RS Aggarwal 2019,2020 solution class 7 chapter 15 Property of Triangles Exercise 15D

Exercise 15D

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Question 1:

Find the length of the hypotenuse of a right triangle, the other two sides of which measure 9 cm and 12 cm.

Answer 1:

Suppose the length of the hypotenuse is a cm.

Then, by Pythagoras theorem:

$a^2$ = $9^2 + {12}^2$
=> $a^2$ = 81 + 144
=> $a^2$ =225
=>  a = $\sqrt{225}$
=> a= 15

Hence, the length of the hypotenuse is 15 cm.

Question 2:

The hypotenuse of a right triangle is 26 cm long. If one of the remaining two sides is 10 cm long, find the length of the other side.

Answer 2:

 Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

${26}^{2 }= {10}^2 + a^2$
$$\begin{gathered} \Rightarrow a^{2 } = 676 - 100 \\ \Rightarrow a^{2 }=576 \\ \Rightarrow a^{ }=\sqrt{576 } \\ \Rightarrow a=24 \end{gathered}$$

Hence, the length of the other side is 24 cm.

Question 3:

The length of one sides of a right triangle is 4.5 cm and the length of its hypotenuse is 7.5 cm. Find the length of its third side.

Answer 3:

Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

$$\begin{gathered} 4.5^{2 }+ a^2 = 7.5^2 \\ \Rightarrow a^{2 } = 56.25 - 20.25 \\ \Rightarrow a^{2 } = 36 \\ \Rightarrow a^{ } = \sqrt{36} \\ \Rightarrow a^{ } =6 \end{gathered}$$

Hence, the length of the other side of the triangle is  6 cm.

Question 4:

The two legs of a right triangle are equal and the square of its hypotenuse is 50. Find the length of each leg.

Answer 4:

Suppose the length of the two legs of the right triangle are a cm and a cm.
Then, by Pythagoras theorem:
$$\begin{gathered} a^2 +a^2 = 50 \\ \Rightarrow 2a^2 =50 \\ \Rightarrow a^2 =25 \\ \Rightarrow a =\sqrt{25} \\ \Rightarrow a =5 \end{gathered}$$

Hence, the length of each leg is 5 cm.

Question 5:

The sides of a triangle measure 15 cm, 36 cm and 39 cm. Show that it is a right-angled triangle.

Answer 5:

The largest side of the triangle is 39 cm.

$$\begin{gathered} {15}^2 + {36}^2 \\ =225 + 1296 =1521 \end{gathered}$$

Also, ${39}^2 = 1521$
∴ ${15}^2 + {36}^2 = {39}^2$

Sum of the square of the two sides is equal to the square of the third side.

Hence, the triangle is right angled.

Question 6:

In right ∆ABC, teh lengths of its legs are given as a = 6 cm and b = 4.5 cm. Find the length of its hypotenuse.

Answer 6:

Suppose the length of the hypotenuse is c cm.
Then, by Pythagoras theorem:
$$\begin{gathered} a^2 + b^2 = c^2 \\ \Rightarrow c^{2 } = 6^2 + 4.5^2 \\ \Rightarrow c^{2 }=36 + 20.25 \\ \Rightarrow c^{2 }= 56.25 \\ \Rightarrow c=\sqrt{56.25} \\ \Rightarrow c=7.5 \end{gathered}$$

Hence, the length of its hypotenuse is 7.5 cm.

Question 7:

The lengths of the sides of triangles are given below. Which of them are right-angled?

(i) a = 15 cm, b = 20 cm and c = 25 cm
(ii) a = 9 cm, b = 12 cm and c = 16 cm
(iii) a = 10 cm, b = 24 cm and c = 26 cm

Answer 7:

(i) Largest side, c = 25 cm

We have:
$a^2 +$$b^2$ = 225 + 400 = 625

Also, $c^2$ = 625

∴ $a^2 +$$b^2$ = $c^2$
Hence, the given triangle is right angled using the Pythagoras theorem.

(ii)  Largest side, c = 16 cm
We have:
$a^2 +$$b^2$  = 81 + 144 = 225

Also,  $c^2$ = 256

Here, $a^2 +$$b^2$ $\ne$$c^2$

Therefore, the given triangle is not right angled.

(iii)  Largest side, c = 26 cm

We have:
$a^2 +$$b^2$  = 100 + 576= 676

Also, $c^2$ = 676

∴ $a^2 +$$b^2$ = $c^2$
Hence, the given triangle is right angled using the Pythagoras theorem.

