sqrt{s (s - a) (s - b) (s - c)} s = \frac{a + b + c}{2} = \frac{13 + 13 + 10}{2} = \frac{36}{2} = 18 cm Area of the triangle = \sqrt{18 (18 - 13) (18 - 13) (18 - 10)} = \sqrt{2 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2} = 60 sq . cm Also, Area of the triangle = \frac{1}{2} \times base \times height \Rightarrow 60 = \frac{1}{2} \times 10 \times height \Rightarrow height = \frac{60}{5} \Rightarrow height = 12 cm$

The centroid is located two third of the distance from any vertex of the triangle.

$∴ Distance between the vertex and the centroid = \frac{2}{3} \times 12 = 8 cm$

Hence, the distance between the vertex opposite the base and the centroid is 8 cm.

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Question 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(â–¢ABCD)

Answer:

According to Pythagoras theorem,
In âˆ†ABD

${AB}^{2} = {AD}^{2} + {DB}^{2} \Rightarrow {(25)}^{2} = {(15)}^{2} + {BD}^{2} \Rightarrow 625 = 225 + {BD}^{2} \Rightarrow {BD}^{2} = 625 - 225 \Rightarrow {BD}^{2} = 400 \Rightarrow BD = 20$

Now,
$Area of the triangle ABD = \sqrt{s (s - a) (s - b) (s - c)} s = \frac{a + b + c}{2} = \frac{20 + 25 + 15}{2} = \frac{60}{2} = 30 Area of the triangle = \sqrt{30 (30 - 25) (30 - 20) (30 - 15)} = \sqrt{30 \times 5 \times 10 \times 15} = 150 sq . units Also, Area of the triangle = \frac{1}{2} \times base \times height \Rightarrow 150 = \frac{1}{2} \times 25 \times DP \Rightarrow DP = \frac{300}{25} \Rightarrow DP = 12$

Therefore, height of the trapezium = 12.

Now,
According to Pythagoras theorem,
In âˆ†ADP

${AD}^{2} = {AP}^{2} + {DP}^{2} \Rightarrow {(15)}^{2} = {(12)}^{2} + {AP}^{2} \Rightarrow 225 = 144 + {AP}^{2} \Rightarrow {AP}^{2} = 225 - 144 \Rightarrow {AP}^{2} = 81 \Rightarrow AP = 9$

∴ AP = QB = 9

∴ CD = PQ = 25 − (9 + 9) = 7
$Area of Trapezium = \frac{1}{2} \times Sum of parallel sides \times Height = \frac{1}{2} \times (25 + 7) \times 12 = \frac{1}{2} \times 32 \times 12 = 32 \times 6 = 192 sq . units$

Hence, A(â–¢ABCD) = 192 sq. units.

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Question 16:

In the given figure, âˆ†PQR is an equilateral triangle. Point S is on seg QR such that
QS = $\frac{1}{3}$ QR.
Prove that : 9 PS²= 7 PQ²

Answer:

Let the side of equilateral triangle âˆ†PQR be x.
PT be the altitude of the âˆ†PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.
∴ QT = TR = $\frac{1}{2} QR = \frac{x}{2}$

Given: QS = $\frac{1}{3}$ QR = $\frac{x}{3}$

$∴ ST = QT - QS = \frac{x}{2} - \frac{x}{3} = \frac{x}{6}$

According to Pythagoras theorem,
In âˆ†PQT

${PQ}^{2} = {QT}^{2} + {PT}^{2} \Rightarrow {(x)}^{2} = {(\frac{x}{2})}^{2} + {PT}^{2} \Rightarrow x^{2} = \frac{x^{2}}{4} + {PT}^{2} \Rightarrow {PT}^{2} = x^{2} - \frac{x^{2}}{4} \Rightarrow {PT}^{2} = \frac{3 x^{2}}{4} \Rightarrow PT = \frac{\sqrt{3} x}{2}$

In âˆ†PST

${PS}^{2} = {ST}^{2} + {PT}^{2} \Rightarrow {PS}^{2} = {(\frac{x}{6})}^{2} + {(\frac{\sqrt{3} x}{2})}^{2} \Rightarrow {PS}^{2} = \frac{x^{2}}{36} + \frac{3 x^{2}}{4} \Rightarrow {PS}^{2} = \frac{x^{2} + 27 x^{2}}{36} \Rightarrow {PS}^{2} = \frac{28 x^{2}}{36} \Rightarrow {PS}^{2} = \frac{7 x^{2}}{9} \Rightarrow 9 {PS}^{2} = 7 {PQ}^{2}$

Hence, 9 PS²= 7 PQ².

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Question 17:

Seg PM is a median of âˆ†PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

Answer:

In âˆ†PQR, point M is the midpoint of side QR.

$QM = MR = \frac{1}{2} QR$

${PQ}^{2} + {PR}^{2} = 2 {PM}^{2} + 2 {QM}^{2} (by Apollonius theorem) \Rightarrow {(40)}^{2} + {(42)}^{2} = 2 {(29)}^{2} + 2 {QM}^{2} \Rightarrow 1600 + 1764 = 1682 + 2 {QM}^{2} \Rightarrow 3364 - 1682 = 2 {QM}^{2} \Rightarrow 1682 = 2 {QM}^{2} \Rightarrow {QM}^{2} = 841 \Rightarrow QM = 29 \Rightarrow QR = 2 \times 29 \Rightarrow QR = 58$

Hence, QR = 58.

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Question 18:

Seg AM is a median of âˆ†ABC. If AB = 22, AC = 34, BC = 24, find AM

Answer:

In âˆ†ABC, point M is the midpoint of side BC.

$BM = MC = \frac{1}{2} BC = 12$

${AB}^{2} + {AC}^{2} = 2 {AM}^{2} + 2 {BM}^{2} (by Apollonius theorem) \Rightarrow {(22)}^{2} + {(34)}^{2} = 2 {AM}^{2} + 2 {(12)}^{2} \Rightarrow 484 + 1156 = 2 {AM}^{2} + 288 \Rightarrow 1640 - 288 = 2 {AM}^{2} \Rightarrow 1352 = 2 {AM}^{2} \Rightarrow {AM}^{2} = 676 \Rightarrow AM = 26$

Hence, AM = 26.

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