A 15 , m long bamboo leans on a wall to reach | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the length of the bamboo be the hypotenuse $c = 15 \, m$.Let the height reached on the wall be one leg $a = 12 \, m$.Let the distance from the

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let the length of the bamboo be the hypotenuse $c = 15 \, m$.Let the height reached on the wall be one leg $a = 12 \, m$.Let the distance from the base of the wall be the other leg $b$.According to the Pythagorean theorem,$a^2 + b^2 = c^2$.Substituting the values,we get $(12)^2 + b^2 = (15)^2$.$144 + b^2 = 225$.$b^2 = 225 - 144 = 81$.$b = \sqrt{81} = 9 \, m$.Thus,the lower end of the bamboo is $9 \, m$ away from the base of the wall.

$A$ $15 \, m$ long bamboo leans on a wall to reach the height of $12 \, m$ on the wall. Then,the lower end of the bamboo is $\ldots \ldots \, m$ away from the base of the wall.

Similar Questions

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $BD = 2.5$,$AB = 5$ and $AC = 4.2$,find $DC$.

If in two triangles $DEF$ and $PQR$,$\angle D = \angle Q$ and $\angle R = \angle E$,then which of the following is not true?

The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{2}{3}$ and $AC = 25$,find $NC$.

The perimeter of rectangle $ABCD$ is $28$. If $AC = 10$ and $AB < BC$,find the measures of its sides.

Vedclass Test Series

Exam Paper Generator

Online Exam Module