Question 8:

In a ∆ABC, ∠B = 35° and ∠C = 55°. Write which of the following is true:

(i) AC² = AB² + BC²
(ii) AB² = BC² + AC²
(iii) BC² = AB² + AC²

Answer 8:

We have:
∠B = 35° and ∠C = 55°

∴ ∠B = 180 - 35 -55 = 90°    (since sum of the angles of any triangle is 180°)

We know that the side opposite to the right angle is the hypotenuse.

By Pythagoras theorem:
BC² = AB² + AC²

Hence, (iii) is true.

Question 9:

A 15-m-long ladder is placed against a wall to reach a window 12 m high. Find the distance of the foot of the ladder from the wall.

Answer 9:

By Pythagoras theorem in $△$ABC:
$A{B}^2 = A{C}^2 + B{C}^2$
$$\begin{gathered} {15}^2 = x^2 +{12}^2 \\ \Rightarrow x^2 = 225 - 144 \\ \Rightarrow x^2= 81 \\ \Rightarrow x^2=9^2 \\ \Rightarrow x=9 \end{gathered}$$
∴ x = 9 cm

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

Question 10:

A 5-m-long ladder whan set against the wall of a house reaches a height of 4.8 m. How far is the foot of the ladder from the wall?

Answer 10:

Suppose the foot of the ladder is x m far from the wall.

Let the ladder is represented by AB, the height at which it reaches the wall be AC and the distance between the foot of ladder and wall be BC.

Then, by Pythagoras theorem:
$$\begin{gathered} A{B}^2 =A{C}^2 + B{C}^2 \\ \Rightarrow 5^2 =4.8^2 + x^2 \\ \Rightarrow x^{2 }=25 -23.04 \\ \Rightarrow x^{2 }=1.96 \\ \Rightarrow x^{2 }=(1.4{)}^2 \\ \Rightarrow x=1.4 \end{gathered}$$


Hence, the foot of the ladder is 1.4 m far from the wall.

Question 11:

A tree is broken by the wind but does not separate. If the point from where it breaks is 9 m above the ground and its top touches the ground at a distance of 12 m from its foot,  find out the total height of the tree before it broke.

Answer 11:

Let BD be the height of the tree broken at point C and suppose CD take the position CA




Now as per given conditions we have AB = 9 m , BC = 12 m

By Pythagoras theorem:
${\mathrm{AC}}^2 ={\mathrm{AB}}^{2 } + {\mathrm{BC}}^2$

$$\begin{gathered} \Rightarrow {\mathrm{AC}}^2 = {12}^2 + 9^2 \\ \Rightarrow {\mathrm{AC}}^2= 144 + 81 \\ \Rightarrow {\mathrm{AC}}^2= 225 \\ \Rightarrow {\mathrm{AC}}^2={15}^2 \\ \Rightarrow \mathrm{AC}=15 \end{gathered}$$

Length of the tree before it broke = AC + AB
                                                       =  15 + 9
                                                        = 24 m

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Question 12:

Two poles, 18 m and 13 m high, stand upright in a playground. If their feet are 12 m apart, find the distance between their tops.

Answer 12:

Suppose, the two poles are AB and CD, having the length of 18 m and 13 m, respectively.
Distance between them, BD, is equal to 12 m.
We need to find AC.

From C, draw CE$\perp$AB.

AE=AB-EB
= AB-CD   (CD = EB)
= 18-13
= 5 cm
EC = BD = 12 m

Now, by Pythagoras theorem in $△AEC$:
$$\begin{gathered} {\mathrm{AC}}^2 = {\mathrm{AE}}^2 + {\mathrm{EC}}^2 \\ \Rightarrow {\mathrm{AC}}^2 = 5^2 + {12}^2 \\ \Rightarrow {\mathrm{AC}}^2 = 25+ 144 \\ \Rightarrow {\mathrm{AC}}^2 = 169 \\ \Rightarrow {\mathrm{AC}}^2={13}^2 \\ \Rightarrow \mathrm{AC}=13 \end{gathered}$$

Hence, the distance between their tops is 13 m.

Question 13:

A man goes 35 m due west and then 12 m due north. How far is he from the starting point?

Answer 13:

Suppose the man starts at point A and goes 35 m towards west, say AB. He then goes 12 m north, say BC.



