In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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

Question

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem (or pr

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem (or properties of right triangles with an altitude to the hypotenuse):$AB^{2} = AM 	imes AC$$BC^{2} = CM 	imes AC$Subtracting the two equations:$AB^{2} - BC^{2} = (AM 	imes AC) - (CM 	imes AC)$$AB^{2} - BC^{2} = AC(AM - CM)$Given $AB^{2} - BC^{2} = 260$ and $AM - CM = 10$:$260 = AC(10)$$AC = \frac{260}{10} = 26$Thus,$AC = 26$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AM - CM = 10$ and $AB^{2} - BC^{2} = 260$,find $AC$.

Similar Questions

The length of a diagonal of a square is $6$. Then,its area is............

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $AC$. If $AD = 9$ and $CD = 4$,find $BD$.

It is given that $\triangle DEF \sim \triangle RPQ.$ Is it true to say that $\angle D = \angle R$ and $\angle F = \angle P?$ Why?

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude. If $PQ = 40$ and $QR = 30$,find the area of $\Delta QMR$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB^2 + AC^2 = 122$ and $AD = 6$,find $BC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module