**8th grade math worksheet 1 answer :**

Here we are going to see 10 practice questions of mixed topics. We have been preparing this set of questions in order to test how you understood the topics.

**Question 1 :**

What is the value of square root of 81?

(A) 81 (B) 3 (C) 9

**Solution :**

√81 = √(3 ⋅ 3 ⋅ 3 ⋅ 3)

= 3 **⋅ **3

= 9

Let us see the solution of next problem on "8th grade math worksheet 1 answer".

**Question 2 :**

Find the area of a right angled triangle, if the radius of the circumcircle is 5 cm and altitude to the hypotenuse is 4 cm.

(A) 20 cm^{2 }(B) 10 cm^{2 }(C) 30 cm^{2}

**Solution :**

Let ABC be the right angled triangle right angled at B. Let O be the center of the circumcircle.

Then, O is the midpoint of the hypotenuse AC.

Hence OA = OB = OC (radius of circle 5 cm)

Hypotenuse AC = Diameter of the circle

= 2 radius of cicumcircle

= 2 **⋅** 5 = 10 cm

Let BM be the perpendicular from B on AC.

MB = 4 cm

Area of right angled triangle ABC

= (1/2) **⋅** Base **⋅** Height

= (1/2) **⋅** 10 **⋅** 4

= 20 cm^{2}

**Question 3 :**

The volume of cubical box is 46.656 cubic meters. Find the length of the side of the cubical box.

(A) 3.6 m^{ }(B) 5.2 m^{ }(C) 7.8 m

**Solution :**

Volume of cubical box = 46.656 cubic meters

a^{3} = 46.656

a = 3.6

Hence side length of cubical box = 3.6 meter

**Question 4 :**

The sum of the digits of a two digit number is 15 and if 9 is added to the number the digits are interchanged. Find the required number.

(A) 78 (B) 87 (C) 96

**Solution :**

Let "xy" be the two digit number

Sum of the two digit number = 15

x + y = 15 -----(1)

If 19 is added to the number the digits are interchanged.

xy + 9 = yx

10x + 1y + 9 = 10y + 1x

10x - x + 1y - 10y = -9

9x - 9y = -9 -----(2)

(1)⋅9 ==> 9x+9y = 135 9x-9y = -9 ___________ 18x = 126 x = 7 |
Applying the value of x in the first equation, we get 7 + y = 15 Subtract 7 on both sides, y = 15 - 7 y = 8 |

Hence the required two digit number is 78.

**Question 5 :**

In a circle with center O and radius 17 cm, PQ is a chord at a distance of 8 cm from the center. Find the length of the chord.

(A) 30 cm (B) 15 cm (C) 24 cm

**Solution :**

In a right triangle OCQ,

OQ^{2} = OC^{2} + CQ^{2}

17^{2} = 8^{2} + CQ^{2}

289 - 64 = CQ^{2}

CQ = √225 ==> CQ = 15

PQ = 2 CQ

= 2(15) = 30 cm

Hence the length of chord is 30 cm.

**Question 6 :**

The radius and height of a cylinder are in the ratio 5:7 and its volume is 550 cm³ find its radius.

(A) 5 cm (B) 6 cm (C) 7 cm

**Solution :**

r : h = 5 : 7

r/h = 5/7

r = 5h/7

Volume of cylinder = 550 cm³

∏r^{2} h = 550

(22/7) **⋅ (**5h/7)^{2} **⋅ **h = 550

h^{3} = 550 **⋅ **(7/22) **⋅ **(49/25)

h^{3} = 343

h = 7

r = 5

**Question 7 :**

The mean weight of 4 boys is 56 kg and that of 6 girls in 46 kg. find the combined mean weight of 10 students.

(A) 40 kg (B) 36 kg (C) 50 kg

**Solution :**

Mean weight of 4 boys = 56 kg

Weight of 4 boys = 56 (4) = 224

Mean weight of 6 girls = 46 kg

Weight of 6 girls = 46 (6) = 276

Total weight = 224 + 276

= 500

Average weight = 500/10 = 50 kg

**Question 8 :**

The arithmetic mean of 8, 10, x and 12 is 9. Find the value of x.

(A) 6 (B) 8 (C) 9

**Solution :**

Arithmetic mean = 9

(8 + 10 + x + 12)/4 = 9

30 + x = 9(4)

30 + x = 36

Subtract 30 on both sides

x = 36 - 30

x = 6

**Question 9 :**

Find the number of sides of regular polygon whose exterior angle measure of 45 degree.

**Solution :**

Sum of the exterior angles of regular polygon = 360°

But each exterior angle = 45°

number of sides of regular polygon = 360° / 45° = 8.

**Question 10 :**

Simplify the expressions and evaluate them as directed.

x (x-3) + 2 for x = 1

**Solution :**

x (x - 3) + 2

x = 1

= 1 ( 1 - 3) + 2

= 1 (-2) + 2

= -2 + 2

= 0

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