Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर १
Construction: Through T, draw seg AB II side SR such that A-T-B, P- A-S and Q-B-R.
Proof:
seg PS || seg QR ...(Opposite sides of rectangle)
∴ seg AS || seg BR ...(P-A-S and Q-B-R)
also seg AB || seg SR ...(Construction)
∴ `square` ASRB is a parallelogram ...((By definition)
∠ASR = 90° ...(Angle of rectangle PSRQ)
∴ `square` ASRB is a rectangle ...(A parallelogram is a rectangle, if one of its angles is a right angle.)
∠SAB = ∠ABR = 90° ...(Angle of a rectangle)
∴ seg TA ⊥ side PS and seg TB ⊥ side QR ...(1)
AS = BR ...(2) (Opposite sides of rectangle are equal)
Similarly, we can prove AP= BQ ...(3)
In ΔTAS,
∠TAS = 90° ...[From (1)]
∴ by Pythagoras theorem,
TS2 = TA2 + AS2 ...(4)
In ΔTBQ,
∠TBQ = 90° ...[From (1)]
∴ by Pythagoras theorem,
TQ2 = TB2 + BQ2 ...(5)
Adding (4) and (5), we get,
TS2 + TQ2 = TA2 + AS2 + TB2 + BQ2 ...(6)
In ΔTAP,
∠TAP = 90° ...[From (1)]
∴ by Pythagoras theorem,
TP2 = TA2 + AP2 ...(7)
In ΔTBR,
∠TBR = 90° ...[From (1)]
∴ by Pythagoras theorem,
TR2 = TB2 + BR2 ...(8)
Adding (7) and (8), we get
TP2 + TR2 = TA2 + AP2 + TB2 + BR2
∴ TP2 + TR2 = TA2 + BQ2 + TB2 + AS2 ...(9) [From (2) and (3)]
∴ from (6) and (g), we get,
TS2 + TQ2 = TP2 + TR2
उत्तर २
Construction: Draw diagonals PR and QS and let them intersect at 0. Draw seg TO.
Proof: `square` PQRS is a rectangle ...(Given)
∴ PR=QS ...(Diagonals of rectangle are congruent)
Multiplying both the sides by `1/2`, we get,
`1/2`PR = `1/2`QS ...(1)
But, `1/2`PR = OP = OR ...(2)
and `1/2`QS = OS = OQ ...(3) [Diagonals of rectangle bisect each other]
∴ OP = OR = OS = OQ ...(4) [From (1), (2) and (3)]
In ΔTSQ,
seg TO is the median ...(By definition)
∴ by Apollonius theorem,
TS2 + TQ2 = 2TO2 + 2OQ2 ...(5)
In ΔPTR,
seg TO is the median ...(By definition)
∴ by Apollonius theorem,
TP2 + TR2 = 2TO2 + 2OR2
∴ TP2 + TR2 = 2TO2 + 2OQ2 ...(6) [From (4)]
∴ from (5) and (6), we get,
TS2 + TQ2 = TP2 + TR2
Notes
Students can refer to the provided solutions based on their preferred marks.
संबंधित प्रश्न
From a point O in the interior of a ∆ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove
that :
`(i) AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2`
`(ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2`
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides.
Identify, with reason, if the following is a Pythagorean triplet.
(4, 9, 12)
Find the side and perimeter of a square whose diagonal is 10 cm.
In right angle ΔABC, if ∠B = 90°, AB = 6, BC = 8, then find AC.
In the given figure, angle BAC = 90°, AC = 400 m, and AB = 300 m. Find the length of BC.
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
In the right-angled ∆PQR, ∠ P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
In the given figure, ∠T and ∠B are right angles. If the length of AT, BC and AS (in centimeters) are 15, 16, and 17 respectively, then the length of TC (in centimeters) is ______.
In figure, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.
In the adjoining figure, a tangent is drawn to a circle of radius 4 cm and centre C, at the point S. Find the length of the tangent ST, if CT = 10 cm.
Jayanti takes shortest route to her home by walking diagonally across a rectangular park. The park measures 60 metres × 80 metres. How much shorter is the route across the park than the route around its edges?
