Advertisements
Advertisements
प्रश्न
A ladder, 6.5 m long, rests against a vertical wall. If the foot of the ladder is 2.5 m from the foot of the wall, find up to how much height does the ladder reach?
Advertisements
उत्तर
Given :
Length of ladder = 6.5 m
Length of the foot of the wall = 2.5 m
To find: Height AC According to Pythagoras Theorem,
(BC)2 = (AB)2 + (AC)2
(6.5)2 = (2.5)2 + (AC)2
42.25 = 6.25 + AC2
AC2 = 42.25 – 6.25 = 36 m
AC = `sqrt(6xx6)` = 6 m
∴ Height of wall = 6 m
APPEARS IN
संबंधित प्रश्न
A man goes 10 m due east and then 24 m due north. Find the distance from the starting point
Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2
Find the Pythagorean triplet from among the following set of numbers.
3, 4, 5
If in a ΔPQR, PR2 = PQ2 + QR2, then the right angle of ∆PQR is at the vertex ________
If length of sides of a triangle are a, b, c and a2 + b2 = c2, then which type of triangle it is?
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)`
If ΔABC ~ ΔPQR, `("ar" triangle "ABC")/("ar" triangle "PQR") = 9/4` and AB = 18 cm, then the length of PQ 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.
