Triangles and its Properties Ex-6.5

NCERT solutions for Maths Triangles and its Properties 

NCERT Solutions for Class 7 Maths Triangles and its Properties

Class –VII Mathematics (Ex. 6.5)

Question 1.PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer:

Given: PQ = 10 cm, PR = 24 cm

Let QR be $x$ cm.

In right angled triangle QPR,

${\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}$

[By Pythagoras theorem]

$\Rightarrow$ ${\left( {QR} \right)^2} = {\text{ }}{\left( {PQ} \right)^2} + {\text{ }}{\left( {PR} \right)^2}$

$\Rightarrow$ ${x^2} = {\left( {10} \right)^2} + {\left( {24} \right)^2}$

$\Rightarrow$ ${x^2}$ = 100 + 576 = 676

$\Rightarrow$ $x = \sqrt {676}$ = 26 cm

Thus, the length of QR is 26 cm.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 2.ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer:

Given: AB = 25 cm, AC = 7 cm

Let BC be $x$ cm.

In right angled triangle ACB,

${\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}$

[By Pythagoras theorem]

$\Rightarrow$ ${\left( {AB} \right)^2} = {\text{ }}{\left( {AC} \right)^2} + {\text{ }}{\left( {BC} \right)^2}$

$\Rightarrow$ ${\left( {25} \right)^2} = {\left( 7 \right)^2} + {x^2}$

$\Rightarrow$ 625 = 49 + ${x^2}$

$\Rightarrow$ ${x^2}$ = 625 – 49 = 576

$\Rightarrow$ $x = \sqrt {576}$ = 24 cm

Thus, the length of BC is 24 cm.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 3.A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance $a.$ Find the distance of the foot of the ladder from the wall.

Answer:

Let AC be the ladder and A be the window.

Given: AC = 15 m, AB = 12 m, CB = $a$ m

In right angled triangle ACB,

${\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}$

[By Pythagoras theorem]

$\Rightarrow$ ${\left( {AC} \right)^2} = {\text{ }}{\left( {CB} \right)^2} + {\text{ }}{\left( {AB} \right)^2}$

$\Rightarrow$ ${\left( {15} \right)^2} = {\left( a \right)^2} + {\left( {12} \right)^2}$

$\Rightarrow$ 225 = ${a^2}$ + 144

$\Rightarrow$ ${a^2}$ = 225 – 144 = 81

$\Rightarrow$ $a = \sqrt {81}$ = 9 cm

Thus, the distance of the foot of the ladder from the wall is 9 m.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 4.Which of the following can be the sides of a right triangle?

1. 2.5 cm, 6.5 cm, 6 cm
2. 2 cm, 2 cm, 5 cm
3. 1.5 cm, 2 cm, 2.5 cm

In the case of right angled triangles, identify the right angles.

Answer:

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

${\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}$

(i) 2.5 cm, 6.5 cm, 6 cm

In $\Delta {\text{ABC}},$ ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}$

L.H.S. = ${\left( {6.5} \right)^2}$ = 42.25 cm

R.H.S. = ${\left( 6 \right)^2} + {\left( {2.5} \right)^2}$ = 36 + 6.25 = 42.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

(ii) 2 cm, 2 cm, 5 cm

${\left( 5 \right)^2} = {\left( 2 \right)^2} + {\left( 2 \right)^2}$

L.H.S. = ${\left( 5 \right)^2}$ = 25

R.H.S. = ${\left( 2 \right)^2} + {\left( 2 \right)^2}$ = 4 + 4 = 8

Since, L.H.S. $\ne$ R.H.S.

