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.

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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