Pythagoras Theorem

The area of a right-angled triangle is $42 {\mathrm{cm}}^2$. If its perpendicular is equal to $10 \mathrm{cm}$, find its base.

In $∆\mathrm{ABC}$, $\angle ABC=90°$ and $D$ is a point on $BC$ such that $\angle BDA=30°$ and $\angle BCA=20°$. What is $\angle CAD$?

For a triangle, one of its sides is the diameter of its circumcircle. The triangle is _____.

What is hypotenuse?

In the given figure, if $BC=3 \mathrm{cm}$, $AC=5 \mathrm{cm}$, $D$ is the midpoint of $AB$, $\angle B=90°$ and then find the value of $DC$.

$ABCD$ is a rhombus, in which the lengths of the diagonals $AC \mathrm{and} BD \mathrm{are} 6 \mathrm{cm} \mathrm{and} 8 \mathrm{cm}$ respectively, Find $k$, if the perimeter of the rhombus $ABCD$ is $k \mathrm{cm}$.

Calculate $x$.

In the adjacent figure, $DC\perp AC, AB\perp AC$. If $AC=12 cm, DC=AB=8 cm$ and $DB$ bisects $AC$ at $O$, then determine the length of $DB$.

Find $p^2$.

Screen size is defined by the distance between two diagonally opposite corners of screen. A manufacturer can make rectangular display screens as per clients' demands. A client purchased a display screen of size $\sqrt{70}$ units from the manufacturer last year. For an upgrade, he wants the same type of screen with larger display. What are the possible dimensions of the screen purchased by the client last year?

Which of the following can be the sides of a right triangle?

In the figure given below, find the area of the triangle.

In the adjoining figure, $PQ\perp QR$ and $QR\perp RS$ also $QR= 6 \mathrm{cm}, PQ= RS= 4 \mathrm{cm}$. Find the length of $PS$, if $PS$ bisects $QR$ at $O$.

Triangle ABC is right-angled at A and AD is perpendicular to BC. If BD $=7.5 \mathrm{cm}$ and DC $=10 \mathrm{cm}$, then what is AD equal to?

$5$ examples of application of Pythagoras Theorem in real life.

Write atleast $10$ Pythagorean triplets.

If length of $AP$ is given as $x$ and $AB=a,BC=a,CA=b$

Use trigonometry to express $x$ in terms of $A\text{ and }b$. Hence show that $a^2=b^2+c^2-2bc\cos A$ .

If length of $AP$ is given as $x$ and $AB=a,BC=a,CA=b$

Use the Pythagorean theorem to write two expressions for $CP$. Hence show that $b^2-x^2=a^2-(c-x{)}^2$.

Label length $AP$ as $x$. Hence label length $PB$ in terms of $c \text{and} x.$

Copy the triangle $ABC,$ with the altitude through $C$ meeting $AB$ at right angles at point $P.$