Question

Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle ? Give reason.
(3) Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.
(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.
(6) In ∆PQR; PQ = $\sqrt{8}$, QR = $\sqrt{5}$, PR = $\sqrt{3}$. Is ∆PQR a right angled triangle ? If yes, which angle is of 90°?

Answer 1

(1)


Since, ABC is an equilateral triangle, AD is the perpendicular bisector of BC.

Now, According to Pythagoras theorem,
In ∆ABD

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2 \\ \Rightarrow {\left(2a\right)}^2={\mathrm{AD}}^2+a^2 \\ \Rightarrow 4a^2-a^2={\mathrm{AD}}^2 \\ \Rightarrow {\mathrm{AD}}^2=3a^2 \\ \Rightarrow \mathrm{AD}=\sqrt{3}a \end{gathered}$$

Hence, the height of an equilateral triangle is $\sqrt{3}a$.


(2) In the triplet (7, 24, 25),
7² = 49, 24² = 576, 25² = 625 and 49 + 576 = 625

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ Sides 7 cm, 24 cm, 25 cm form a right angled triangle.


(3)


According to Pythagoras theorem,

In ∆ABC

$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2 \\ \Rightarrow {\left(60\right)}^2+{\left(11\right)}^2={\mathrm{AC}}^2 \\ \Rightarrow 3600+121={\mathrm{AC}}^2 \\ \Rightarrow {\mathrm{AC}}^2=3721 \\ \Rightarrow \mathrm{AC}=61\mathrm{cm} \end{gathered}$$

Hence, the length of a diagonal of the rectangle is 61 cm.


(4) In a right angled triangle,

According to Pythagoras theorem.

$$\begin{gathered} {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Height}\right)}^2 \\ ={\left(9\right)}^2+{\left(12\right)}^2 \\ =81+144 \\ =225 \\ \therefore \mathrm{Hypotenuse}=15\mathrm{cm} \end{gathered}$$

Hence, the length of the hypotenuse is 15 cm.


(5) It is given that, a side of an isosceles right angled triangle is x.

Then, the other side of the triangle is also x.

According to Pythagoras theorem.

$$\begin{gathered} {\left(\mathrm{Hypotenuse}\right)}^2=x^2+x^2 \\ =2x^2 \\ \therefore \mathrm{Hypotenuse}=\sqrt{2}x \end{gathered}$$

Hence, its hypotenuse is $\sqrt{2}x$.


(6) In ∆PQR, PQ = $\sqrt{8}$, QR = $\sqrt{5}$, PR = $\sqrt{3}$.
($\sqrt{8}$)² = 8, ($\sqrt{5}$)² = 5, ($\sqrt{3}$)² = 3 and 3 + 5 = 8

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ ∆PQR form a right angled triangle, where angle R is of 90°.