The DAV Maths Book Class 7 Solutions Pdf and **DAV Class 7 Maths Chapter 8 Worksheet 4** Solutions of Triangle and Its Properties offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 8 WS 4 Solutions

Question 1.

The hypotenuse of a right triangle is 17 cm long. If one of the remaining two sides is of length 8 cm, find the length of the third side. A

Answer:

In right ΔABC, ∠B = 90°

∴ From Pythagoras Theorem,

AB^{2} + BC^{2} = AC^{2}

⇒ AB^{2} + (8)^{2} = (17)^{2}

⇒ AB^{2} + 64 = 289

⇒ AB^{2} = 289 – 64 = 225

⇒ AB = 15cm

Hence the third side is 15 cm.

Question 2.

The length of the hypotenuse of a right triangle is 13 cm. If one of the side of the triangle be 5cm long, find the length of the other side.

Answer:

In the figure given alongside,

ABC is a right triangle in which

∠B = 90°

∴ From Pythagoras theorem, we get

AB^{2} + BC^{2} = AC^{2}

⇒ AB^{2} + (5)^{2} = (13)^{2}

⇒ AB^{2} + 25 = 169

⇒ AB^{2} = 169 – 25

⇒ AB^{2} = 144

AB = 12 cm

Hence the required side = 12 cm

Question 3.

The length of two sides of a right triangle are equal. The square of hypotenuse is 800 cm^{2}. Find the length of each side.

Answer:

In right triangle ABC, ∠B = 90°

∴ AC^{2} = 800

AB = AC (given)

∴ From Pythagoras theorem, we get

AB^{2} + BC^{2} = AC^{2}

⇒ AB^{2} + AB^{2} = AC^{2}

⇒ 2AB^{2} = AC^{2}

⇒ 2AB^{2} = 800

⇒ AB^{2} = \(\frac{800}{2}\) = 400

∴ AB = \(\sqrt{400}\) = 20 cm

Hence AB = BC = 20 cm

Question 4.

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Answer:

In the figure given alongside AB = 6m, CD = 11m and BD = 12m

Draw AE ∥ BD

CE = CD – ED

= CD – AB = 11m – 6m = 5m

and BD = AE = 12 m

From Pythagoras theorem, in right triangle AEC

AE^{2} + CE^{2} = AC^{2}

⇒ (12)^{2} + (5)^{2} = AC^{2}

⇒ 144 + 25 = AC^{2}

⇒ 169 = AC^{2}

∴ AC = \(\sqrt{169}\) = 13 m

Hence the distance between their tops is 13 m.

Question 5.

Find the length of the diagonal of a rectangle whose sides are 15 cm and 8 cm.

Answer:

Here is the given figure AB = 15 cm, BC = 8 cm

In right triangle ABC,

AB^{2} + BC^{2} = AC^{2} (From Pythagoras theorem)

⇒ (15)^{2} + (8)^{2} = AC^{2}

⇒ 225 + 64 = AC^{2}

⇒ 289 = AC^{2}

AC = \(\sqrt{289}\) = 17 cm

Hence the length of the diagonal = 17 cm.

Question 6.

The foot of a ladder is 6 m away from a wall and its top reaches a window 8m above the ground. If ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

Answer:

From Fig. (i),

AC^{2} = AB^{2} + BC^{2} (From Pythagoras theorem)

AC^{2} = (8)^{2} + (6)^{2}

AC^{2} = 64 + 36 = 100

AC = \(\sqrt{100}\) = 10 m

Now in Fig. (ii),

DE = 10 m, EF = 8 m

∴ DE^{2} + EF^{2} = DF^{2} (from Pythagoras theorem)

⇒ DE^{2} + (8)^{2} = (10)^{2}

⇒ DE^{2} + 64 = 100

⇒ DE^{2} = 100 – 64 = 36

∴ DE = \(\sqrt{36}\) = 6 m

Hence the required height is 6 m.

Question 7.

A right angled triangle is isosceles. If the square of the hypotenuse is 50 m2, what is the length of each of its sides?

Answer:

From the given figure

AC^{2} = 50 m^{2}

AB = BC

∴ From Pythagoras theorem, we get

AB^{2} + BC^{2} = AC^{2}

⇒ AB^{2} + AB^{2} = 50

⇒ 2AB^{2} = 50

⇒ AB^{2} = \(\frac{50}{2}\) = 25

∴ AB = \(\sqrt{25}\) = 5 m

Hence AB = BC = 5m.

Question 8.

If the sides of a triangle are 3m, 4m and 6m long, determine whether the triangle is a right angled triangle.

Answer:

Here, (6)^{2} = 36

(4)^{2} = 16 and (3)^{2} = 9

16 + 9 ≠ 36

Hence the given triangle is not a right triangle.

Question 9.

Verify whether the following triplets are Pythagorean or not.

(i) (8,9,10)

Answer:

8,9,10

(10)^{2} = 100, (9)^{2} = 81, (8)^{2} = 64

100 ≠ 81 + 64

100 ≠ 145

Hence it is not a Pythagorean triplet.

(ii) (5,7,12)

Answer:

5, 7, 12

(12)^{2} = 144, (7)^{2} = 49, (5)^{2} = 25

144 ≠ 49 + 25

144 ≠ 74

Hence it is not a Pythagorean triplet.

(iii) (3,4,5)

Answer:

3, 4, 5

(5)^{2} = 25, (4)^{2} = 16, (3)^{2} = 9

25 = 16 + 9

25 = 25

Hence it is a Pythagorean triplet.

(iv) (5,12,13)

Answer:

5, 12, 13

(13)^{2} = 169, (12)^{2} = 144, (5)^{2} = 25

169 = 144 + 25

169 = 169

Hence it is a Pythagorean triplet.

(v) (9, 12, 15)

Answer:

9, 12, 15

(15)^{2} = 225, (12)^{2} = 144, (9)^{2} = 81

225 = 144 + 81

225 = 225

Hence it is a Pythagorean triplet.