PRACTICE PROBLEMS ON SPECIAL RIGHT TRIANGLES FOR SAT

Problem 1 :

In the figure given below, find the value of x.

Solution :

In triangle ABD,

BD2 = AB+ AD2

BD2 = √7+ 32

BD2 = + 9

BD2 = 16

BD = 4

In triangle BDC,

BC2 = BD+ DC2

x2 = 4+ 32

x2 = 16 + 9

x2 = 25

x = 5

Problem 2 :

In the figures below, find the values of x and y.

Solution :

(a) Since it is 45-45-90 triangle is an isosceles triangle, the value of x is 2.

In 45-45-90 right triangle, hypotenuse side = √2 ⋅ leg.

y = 2√2

(b) In a 30-60-90 right triangle,

longer leg = √3 ⋅ shorter leg

3 = √3 x

x = 3/√3

hypotenuse = 2⋅ shorter leg

y = 2x

y = 2(3/√3)

y = 2√3

Problem 3 :

In the figure above, if AD = BD = 2√3, what is the length of AB?

Solution :

AD = BD = 2√3  (Given)

ADC = 30° + 30° (Using exterior angle theorem)

ADC = 60°

In triangle ADC,

CAD + CDA + ACD = 180°

CAD + 60° + 90° = 180°

CAD = 180°-150°

CAD = 30°

In triangle ACD, sine CAD = 30°, its is a 30-60-90 triangle.

AD is hypotenuse and CD is the shorter leg.

hypotenuse (AD) = 2√3 (2 times the shorter leg)

CD = √3

The longer leg = (√3 times the shorter leg)

The longer leg(AC) = √3 √3

AC = 3

In triangle ABC,

BC = CD + BD

BC = √3 + 2√3 = 3√3

AB2 = AC2 + BC2

AB2 = 32 + (3√3)2

AB2 = 9 + 27

AB2 = 36

AB = 6

So, length of AB is 6.

Problem 4 :

In the figure given above AB = 6, BC = 8 and CD = 5. What is the length of AD?

Solution :

In triangle ABC,

AC2 = AB2 + BC2

AC2 = 62 + 82

AC2 = 36 + 64

AC = 10

In triangle ADC,

AC2 = AD2 + DC2

102 = AD2 + 52

AD2 = 100 - 25

AD = √75

AD = 5√3

Problem 5 :

In the above triangle ABC, BD = √3. What is the perimeter of triangle ABC.

Solution :

In triangle BDC,

DBC  =  30

The side which is opposite to smaller angle is shorter, then, let DC = x.

BC = (2 times the shorter leg) = 2x

BC2 = BD+ DC2

(2x)2 = √3+ x2

3x3

x2 = 1

BC = 2(1)

= 2

DC = 1

In triangle ABD,

BD = √3 (Side which is opposite to smaller angle)

AB = 2√3 

AB2 = AD+ BD2

(2√3)2 =AD2+(√3)2

4(3) - 3 AD2

12 - 3 = AD2

9 = AD2

3 = AD

CA = AD + DC

= 3 + 1

CA = 4

Perimeter of triangle ABC = AB + BC + CA

2√3 + 2 + 4

2√3 + 6

Problem 6 :

In the triangle above,  C and BD bisects AC. What is the perimeter of triangle ABC ?

Solution :

Since BD is the bisector of AC,

AD = DC = 7

BD = 4√2

In triangle BDC,

BC2 = BD+ DC2

BC2 = (4√2)+ 72

BC2 = 16(2) + 49

BC2 = 81

BC = 9

AB = 9

Perimeter of triangle ABC = AB + BC + CA

= 9 + 9 + 14

= 32

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Videos (Part 7 - Calculator)

    Jun 13, 24 03:21 AM

    satmathrectangle.png
    SAT Math Videos (Part 7 - Calculator)

    Read More

  2. SAT Math Videos - Part 8 (Calculator)

    Jun 13, 24 02:35 AM

    SAT Math Videos - Part 8 (Calculator)

    Read More

  3. Velocity and Acceleration

    Jun 13, 24 02:12 AM

    velocityandacceleration1
    Velocity and Acceleration

    Read More