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### Practice Set 2.2 Pythagoras Theorem Class 10th Mathematics Part 2 MHB Solution

Practice Set 2.2

1. In Î”PQR, point S is the midpoint of side QR. If PQ = 11,PR = 17, PS = 13,find QR.…
2. In Î”ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C…
3. In the figure 2.28 seg PS is themedian of Î”PQR and PT ⊥ QR. Prove that, (1) pr^2 = ps^2…
4. In Î”ABC, point M is the midpointof side BC. If, AB^2 + AC^2 = 290 cm^2 ,AM = 8 cm, find…
5. In figure 2.30, point T is in theinterior of rectangle PQRS, Prove that, TS^2 + TQ^2 =…

###### Practice Set 2.2
Question 1.

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

Given PS = 13, PQ = 11,PR = 17

By Apollonius's Theorem,

PS2 =

⇒ 169 =

⇒  = 36

⇒ QR2 = 144

QR = 12

Question 2.

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

The figure is given below:

According to Pythagoras theorem,

Median2 =

⇒ Median2 =

⇒ Median2 = 40

Median = 2√10

Thus the median is 2√10

Question 3.

In the figure 2.28 seg PS is the median of Î”PQR and PT ⊥ QR. Prove that,
(1)

(2)

Answer: (i) In Î”PTR, PT ⊥ TR, By Pythagoras Theorem we have

Perpendicular2 + Base2 = Hypotenuse2

⇒ PT2 + TR2 = PR2 [1]

Similarly, In Î”PTS

PT2 + TS2 = PS2

⇒ PT2 = PS2 – ST[2]

Using [2] in [1], we have

⇒ PS2 – ST2 + (ST + SR)2 = PR2

⇒ PS2 – ST2 + ST2 + SR2 + (2 × ST × SR) = PR2

Now, Since PS is a median we can write

(ii)

In Î”PQT, By Pythagoras we have

PQ2 = PT2 + QT2

⇒ PQ2 = PS2 – ST2 + (QS – ST)2 [From 2]

⇒ PQ2 = PS2 – ST2 + QS2 + ST2 – (2 × QS × ST)

Question 4.

In Î”ABC, point M is the midpointof side BC.

If, AB2 + AC2 = 290 cm2,AM = 8 cm, find BC.

Given AB2 + AC2 = 290 cm2,AM = 8 cm, BM = MC

According to formula,

AM2 =

⇒ 64 =

⇒ 64 –

⇒ BC2 = 324

BC = 18.

Thus BC = 18 cm.

Question 5.

In figure 2.30, point T is in theinterior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2(As shown in the figure, drawseg AB || side SR and A – T – B)

From figure,

In ∆PAT, ∠PAT = 900

TP2 = AT2 + PA2 …1

In ∆AST, ∠SAT = 900

TS2 = AT2 + SA2 …2

In ∆QBT, ∠QBT = 900

TQ2 = BT2 + QB2 …3

In ∆BTR, ∠RBT = 900

TR2 = BT2 + BR2 …4

TS2 + TQ2 = AT2 + SA2 + BT2 + QB2 [Adding 2 and 3]

⇒ TS2 + TQ2 = AT2 + PA2 + BT2 + BR2 [SA = BR, QB = AP]

⇒ TS2 + TQ2 = TP2 + TR2 [From 1 and 4]

PROVED.