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CAT 2025 Lesson : Lines & Triangles - Right-angled Triangles

3.7 Right-angled Triangles

Questions where we have to apply properties of right-angled triangles are very common in MBA entrance exams. Lot of questions are solved by dropping a perpendicular/altitude and then applying Pythagoras theorem.

Tip: If a question seems unapproachable, draw an altitude and check if this helps.

Theorems and properties associated with Right angled triangles:

Definition & Features	Figure
Pythagoras theorem: Hypotenuse is the longest side in a right angled triangle and its square is equal to the sum of the squares of the other 2 sides. AB $^2$ + BC $^2$ $=$ AC $^2$ AB and BC are altitudes. B is the orthocentre, where the altitudes meet. ∴ Area of $\triangle$ ABC $= \dfrac{1}{2} \times$ AB $\times$ BC.
As the angle subtended by the diameter of a circle is $90\degree$ , AC is the diameter of the circle that circumscribes $\triangle$ ABC. D is the mid-point of the hypotenuse and the centre of the circumcircle. ∴ Radius of the circumcircle $= \dfrac{\mathrm {AC}}{2} =$ AD $=$ BD $=$ CD Angle subtended by a chord in the centre is twice that of its inscribed angle. ∴ $\angle$ BDC $= 2 \times$ $\angle$ BAC and $\angle$ BDA $= 2 \times$ $\angle$ BCA
Inradius $=$ Semiperimeter $-$ Hypotenuse $= \mathrm{\dfrac{(AB + BC + CA)}{2} - CA}$ Explanation: In $\triangle$ ABC, D, E and F are the tangential points. As tangents from the same external point are equal, AE $=$ AF $= x$ , BD $=$ BF $= y$ , CD $=$ CE $= z$ . As, BDOF is a square, $y =$ OD, the inradius. Semi-perimeter $= \dfrac{ 2x + 2y + 2z}{2} = x + y + z$ Hypotenuse $= x + z$ Inradius $=$ Semiperimeter $-$ Hypotenuse $= y$
Where P is the perpendicular altitude drawn from B to AC, 1) BP $\times$ AC $=$ AB $\times$ BC 2) AB $^2$ $=$ AP $\times$ AC 3) CB $^2$ $=$ CP $\times$ CA 4) PB $^2$ $=$ PA $\times$ PC In this $\triangle$ ABC, BD is the median. The median and altitude drawn to the hypotenuse are two distinct lines in scalene right-angled triangles.

To summarise, where c is the length of the hypotenuse and, a and b are the lengths of the other two sides, Inradius

= \bm{ \dfrac{ a + b - c}{2}}

and Circumradius

= \bm{\dfrac{c}{2}}

.

Note in a (3, 4, 5) triple, inradius

=

1 and circumradius

=

2.5.

Example 17

\triangle

ABC,

\angle

=

90\degree

, AB

=

6 cm, BC

=

8 cm and D is a point on AC such that BC

\bot

AD. Then, what are the values of BD, AD, inradius and circumradius?

Solution

Using the above properties,
AC = BP

\times

=

\times

BC
⇒ BD

= \dfrac{6 \times 8}{10} = 4.8

cm

AB

^2

=

\times

AC
36 = AD

\times

10
⇒ AD

=

3.6 cm

Semiperimeter =

\dfrac{ 6 + 8 + 10}{2} = 12

Inradius

=

Semiperimeter

-

Hypotenuse

=

-

=

2 cm
Circumradius

= \dfrac{\mathrm{AC}}{2} = \dfrac{10}{2} = 5

cm

Answer:

4.8; 3.6; 2; 5

Note: We earlier observed that the inradius and circumradius of (3, 4, 5) triple is 1 and 2.5 respectively. Therefore, for a derived triple of (6, 8, 10), they will be twice the values – 2 and 5 respectively.

3.7.1 Pythagorean Triples

A Pythagorean triple is a set of 3 integers that can be the lengths of the sides of a right-angled triangle. The largest value here, will be that of the hypotenuse. Below are two forms in which triplets occur.

Where c is the length of the hypotenuse and, a and b are the lengths of the other two sides, such that

a \lt b \lt c

,

Case 1: Where

a

is odd,

a^2 = b + c

and c = b + 1. For instance,

$a$	$a^2 = b + c$	$b$	$c$	$(a, b, c)$
3	9	4	5	(3, 4, 5)
5	25	12	13	(5, 12, 13)
7	49	24	25	(7, 24, 25)
9	81	40	41	(9, 40, 41)

Case 2: Where

a

is even,

\dfrac{a^2}{2} = b + c

and

c = b + 2

. For instance,

$a$	$\dfrac{a^2}{2} = b + c$	$b$	$c$	$(a, b, c)$
4	8	3	5	(4, 3, 5)
6	18	8	10	(6, 8, 10)
8	32	15	17	(8, 15, 17)
10	50	24	26	(10, 24, 26)

Note the following
1) The minimum value of a side in a pythagorean triple is 3.
2) Only when

a = 4

a \gt b

. In all other cases,

a \lt b

.
3) Pythagorean triples that do not fall under the above 2 cases also exist.

Derived Triples are multiples of primitive Pythagorean triples. The following are the smallest base triples that appear often in entrance tests.

(3, 4, 5)	(5, 12, 13)	(7, 24, 25)	(8, 15, 17)

If each number in a base triple is multiplied by the same integer, we get a derived Pythagorean triple. If all the terms of the Primitive Triple (3, 4, 5) are multiplied by 2, we get another pythagorean triple (6, 8, 10).

∴ Derived triples of (3, 4, 5) are (6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), and so on.

3.7.2 Ratio of sides

Where the angles of a right-angled triangle are
1)

30\degree - 60\degree - 90\degree

, then sides opposite them are in the ratio

1 : \sqrt{3}: 2

.
2)

45\degree - 45\degree - 90\degree

, then sides opposite them are in the ratio 1 : 1:

\sqrt{2}

.

Note: This is derived from trigonometry, which is covered in length in Coordinate & Trigonometry lesson.

Example 18

In the following figure,

\triangle

ABC and

\triangle

ADE are right-angled triangles. If

\angle

BAC

= 30\degree

, BC

=

15, DE

=

7 and both AD and AE are integers. Then, EC = ?

Solution

\triangle

ABC is a

30\degree - 60\degree - 90\degree

triangle where sides should be in ratio

1 : \sqrt{3}: 2

. As BC

=

15, AC

= 2 \times 15 = 30

As right-angled

\triangle

ADE has integral sides where DE

=

7, only the pythagorean triplet (7, 24, 25) applies.

∴ AE

=

25
EC

=

-

=

-

=

5

Answer:

5

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