Altitude of a Triangle

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is $100 \mathrm{m}$, then the altitude of the triangle is?

In $∆\mathrm{ABC}$, $\angle A<\angle B$, the altitude to the base divides vertex angle $C$ into two parts $C_1$ and $C_2$ adjacent to $BC$, then which one of the relation can be obtained?

Let $ABC$ be an equilateral triangle and $AX$, $BY$, $CZ$ be the altitudes. Then the right statement out of the four given responses is:

In triangle $\mathrm{PQR},\mathrm{PS}$ is the altitude to $\mathrm{QR}$. Which of the following must be true?

The number of altitudes of a triangle is

If the length of the sides of a triangle are $26$ cm, $28$ cm and $30$ cm. Then find the length of the altitude drawn on side having length $30$ cm.

In a right angle triangle PQR right-angled at Q. If the length of PQ and QR are respectively $40$ cm and $9$ cm, then find the length of altitude QT drawn on PR?

If the length of the two altitudes of a $∆PQR$ are $9$$cm$ and $12$$cm$, then what would be the possible length of the unknown altitude?

In $∆ABC, AB=AC$ and $AL$ is perpendicular to $BC$ at $L$. In $∆DEF, DE=DF$ and $DM$ is perpendicular to $EF$ at $M$. If $(area of ∆ABC):(area of ∆DEF)=9.25$, then $\frac{DM+AL}{DM-AL}$ is equal to:

Name the orthocentre of $△PQR$.

Identify the type of segment required in the below triangle:

$AD=$

$△ABC$ is a right angled triangle. If $CD=k \mathrm{cm}$ then find $k$.

If the altitude to side $AC$ of triangle with side $AB = 20 \mathrm{cm}, AC = 20 \mathrm{cm}, BC = 30 \mathrm{cm}$ is $7.5\sqrt{k }\mathrm{cm}$ then find $k$.

Given: $BX$ is the altitude, $\angle BAX=40°$. If $\angle 1=k°$ then find $k$.

Given: $AX$ is the altitude, $\angle CAB\cong \angle B, \angle C=50°$. If $\angle B$ is $k°$ then find $k$

The corresponding sides of two similar triangle are in the ratio $1\mathbf{:}5$, then the ratio of their altitude will be:  

$\mathrm{BL}$ and  $\mathrm{CM}$ are medians of $\mathrm{\Delta ABC}$ right-angled at $\mathrm{A}$ and $\mathrm{BC}$$=5 \mathrm{cm}$.If $\mathrm{BL}=\frac{3\sqrt{5}}{2} \mathrm{cm}$. Find the length of $\mathrm{CM}$.

Where does the orthocentre lie in the case of a right angled triangle?

Where does the orthocentre lie in the case of an obtuse-angled triangle?

Where does the orthocentre lie in the case of an acute-angled triangle?