In this chapter, we are going to discuss important right angle triangle theorems and properties along with their proofs.
Basic Formulas
△ABC is a right triangle where ∠C=90°, and lengths of the sides BC, CA, and AB are a,b, and c respectively.
∠A=x° and ∠B=y°
So, x+y=90 ………(i)
![Right Angle triangle Theorems (50 Powerful exam hack) 2 Right angle triangle properties](https://logicxonomy.com/wp-content/uploads/2023/02/20230218_090603_0000-1024x576.webp)
Draw CH⟂AB where the height CH=h
In the right △AHC,
∠ACH=90°-∠HAC=y° …..From equation (i)
![Right Angle triangle Theorems (50 Powerful exam hack) 3 Right angle triangle](https://logicxonomy.com/wp-content/uploads/2023/02/20230218_090633_0000-1024x576.webp)
Similarly, In the right △BHC,
∠BCH=90°-∠HBC=x° …..From equation (i)
Let, AH=m and BH=n
Now, △AHC~△CHB (Similar Triangles)
\frac{CH}{AH}=\frac{BH}{CH}.
\frac{h}{m}=\frac{n}{h}.
h^2=m\times n …….(iii)
△CHB~△ACB (Similar Triangles)
\frac{AH}{AC}=\frac{AC}{AB}.
\frac{m}{b}=\frac{b}{c}.
b^2=m\times c …….(iv)
△AHC~△ACB (Similar Triangles)
\frac{BH}{BC}=\frac{BC}{AB}.
\frac{n}{a}=\frac{a}{c}.
a^2=n\times c …….(v)
Add equations (iv) and (v):
a^2+b^2=c\times (m+n).
a^2+b^2=c^2 ………(vi)
In the right △AHC,
Sin(A)=\frac{CH}{AC}.
Sin(x)=\frac{h}{b} ………(vii)
In the right △BHC,
Sin(B)=\frac{CH}{BC}.
Sin(y)=\frac{h}{a}.
Sin(90°-x)=\frac{h}{a}.
Cos(x)=\frac{h}{a} ………(viii)
From equations (vii) and (viii):
Sin^{2}x+Cos^{2}x=1 (Trigonometric identity)
\frac{h^2}{b^2}+\frac{h^2}{a^2}=1.
\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{h^2} ………(ix)
The Area of the right angle triangle= \frac{1}{2}\times Base\times Height.
\frac{1}{2}\times b\times a=\frac{1}{2}\times c\times h.
a\times b=c\times h ………..(x)
Trisection of hypotenuse
In the given figure, the hypotenuse of a right-angled triangle is trisected by two circles. If the lengths of these segments are a, b, and c respectively. Then prove:
\frac{a}{b}\times\sqrt{\frac{c}{a+b}}=2
![Right Angle triangle Theorems (50 Powerful exam hack) 5 Trisection of Hypotenuse](https://logicxonomy.com/wp-content/uploads/2023/02/20230220_083626_0000-1024x576.webp)
Let the lengths of the sides PQ, QR, and PR of △PQR be y, x, and z respectively. The length of the radius of the smaller semicircle is r.
Let ∠QPR=θ
Therefore ∠QRP=90°-θ
Draw ST and QU
Then ∠STP and ∠QUR=90°
Theorem: The angle subtended by a diameter of a circle at any point on its boundary is 90°
![Right Angle triangle Theorems (50 Powerful exam hack) 6 Right angle triangle problems](https://logicxonomy.com/wp-content/uploads/2023/02/20230220_083718_0000-1024x576.webp)
So, ∠PST=90°-θ and ∠UQR=θ
PQ is the tangent and PR is the secant of the larger circle, so by theorem
PQ2=PU×PR
y2=(a+b)×z
(a+b)=\frac{y^2}{z}……..(i)
△QUR~△PQR (Similar Triangles)
\frac{UR}{RQ}=\frac{RQ}{PR}.
\frac{c}{x}=\frac{x}{z}.
c=\frac{x^2}{z} …….(ii)
△PST~△PQU (Similar Triangles)
\frac{PT}{TU}=\frac{PS}{SQ}.
\frac{a}{b}=\frac{2r}{y-2r} …….(iii)
Now draw VW, here V and W are the centers of the respective circles.
![Right Angle triangle Theorems (50 Powerful exam hack) 7 Right angle triangle short tricks](https://logicxonomy.com/wp-content/uploads/2023/02/20230220_083747_0000-1024x576.webp)
In the right triangle △VQW, apply Pythagoras’ Theorem:
VW2=VQ2+QW2
(r+\frac{x}{2})^2=(y-r)^2+(\frac{x}{2})^2.
r^2+\frac{x^2}{4}+rx=y^2+r^2-2yr+\frac{x^2}{4}.
rx=y^2-2yr.
r=\frac{y^2}{x+2y}.
