1. ABC is an isosceles triangle in which angle B = angle C and angle A is greater than the angle B. Find the measure of angle A.
Solution: An isosceles triangle has two angles equal in size. In this problem A is greater than B, therefore, angles B and C are equal in size. Since angle A is 30 greater than angle B then A = B + 30 o. The sum of all angles in a triangle is equal to 180 o.
(B+30) + B + B = 180
Solve the above equation for B.
B = 50 o
Therefore, A = B + 30 = 80 o
2. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution: ABCD is a rhombus in which diagonals AC and BD intersect at point O.
To Prove: AB2 + BC2 + CD2 + AD2 = AC2 + BD2
In ∆ AOB; AB2 = AO2 + BO2
In ∆ BOC; BC2 = CO2 + BO2
In ∆ COD; CD2 = CO2 + DO2
In ∆ AOD; AD2 = DO2 + AO2
Adding the above four equations, we get;
AB2 + BC2 + CD2 + AD2
= AO2 + BO2 + CO2 + BO2 + CO2 + DO2 + DO2 + AO2
Or, AB2 + BC2 + CD2 + AD2 = 2(AO2 + BO2 + CO2 + DO2)
Or, AB2 + BC2 + CD2 + AD2 = 2(2AO2 + 2BO2)
(Because AO = CO and BO = DO)
Or, AB2 + BC2 + CD2 + AD2 = 4(AO2 + BO2) ………(1)
Now, let us take the sum of squares of diagonals;
AC2 + BD2 = (AO + CO)2 + (BO + DO)2
= (2AO)2 + (2BO)2
= 4AO2 + 4BO2 ……(2)
From equations (1) and (2), it is clear;
AB2 + BC2 + CD2 + AD2 = AC2 + BD2
3. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(a) 7 cm, 24 cm, 25 cm
Solution: In a right triangle, the longest side is the hypotenuse. We also know that according to Pythagoras theorem:
Hypotenuse2 = Base2 + Perpendicular2
Let us check if the given three sides fulfill the criterion of Pythagoras theorem.
252 = 242 + 72
Or, 625 = 576 + 49
Or, 625 = 625
Here; LHS = RHS
Hence; this is a right triangle.
(b) 3 cm, 8 cm, 6 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
82 = 62 + 32
Or, 64 = 36 + 9
Or, 64 ≠ 45
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(c) 50 cm, 80 cm, 100 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
1002 = 802 + 502
Or, 10000 = 6400 + 2500
Or, 10000 ≠ 8900
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(d) 13 cm, 12 cm, 5 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
132 = 122 + 52
Or, 169 = 144 + 25
Or, 169 = 169
Here; LHS = RHS
Hence; this is a right triangle.
Solution: In triangles PMQ and RMP
∠ PMQ = ∠ RMP (Right angle)
∠ PQM = ∠ RPM (90 – MRP)
Hence; PMQ ~ RMP (AAA criterion)
[latex]\frac{PM}{QM}= \frac{MR}{PM}[/latex]
hence [latex]PM^{2}=QM\times RM[/latex]
5. In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(a) AB2 = BC. BD
Solution: In triangles ACB and DAB
∠ ACB = ∠ DAB (Right angle)
∠ CBA = ∠ ABD (common angle)
Hence; ACB ~ DAB
[latex]\frac{AB}{BC}= \frac{BD}{AB}[/latex]
hence [latex]AB^{2}=BC\times DC[/latex]
(b) AC2 = BC. DC
Solution: In triangles ACB and DCA
∠ ACB = ∠ DCA (right angle)
∠ CBA = ∠ CAD
Hence; ACB ~ DCA
[latex]\frac{AC}{BC}= \frac{DC}{AC}[/latex]
hence [latex]AC^{2}=BC\times DC[/latex]
(c) AD2 = BD. CD
Solution: In triangles DAB and DCA
∠ DAB = ∠ DCA (right angle)
∠ ABD = ∠ CAD
Hence; DAB ~ DCA
[latex]\frac{AD}{BD}= \frac{CD}{AD}[/latex]
hence [latex]AD^{2}=BD\times DC[/latex]