NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.5

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

Ans: Given: $PQ = 10$cm, $PR = 24$cm

Let $QR$ be $x$cm.

In right angled triangle $QPR$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

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

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

${\left( x \right)^2} = 100 + 576 = 676$

$x = \sqrt {676}  = 26cm$

Thus, the length of  $QR$ is $26cm$.

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

Ans: Given: $AB = 25$cm, $AC = 7$cm

Let $BC$ be $x$cm.

In right angled triangle $ACB$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

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

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

$625 = 49 + {x^2}$

${x^2} = 625 - 49 = 576$

$x = \sqrt {576}  = 24cm$

Thus, the length of  $BC$is $24cm$.

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.

Ans: Let $AC$ be the ladder and $A$ be the window.

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

Let $CB$ be $a$m.

In right angled triangle $ACB$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

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

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

$225 = {a^2} + 144$

${a^2} = 225 - 144 = 81$

$x = \sqrt {81}  = 9m$

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

4. Which of the following can be the sides of a right triangle? In the case of right angled triangles, identify the right angles.

(i) $2.5$cm, $6.5$cm, $6$cm

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

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

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

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

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

Since, $L.H.S. = R.H.S.$ 

Therefore, the given sides form a 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

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

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$ 

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

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

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

Since, $L.H.S. \ne R.H.S.$ 

Therefore, the given sides do not form a right angled triangle.

(iii) $1.5$cm, $2$cm, $2.5$cm

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

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

In $\vartriangle PQR$,               ${\left( {\operatorname{P} R} \right)^2} = {\left( {PQ} \right)^2} + {\left( {RQ} \right)^2}$ 

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

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

Since, $L.H.S. = R.H.S.$ 

Therefore, the given sides form a right angled triangle.

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

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.

Ans: 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 $\vartriangle ABC$ is a right angled triangle, right angled at $B$.

$AB = 12$m and $BC = 5$m

Using Pythagoras theorem, In $\vartriangle ABC$

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

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

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

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

$AC = 13m$

The total height of the tree is sum of sides $AC$ and $CB$ that is$AC + CB = 13 + 5 = 18m$.

6. Angles $Q$ and $R$ of a $\vartriangle PQR$ are ${25^ \circ }$ and ${65^ \circ }$.

Write which of the following is true:

(i) $P{Q^2} + Q{R^2} = R{P^2}$ 

(ii) $P{Q^2} + R{P^2} = Q{R^2}$

(iii) $R{P^2} + Q{R^2} = P{Q^2}$

Ans: In $\vartriangle PQR$,

$\angle PQR + \angle QRP + \angle RPQ = {180^ \circ }$                 By Angle sum property of a [$\vartriangle $]

${25^ \circ } + {65^ \circ } + \angle RPQ = {180^ \circ }$              $ \Rightarrow $          ${90^ \circ } + \angle RPQ = {180^ \circ }$ 

$\angle RPQ = {180^ \circ } - {90^ \circ } = {90^ \circ }$

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

$\therefore {\text{       Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$       (By Pythagoras theorem)

$Q{R^2} = P{Q^2} + R{P^2}$

Hence, Option (ii) is correct.

7. Find the perimeter of the rectangle whose length is \[40\]cm and a diagonal is $41$cm.

Ans: Given diagonal $PR = 41$cm, length \[PR = 40\]cm

Let breadth $\left( {QR} \right)$ be $x$cm.

Now, in right angled triangle $PQR$,

${\left( {PR} \right)^2} = {\left( {RQ} \right)^2} + {\left( {PQ} \right)^2}$                         (By Pythagoras theorem)

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

$1681 = {\left( x \right)^2} + 1600$

${x^2} = 1681 - 1600$

${x^2} = 81$

$x = \sqrt {81}  = 9cm$ 

Therefore the breadth of the rectangle is $9$cm.

${\text{Perimeter of rectangle  = 2}}\left( {{\text{length + breadth}}} \right)$ 

$ = 2\left( {9 + 49} \right)$ 

$ = 2 \times 49 = 98cm$ 

Hence the perimeter of the rectangle is $98$cm.

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

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

Since the diagonals of the rhombus bisect at right angle to each other so it divides the diagonals into equal halves.

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

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

Now, In right angle triangle $DOC$,

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

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

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

$DC = \sqrt {289}  = 17cm$ 

${\text{Perimeter of rhombus  = }}4 \times {\text{Side}}$ 

$ = 4 \times 17 = 68cm$ 

Thus, the perimeter of rhombus is $68$ cm.

Conclusion

In conclusion, Exercise 6.5 of Chapter 6 in Class 7 Maths is crucial for mastering the properties of right-angled triangles, especially the Pythagoras Theorem. Through this exercise, students gain practical experience in applying the theorem to find missing sides and verify given triangles. By working through these problems, learners not only strengthen their understanding of geometric principles but also build a solid foundation for more advanced mathematical concepts. Mastery of these skills is essential for academic success in geometry and related fields.

From the analysis of previous exams, the number of questions asked from Class 7 Maths Chapter 6 Exercise 6.5 typically ranges from 2 to 4 questions per exam. These questions cover various aspects of the chapter, including types of triangles, properties of triangles, the Pythagoras theorem, and application-based problems involving angle sum property and exterior angle property​

Class 7 Maths Chapter 6: Exercises Breakdown

CBSE Class 7 Maths Chapter 6 Other Study Materials

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.