Questions on triangle (Pythagoras theorem)

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.

4. Question 2: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM. MR.

Solution: In triangles PMQ and RMP

∠ PMQ = ∠ RMP (Right angle)

∠ PQM = ∠ RPM (90 – MRP)

Hence; PMQ ~ RMP (AAA criterion)

$\frac{PM}{QM}= \frac{MR}{PM}$

hence $PM^{2}=QM\times RM$

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

$\frac{AB}{BC}= \frac{BD}{AB}$

hence $AB^{2}=BC\times DC$

(b) AC² = BC. DC

Solution: In triangles ACB and DCA

∠ ACB = ∠ DCA (right angle)

∠ CBA = ∠ CAD

Hence; ACB ~ DCA

$\frac{AC}{BC}= \frac{DC}{AC}$

hence $AC^{2}=BC\times DC$

Solution: In triangles DAB and DCA

∠ DAB = ∠ DCA (right angle)

∠ ABD = ∠ CAD

Hence; DAB ~ DCA

$\frac{AD}{BD}= \frac{CD}{AD}$

hence $AD^{2}=BD\times DC$

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