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?

Identify the base of a  $△\mathrm{ABC}$ right-angled at $\mathrm{B}$. ($\mathrm{AB}/\mathrm{AC}/\mathrm{BC}$)

The number of altitudes of a triangle is

The area of a right-angled triangle is $40$ times its base. What is its height?

Height and base of a right triangle are $24 \mathrm{cm}$ and $18 \mathrm{cm}$, find the length of its hypotenuse.

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?

Which line segment shows the base that corresponds to the given height of the triangle?

Which line segment shows the height that corresponds to the given base of the triangle?

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:

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 of a triangle is $k \mathrm{cm}$, its area is $120 {\mathrm{cm}}^2$ and its base is $6 \mathrm{cm}$ then find $k$.

If the altitude of an equilateral triangle when its equal sides are given as $10 \mathrm{cm}$ is $k\sqrt{3} \mathrm{cm}$ 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 $k$