 Triangles Pythagoras Theorem NCERT Exercise 6.5 part one Class Ten Mathematics

# Triangle

## Exercise 6.5 Part 1

Question 1: 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.

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.

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)

So, (PM)/(QM)=(MR)/(PM)

Or, PM^2=QM.MR proved

Question 3: In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that (a) (AB)^2 = (BC)xx(BD)

Solution: In triangles ACB and DAB

∠ ACB = ∠ DAB (Right angle)

∠ CBA = ∠ ABD (common angle)

Hence; ACB ∼ DAB

So, (AB)/(BC)=(BD)/(AB)

Or, AB^2=BD.BC proved

(b) AC2 = BC. DC

Solution: In triangles ACB and DCA

∠ ACB = ∠ DCA (right angle)

Hence; ACB ~ DCA

So, (AC)/(BC)=(DC)/(AC)

Or, AC^2=BD.CD proved

Solution: In triangles DAB and DCA

∠ DAB = ∠ DCA (right angle)

Hence; DAB ∼ DCA

So, (AD)/(BD)=(CD)/(AD)

Or, AD^2=BD.CD proved

Question 4: ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Solution: In this case; AB is hypotenuse and AC = BC are the other two sides

According to Pythagoras theorem:

AB^2 = AC^2 + BC^2

Or, AB^2 = AC^2 + AC^2

Or, AB^2 = 2AC^2 proved

Question 5: ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Solution: This question will be sold in the same way as the earlier question.

In this case; square of the longest side = sum of squares of other two sides

Hence, this is a right triangle.

Question 6: ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution: In case of an equilateral triangle, an altitude will divide the triangle into two congruent right triangles. In the right triangle thus formed, we have;

Hypotenuse = One of the sides of the equilateral triangle = 2a

Perpendicular = altitude of the equilateral triangle = p

Base = half of the side of the equilateral triangle = a

Using Pythagoras theorem, the perpendicular can be calculated as follows:

p^2 = h^2 – b^2

Or, p^2 = (2a)^2 – a^2

Or, p^2 = 4a^2 – a^2 = 3a^2

Or, p = a\sqrt3

Question 7: 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 proved

Question 8: In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (a) OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2

Solution: In ∆ AFO; AF^2 = OA^2 – OF^2

In ∆ BDO; BD^2 = OB^2 – OD^2

In ∆ CEO; CE^2 = OC^2 – OE^2

Adding the above three equations, we get;

AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 proved

(b) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2

Solution: In ∆ AEO; AE^2 = OA^2 – OE^2

In ∆ CDO; CD^2 = OC^2 – OD^2

In ∆ BFO: BF^2 = OB^2 – OF^2

Adding the above three equations, we get;

AE^2 + CD^2 + BF^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2

From the previous solution, we also have;

AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2

Comparing the RHS of the above two equations, we get;

AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 proved