Class 10 Mathematics

Triangle

Exercise 6.5 (NCERT) 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.

similar triangles exercise solution

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

similar triangles exercise solution

(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)

∠ CBA = ∠ CAD

Hence; ACB ~ DCA

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

Or, `AC^2=BD.CD` proved

(c) AD2 = BD. CD

Solution: In triangles DAB and DCA

∠ DAB = ∠ DCA (right angle)

∠ ABD = ∠ CAD

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.

similar triangles exercise solution

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

similar triangles exercise solution

(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


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Theorem (Part 1)

Theorem (Part 2)

Exercise 6.1

Exercise 6.2 (Part 1)

Exercise 6.2 (Part 2)

Exercise 6.3 (Part 1)

Exercise 6.3 (Part 2)

Exercise 6.3 (Part 3)

Exercise 6.4

Exercise 6.5 (Part 2)

Exercise 6.6 (Part 1)

Exercise 6.6 (Part 2)