English

A Ladder 15m Long Reaches a Window Which is 9m Above the Ground on One Side of a Street. Keeping Its Foot at the Same Point, the Ladder is Turned to Other Side of the Street to Reach a Window

Advertisements
Advertisements

Question

A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street.

Sum
Advertisements

Solution

let O be the foot of the ladder. Let AO be the position of the ladder when it touches the window at A which is 9m high and CO be the position of the ladder when it touches the window at C which is 12m high.
Using Pythagoras theorem,
In ΔAOB,
BO2 = AO2 - AB2
BO2 = (15m)2 - (9m)2
BO2 = 225m2 - 81m2
BO2 = 144m2
BO2 = (12m)2
BO2 = 12m
Using Pythagoras theorem in ΔCOB,
DO2 = CO2 - CD2
DO2 = (15m)2 - (12m)2
DO2 = 225m2 - 144m2
DO2 = 81m2
DO = 9m
Width of the street 
= DO + BO
= 9m + 12m
= 21m.

shaalaa.com
  Is there an error in this question or solution?
Chapter 17: Pythagoras Theorem - Exercise 17.1

APPEARS IN

Frank Mathematics [English] Class 9 ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 7

RELATED QUESTIONS

Side of a triangle is given, determine it is a right triangle.

`(2a – 1) cm, 2\sqrt { 2a } cm, and (2a + 1) cm`


In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE= AE2 + CD2 + BF2


Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.


The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(A)\[7 + \sqrt{5}\]
(B) 5
(C) 10
(D) 12


In right angle ΔABC, if ∠B = 90°, AB = 6, BC = 8, then find AC.


O is any point inside a rectangle ABCD.
Prove that: OB2 + OD2 = OC2 + OA2.


Choose the correct alternative: 

In right-angled triangle PQR, if hypotenuse PR = 12 and PQ = 6, then what is the measure of ∠P? 


In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.


In the figure below, find the value of 'x'.


Find the Pythagorean triplet from among the following set of numbers.

2, 4, 5


Find the Pythagorean triplet from among the following set of numbers.

2, 6, 7


Find the length of the hypotenuse of a triangle whose other two sides are 24cm and 7cm.


A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.


In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2AD2 + `(1)/(2)"BC"^2`


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.


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 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.


The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is ______.


Two angles are said to be ______, if they have equal measures.


If the areas of two circles are the same, they are congruent.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×