We need to find AC.

By Pythagoras theorem:
${\mathrm{AC}}^2 = {\mathrm{BC}}^2 + {\mathrm{AB}}^2$
$$\begin{gathered} \Rightarrow {\mathrm{AC}}^2 = {35}^2 + {12}^2 \\ \Rightarrow {\mathrm{AC}}^2 =1225 + 144 \\ \Rightarrow {\mathrm{AC}}^2 = 1369 \\ \Rightarrow {\mathrm{AC}}^2 ={37}^2 \\ \Rightarrow \mathrm{AC}=37 \mathrm{m} \end{gathered}$$

Hence, the man is 37 m far from the starting point.

Question 14:

A man goes 3 km due north and then 4 km due east. How far is he away from his initial position?

Answer 14:

Suppose the man starts from A and goes 3 km north and reaches B.
He then goes 4 km towards east and reaches C.

∴ AB = 3 km
    BC = 4 km
We have to find AC.

By Pythagoras theorem:

$$\begin{gathered} \Rightarrow {\mathrm{AC}}^2 = {\mathrm{AB}}^2 + {\mathrm{BC}}^2 \\ \Rightarrow {\mathrm{AC}}^2 = 3^2 + 4^2 \\ \Rightarrow {\mathrm{AC}}^2 =25 \\ \Rightarrow {\mathrm{AC}}^2 =5^2 \\ \Rightarrow \mathrm{AC}=5 \mathrm{km} \end{gathered}$$

Hence, he is 5 km far from the initial position.

Question 15:

Find the length of diagonal of the rectangle whose sides are 16 cm and 12 cm.

Answer 15:

Suppose the sides are x and y of lengths 16 cm and 12 cm, respectively.

Let the diagonal be z cm.

Clearly, the diagonal is the hypotenuse of the right triangle with legs x and y.

By Pythagoras theorem:
$$\begin{gathered} z^2 = x^2 + y^2 \\ \Rightarrow z^2 = {16}^2 + {12}^2 \\ \Rightarrow z^2 =256 + 144 \\ \Rightarrow z^2 =400 \\ \Rightarrow z^2 ={20}^2 \\ \Rightarrow z=20 \end{gathered}$$

Hence, the length of the diagonal is 20 cm.

Question 16:

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

Answer 16:

AB = 40 cm
Diagonal, AC = 41 cm

Then, by Pythagoras theorem in right $△ABC$:
$$\begin{gathered} {\mathrm{AC}}^2 ={\mathrm{AB}}^2 + {\mathrm{BC}}^2 \\ \Rightarrow {\mathrm{BC}}^2 ={41}^2 -{40}^2 \\ \Rightarrow {\mathrm{BC}}^2=1681 - 1600 \\ \Rightarrow {\mathrm{BC}}^2=81 \\ \Rightarrow {\mathrm{BC}}^2=9^2 \\ \Rightarrow \mathrm{BC}=9 \mathrm{cm} \end{gathered}$$


∴ Length = 40 cm
    Breadth = 9 cm

∴ Perimeter of the rectangle = 2(length + breadth)
                                                = 2(40+9)
                                                = 98 cm

Question 17:

Find the perimeter of a rhombus, the lengths of whose diagonals are 16 cm and 30 cm.

Figure

Answer 17:

We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, in right triangle AOB, we have:
AO = 8 cm
BO = 15 cm

By Pythagoras theorem in $∆$AOB:

$$\begin{gathered} A{B}^2=A{O}^2 + B{O}^2 \\ \Rightarrow A{B}^2 =8^2 + {15}^2 \\ \Rightarrow A{B}^2 =64 +225 \\ \Rightarrow A{B}^2 =289 \\ \Rightarrow A{B}^2 ={17}^2 \\ \Rightarrow AB=17 \mathrm{cm} \end{gathered}$$

Now, as we know that all sides of a rhombus are equal.
∴ Perimeter of the rhombus = 4(side)
                                               = 4(17)
                                               = 68 cm

Question 18:

Fill in the blanks:

(i) In a right triangle, the square of the hypotenuse is equal to the ...... of the squares of the other two sides.
(ii) If the square of one side of a triangle s equal to the sum of the squares of the other two sides then the triangle is ...... .
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the ...... is the shortest.

Answer 18:

(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
(ii) If the square of one side of a triangle s equal to the sum of the squares of the other two sides then the triangle is right angled.
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.