**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: **AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}

In ∆ AOB; AB^{2} = AO^{2} + BO^{2}

In ∆ BOC; BC^{2} = CO^{2} + BO^{2}

In ∆ COD; CD^{2} = CO^{2} + DO^{2}

In ∆ AOD; AD^{2} = DO^{2} + AO^{2}

Adding the above four equations, we get;

AB^{2} + BC^{2} + CD^{2} + AD^{2}

= AO^{2} + BO^{2} + CO^{2} + BO^{2} + CO^{2} + DO^{2} + DO^{2} + AO^{2}

Or, AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2(AO^{2} + BO^{2} + CO^{2} + DO^{2})

Or, AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2(2AO^{2} + 2BO^{2})

(Because AO = CO and BO = DO)

Or, AB^{2} + BC^{2} + CD^{2} + AD^{2} = 4(AO^{2} + BO^{2}) ………(1)

Now, let us take the sum of squares of diagonals;

AC^{2} + BD^{2} = (AO + CO)^{2} + (BO + DO)^{2}

= (2AO)^{2} + (2BO)^{2}

= 4AO^{2} + 4BO^{2} ……(2)

From equations (1) and (2), it is clear;

AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}

**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:

Hypotenuse^{2} = Base^{2} + Perpendicular^{2}

Let us check if the given three sides fulfill the criterion of Pythagoras theorem.

25^{2} = 24^{2} + 7^{2}

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.

8^{2} = 6^{2} + 3^{2}

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.

100^{2} = 80^{2} + 50^{2}

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.

13^{2} = 12^{2} + 5^{2}

Or, 169 = 144 + 25

Or, 169 = 169

Here; LHS = RHS

Hence; this is a right triangle.

**4.**Question 2: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM

^{2}= QM. MR.

**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) AB^{2} = 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) AC^{2} = 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) AD^{2} = 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]