Put this value of r in equation (iii)
\frac{a}{b}=\frac{2r}{y-2r}.
After solving, \frac{a}{b}=\frac{2y}{x} ……(iv)
Now put all the values in the given expression:
\frac{a}{b}\times\sqrt{\frac{c}{a+b}}=\frac{2y}{x}\times \sqrt{\frac{\frac{x^2}{z}}{\frac{y^2}{z}}}.
=\frac{2y}{x}\times \frac{x}{y}=2
Hence Proved…
A square inscribed in a Right Triangle
Case I: If One side of the square lies on the hypotenuse of the right triangle.
In the given right triangle △ABC, the lengths of the sides of the triangle are p, q, and r. A square is inscribed in this triangle such that one of its sides is on the hypotenuse then find the area of the Square.
![Right Angle triangle Theorems (50 Powerful exam hack) 8 Square inscribed in right triangle](https://logicxonomy.com/wp-content/uploads/2023/02/20230221_074855_0000-1024x576.webp)
Let, the length of the side of Square is ‘x’ and ∠BCA=θ
![Right Angle triangle Theorems (50 Powerful exam hack) 9 Right Angle Theorems proof](https://logicxonomy.com/wp-content/uploads/2023/02/20230221_074915_0000-1024x576.webp)
△CNK~△CBA (Similar Triangles)
\frac{NK}{AB}=\frac{NC}{BC}.
\frac{x}{p}=\frac{NC}{q}.
NC=\frac{q}{p}.x …….(i)
△AML~△ABC (Similar Triangles)
\frac{LM}{BC}=\frac{AM}{AB}.
\frac{x}{q}=\frac{AM}{p}.
AM=\frac{p}{q}.x …….(ii)
Now, AM:NC=\frac{px}{q} :\frac{qx}{p}=(\frac{p}{q} :\frac{q}{p})\times pq.
AM:NC=p^2:q^2 …….(iii)
The length of the hypotenuse= r
AM+MN+NC=r
\frac{p}{q}.x+x+\frac{q}{p}.x=r.
x.(\frac{p}{q}+1+\frac{q}{p})=r.
x.(\frac{p^2+pq+q^2}{pq})=r.
x=\frac{pqr}{p^2+pq+q^2}.
Now, the area of the square= x2
Case II: If the sides of the square lie on the base and perpendicular of the right triangle
![Right Angle triangle Theorems (50 Powerful exam hack) 10 Area of square inscribed in a Right Triangle](https://logicxonomy.com/wp-content/uploads/2023/02/20230405_090123_0000-1024x576.webp)
The length of AB=p, BC =q, and the length of the sides of Square is ‘x’.
△AFE~△ABC (Similar Triangles)
\frac{AF}{AB}=\frac{FE}{BC}.
\frac{p-x}{p}=\frac{x}{b}.
pb-bx=px.
x=\frac{pb}{p+b}.
Now, the area of the square= x2
![Right Angle triangle Theorems (50 Powerful exam hack) 11 Online Class Logicxonomy](https://logicxonomy.com/wp-content/uploads/2023/03/20230314_070946_0000-1024x1024.webp)
Relation between the heights of the triangles
In the given right angle triangle △ABC in which ∠A=90°, the line segments AD, DP, DR, PQ, and RS are perpendicular to the corresponding sides. Then find the relationship between AD, PQ, and RS.
![Right Angle triangle Theorems (50 Powerful exam hack) 12 Right triangle properties](https://logicxonomy.com/wp-content/uploads/2023/02/20230222_085745_0000-1024x576.webp)
Let, ∠B=θ
APDR is a Rectangle because its all interior angle is 90°
△PQD~△DPA (Similar Triangles)
\frac{QD}{AP}=\frac{PD}{AD}.
PD\times AP=QD\times AD.
Area of the Rectangle APDR= QD×AD ……….(i)
![Right Angle triangle Theorems (50 Powerful exam hack) 13 Similarity in Right Triangle](https://logicxonomy.com/wp-content/uploads/2023/02/20230222_085809_0000-1024x576.webp)
△DSR~△ARD (Similar Triangles)
\frac{DS}{AR}=\frac{DR}{AD}.
AR\times DR=DS\times AD.
Area of the Rectangle APDR= DS×AD ……….(ii)
Compare equations (i) and (ii)
QD×AD=DS×AD
QD=DS
Let, QD=DS=x …………(iii)
![Right Angle triangle Theorems (50 Powerful exam hack) 14 Right angle triangle tricks](https://logicxonomy.com/wp-content/uploads/2023/02/20230222_085830_0000-1024x576.webp)
Draw PM and RN, perpendicular to AD.
So, PM=RN=x
△PMA~△RND (Similar Triangles)
\frac{AM}{ND}=\frac{PM}{RN}=\frac{x}{x}.
AM=ND ……..(iv)
AM=ND=RS
MD=PQ
Now, AD=MD+AM
AD=PQ+RS
Que: Find the area of rectangle APDR if AB=4 cm and AC=3 cm.
