Advertisements
Advertisements
प्रश्न
A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Advertisements
उत्तर
The pictorial representation of the given problem is given below,
Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, AB is the hypotenuse.
Therefore applying the Pythagoras theorem we get,
AB2 = BC2 + CA2
132 = 52 + CA2
CA2 = 132 - 52
CA2 = 144
CA = 12 m
Therefore, the distance of the other end of the ladder from the ground is 12m.
APPEARS IN
संबंधित प्रश्न
In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC2 = AB2 + BC2 – 2BC × BD
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`
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
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.
Which of the following can be the sides of a right triangle?
2.5 cm, 6.5 cm, 6 cm
In the case of right-angled triangles, identify the right angles.
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)
In ΔABC, Find the sides of the triangle, if:
- AB = ( x - 3 ) cm, BC = ( x + 4 ) cm and AC = ( x + 6 ) cm
- AB = x cm, BC = ( 4x + 4 ) cm and AC = ( 4x + 5) cm
In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm;
find the length of side BC.
In triangle ABC, angle A = 90o, CA = AB and D is the point on AB produced.
Prove that DC2 - BD2 = 2AB.AD.
In figure AB = BC and AD is perpendicular to CD.
Prove that: AC2 = 2BC. DC.
Triangle PQR is right-angled at vertex R. Calculate the length of PR, if: PQ = 34 cm and QR = 33.6 cm.
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.
A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB2 + OD2 = OC2 + OA2
In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that: 9BP2 = 9BC2 + 4AC2
In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.
Find the unknown side in the following triangles
From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°
∴ ∠BAC = `square`
∴ ∆ABC is 30° – 60° – 90° triangle
∴ In ∆ABC by property of 30° – 60° – 90° triangle.
∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC
∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`
∴ `square` = 6 and BC = `6sqrt(3)`
In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.
In a right-angled triangle ABC, if angle B = 90°, BC = 3 cm and AC = 5 cm, then the length of side AB is ______.
