HIGH SCHOOL MATH QUESTIONS WITH ANSWERS

Problem 1 :

Find the centroid of the triangle whose vertices are the points A (8 , 4) B (1 , 3) and C (3 , -1).

Solution :

Centroid of the triangle  =  (x1 +x2 + x3)/3, (y1+y2+y3)/3

=  (8+1+3)/3, (4+3-1)/3

=  12/3, 6/3

=  (4, 2)

So, the centroid of the triangle is (4, 2).

Problem 2 :

If the two lines are perpendicular with the slopes m1 and m2 then m1  m2 =

Solution :

If two lines are perpendicular, then the product of their slopes will be equal to -1.

Problem 3 :

Find the coordinates of the orthocenter of the triangle whose vertices are (3, 1)(0, 4) and (-3, 1).

Solution :

Slope of AC  =  [(y2 - y1)/(x2 - x1)]

A (3, 1) and C (-3, 1)

here x1  =  3, x2  =  -3, y1  =  1 and y2  =  1

=  (1 - 1) / (-3 - 3)

=  0 / (-6)

=  0

Slope of the altitude BE  =  -1/ slope of AC

=  -1/0

Equation of the altitude BE :

(y - y1)  =  m (x -x1)

Here B (0, 4) and m  =  -1/0

(y - 4)  =  -1/0 (x - 0)

10 (y - 4)  =  -1(x)

x + 10y - 40  =  0  --------(1)

Slope of BC  =  (y2 - y1)(x2 - x1)]

B (0, 4) and C (-3, 1)

here x₁ = 0,x₂ = -3, y₁ = 4 and y₂ = 1

=  (1 - 4) / (-3 - 0)

=  (-3)/(-3)

=  1

Slope of the altitude AD  =  -1/ slope of AC

=  -1/1

=  -1

Equation of the altitude AD :

(y - y1)  =  m (x - x1)

Here A(3, 1)  m  =  -1

(y - 1)  =  -1 (x - 3)

(y - 1)  =  -x + 3

x + y - 1 - 3  =  0

x + y - 4  =  0

x  =  -y + 4--------(2)

Substituting (2) into (1), we get

-y + 4 + 10y - 40  =  0

9y - 36  =  0

y  =  36/9

y  =  4

By applying y  =  4 in (1), we get

x  =  -4 + 4

x  =  0

Therefore the orthocenter is (0, 4).

Problem 4 :

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, determine the sides of the two squares.

Solution :

Let x and y be the side length of squares.

x2 + y2  =  468  ----(1)

4x - 4y  =  24

x - y  =  6

x  =  6 + y ----(2)

(6+y)2 + y2  =  468

36+y2+12y+y2-468  =  0

2y2 + 12y - 432  =  0

y2 + 6y - 216  =  0

(y - 12)(y + 18)  =  0

y  =  12 and y  =  -18 (not admissible)

If y =  12, then x  =  18

So, the side length of required squares are 12 and 18 respectively.

Problem 5 :

Two concentric circles are of radii 5 cm and 3 cm. Determine the length of the chord of the larger circle which touches the smaller circle.

Solution :

In triangle OCB,

OB2  =  OC2 + BC2

52  =  32 + BC2

BC2  =  25 - 9

BC2  =  16

BC  =  4

Problem 6 :

If sin A = 3/4, then the value of tan A

Solution :

Given,

sin A = 3/4  =  Opposite side/Hypotenuse

Adjacent side2  =   Hypotenuse2 - Opposite side2

Adjacent side2  =   42 - 32

Adjacent side2  =  16 - 9

tan A  =  Opposite side / Adjacent side

tan A  =  3/7

Problem 7 :

In triangle PQR right-angled at Q , PQ = 3 cm and PR = 6 cm. Determine angle QPR

tan p  =  Adjacent side / Hypotenuse

tan p  =  3/6

tan p  =  1/2

Problem 8 :

Simplify the following.

=  2tan 30 / (1 + tan 30)

=  2 (1/√3) / (1 + 1/√3)

=  2/(√3 + 1)

=  2/(√3 + 1) ⋅ [(√3 - 1)/(√3 - 1)]

=  2(√3 - 1)/(√32 - 1)

=  2(√3 - 1)/2

=  √3 - 1

Problem 9 :

Evaluate the following

Solution :

sin60 cos30 + sin30 cos60

=  √3/2 (√3/2) + (1/2)(1/2)

=  3/4 + 1/4

=  4/4

=  1

Problem 10 :

Simplify (sec A + tan A)  (1 - sin A) =

Solution :

(sec A + tan A)  (1 - sin A)

=  (1/cos A + sin A/cos A) (1-sinA)

=  [(1+sinA)/cos A] (1-sinA)

=  (1 - sin2A)/cos A

=  cos2A/cos A

=  cos A

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