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.

similar triangles exercise solution

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.
similar triangles exercise solution

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

similar triangles exercise solution

(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

(c) AD2 = BD. CD

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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