Solution: Here DP⊥AB and DR⊥AC
As we know in the Right triangle: BC×AD=AB×AC
5×AD=3×4
AD=12/5 cm ……..(i)
In Right Triangle ABC⇒ BD: DC= AB²:AC²=16:9
BD+DC=BC
16k+9k=5
k=1/5
BD=16/5 and DC=9/5
In △BDA⇒ BD×DA=AB×DP
16/5×12/5=4×DP
DP=48/25 cm ………(ii)
Similarly in In △ADC⇒ AD×DC=DR×AC
12/5×9/5=3×DR
DR=36/25 cm …….(iii)
Now, the Area of APDR= DP×DR=2.7648 cm²
Altitude, Angle Bisector, and Median
In the given Right angle △ABC, ∠B=90°. BP is the altitude, BQ is the angle bisector and BR is the median of the triangle then Find PQ: QR
![Right Angle triangle Theorems (50 Powerful exam hack) 15 Median of the right angle triangle](https://logicxonomy.com/wp-content/uploads/2023/02/Right-Triangle-1024x576.webp)
Use Pythagoras’ theorem in this right triangle⇒ AC=5 cm
Property of the Median: AR=RC=BR= 2.5 cm (R is the circumcenter of the triangle)
BR= 2.5 cm ……..(i)
Property of the Altitude: 1/2×BC×AB=1/2×AC×BP
12=5×BP
BP=12/5 ……..(ii)
△ARB is an isosceles triangle (AR=BR)
Let, ∠A=θ
So, ∠ABR=θ
∠ABQ=45° (BQ is the angle bisector)
So, ∠RBQ= ∠ABQ – ∠ABR= 45°-θ …….(iii)
In the right angle △BPC, Since ∠C=90°-θ
So, ∠CBP=θ
∠QBP= ∠CBQ – ∠CBP= 45°-θ ………(iv)
From equations (iii) and (iv):
BQ is the angle bisector of ∠RBP, so PQ: QR= BP: BR (Angle Bisector Theorem)
PQ: QR= 12/5: 2.5
PQ: QR= (12/5: 2.5)×5= 12: 12.5
PQ: QR= 24:25
Relation between the radii of circles
In the given right angle △ABC, ∠B=90°. Then the relation between the radii of given circles is: x2=ab
![Right Angle triangle Theorems (50 Powerful exam hack) 16 Relation between the radii of circles](https://logicxonomy.com/wp-content/uploads/2023/02/20230326_122748_0000-1024x576.webp)
Let, ∠A=θ then ∠ABD=90°-θ
∠EBD=∠EFD=90°-θ (BFDE is a Rectangle)
So, △AED~△EDF~△DFC (Similar Triangles)
\frac{a}{AD}=\frac{x}{EF}=\frac{b}{DC} ………..(i)
Let, AB=p
BC=q
AC=r
Then \frac{1}{2}\times r\times BD=\frac{1}{2}\times p\times q.
BD=EF (Diagonals of the Rectangle)
EF=\frac{pq}{r}.
AD=AB\times Cos(A)=p\times \frac{p}{r}=\frac{p^2}{r}.
DC=BC\times Cos(C)=q\times \frac{q}{r}=\frac{q^2}{r}.
Put these values in equation (I):
\frac{a}{AD}=\frac{x}{EF}=\frac{b}{DC}=k (k is a constant)
a=AD.k
x=EF.k
b=DC.k
ab=AD.DC.k2=(EF.k)2=x2
x=\sqrt{ab}The ratio between the sides of a Right Triangle
Que: What is the area of the right triangle whose one side is 58 cm and all the sides are integers?
(a) 840 cm²
(b) 24360 cm²
(c) 1160 cm²
(d) None
Solution: The ratio between the sides of a Right Triangle is:
⇒ (a2-b2): 2ab: (a2+b2)
Here a and b are positive integers
(a2+b2) represents the Hypotenuse of the triangle
If 2ab=58
ab=29
(a,b)=(29,1)
Then the sides⇒ (292-12): 2×29×1: (292+12)= 840: 58: 842
The area of the triangle= \frac{1}{2}\times 840\times 58=24360 cm²
If (a2+b2)=58
(a,b)=(7,3)
Then the sides⇒ (72-32): 2×7×3: (72+32)= 40: 42: 58
The area of the triangle= \frac{1}{2}\times 40\times 42=840 cm²
If (a2-b2)=58
(a+b)(a-b)=58
{(a+b), (a-b)}={(58,1), (29,2)}
If a+b=58 and a-b=1
a=29.5 (Not an integer)
If a+b=29 and a-b=2
a=15.5 (Not an integer)
So, only two areas are possible which are 24360 cm² and 840 cm²
Option (d) is correct.
Shaded Area Problems of Geometry: Click Here