Practice Set 2.2
- In ΔPQR, point S is the midpoint of side QR. If PQ = 11,PR = 17, PS = 13,find QR.…
- In ΔABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C…
- In the figure 2.28 seg PS is themedian of ΔPQR and PT ⊥ QR. Prove that, (1) pr^2 = ps^2…
- In ΔABC, point M is the midpointof side BC. If, AB^2 + AC^2 = 290 cm^2 ,AM = 8 cm, find…
- 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 ΔPQR, point S is the midpoint of side QR. If PQ = 11,PR = 17, PS = 13,find QR.
Answer:
Given PS = 13, PQ = 11,PR = 17
By Apollonius's Theorem,
PS2 = 
⇒ 169 = 
⇒ 
= 36
⇒ QR2 = 144
QR = 12
Question 2.
In ΔABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB
Answer:
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 ΔPQR and PT ⊥ QR. Prove that,
(1) 
(2) 
Answer: (i) In ΔPTR, PT ⊥ TR, By Pythagoras Theorem we have
Perpendicular2 + Base2 = Hypotenuse2
⇒ PT2 + TR2 = PR2 [1]
Similarly, In ΔPTS
PT2 + TS2 = PS2
⇒ PT2 = PS2 – ST2 [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 ΔPQT, 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 ΔABC, point M is the midpointof side BC.
If, AB2 + AC2 = 290 cm2,AM = 8 cm, find BC.
Answer:
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)
Answer:
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.