Therefore, the given sides are not of the right angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

In $\Delta$PQR, ${\left( {{\text{PR}}} \right)^2} = {\left( {{\text{PQ}}} \right)^2} + {\left( {{\text{RQ}}} \right)^2}$

L.H.S. = ${\left( {2.5} \right)^2}$ = 6.25 cm

R.H.S. = ${\left( {1.5} \right)^2} + {\left( 2 \right)^2}$ = 2.25 + 4 = 6.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 5.A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Answer:

Let A’CB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then $\Delta {\text{ABC}}$ is a right angled triangle, right angled at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem, In $\Delta {\text{ABC}}$

${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}$

$\Rightarrow$ ${\left( {{\text{AC}}} \right)^2} = {\left( {12} \right)^2} + {\left( 5 \right)^2}$

$\Rightarrow$ ${\left( {{\text{AC}}} \right)^2} = 144 + 25$

$\Rightarrow$ ${\left( {{\text{AC}}} \right)^2} = 169$

$\Rightarrow$ AC = 13 m

Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 6.Angles Q and R of a $\Delta$PQR are $25^\circ$ and $65^\circ .$

Write which of the following is true:

1. $P{Q^2} + {\text{ }}Q{R^2} = {\text{ }}R{P^2}$
2. $P{Q^2} + {\text{ }}R{P^2} = {\text{ }}Q{R^2}$
3. $R{P^2} + {\text{ }}Q{R^2} = {\text{ }}P{Q^2}$

$25^\circ$ $65^\circ$

Answer:

In $\Delta$PQR,

$\angle$PQR + $\angle$QRP + $\angle$RPQ = $180^\circ$

[By Angle sum property of a $\Delta$ ]

$\Rightarrow$ $25^\circ + 65^\circ + \angle {\text{RPQ = 180}}^\circ$

$\Rightarrow$ $90^\circ + \angle {\text{RPQ = 180}}^\circ$

$\Rightarrow$ $\angle$RPQ = $180^\circ – 90^\circ = 90^\circ$

Thus, $\Delta$PQR is a right angled triangle, right angled at P.

$\therefore$ (Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

$\Rightarrow$ ${\left( {{\text{QR}}} \right)^2} = {\left( {{\text{PR}}} \right)^2} + {\left( {{\text{QP}}} \right)^2}$

Hence, Option (ii) is correct.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 7.Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer:

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be $x$ cm.

Now, in right angled triangle PQR,

${\left( {{\text{PR}}} \right)^2} = {\left( {{\text{RQ}}} \right)^2} + {\left( {{\text{PQ}}} \right)^2}$

[By Pythagoras theorem]

$\Rightarrow$ ${\left( {41} \right)^2} = {x^2} + {\left( {40} \right)^2}$

$\Rightarrow$ 1681 = ${x^2}$ + 1600 $\Rightarrow$ ${x^2}$ = 1681 – 1600

$\Rightarrow$ ${x^2}$ = 81 $\Rightarrow$ $x = \sqrt {81} = 9$ cm

Therefore the breadth of the rectangle is 9 cm.

Perimeter of rectangle = 2(length + breadth)

= 2 (9 + 49)

= 2 x 49 = 98 cm

Hence the perimeter of the rectangle is 98 cm.

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NCERT Solutions for Class 7 Maths Exercise 6.5

Question 8.The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer:

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore, OD = $\frac{{{\text{DB}}}}{2} = \frac{{16}}{2}$ = 8 cm

And OC = $\frac{{{\text{AC}}}}{2} = \frac{{30}}{2}$ = 15 cm

Now, In right angle triangle DOC,

${\left( {{\text{DC}}} \right)^2} = {\left( {{\text{OD}}} \right)^2} + {\left( {{\text{OC}}} \right)^2}$ [By Pythagoras theorem]

$\Rightarrow$ ${\left( {{\text{DC}}} \right)^2} = {\left( 8 \right)^2} + {\left( {15} \right)^2}$

$\Rightarrow$ ${\left( {{\text{DC}}} \right)^2}$ = 64 + 225 = 289

$\Rightarrow$ DC = $\sqrt {289}$ = 17 cm

Perimeter of rhombus = 4 x side

= 4 x 17 = 68 cm

Thus, the perimeter of rhombus is 68 cm.

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