PRACTICE PROBLEMS ON SPECIAL RIGHT TRIANGLE FOR SAT

Problem 1 :

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

Solution :

In triangle ABD,

BD²  =  AB²+AD²

BD²  =  √7²+3²

BD²  =  7+9

BD²  =  16

BD  =  4

In triangle BDC,

BC²  =  BD²+DC²

x²  =  4²+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

If CD  =  x (The side opposite to smaller angle)

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 small, then

let x  =  DC

BC  =  (2 times the shorter leg)  ==>  2x

BC²  =  BD²+DC²

(2x)²  =  √3²+x²

3x² =  3

x²  =  1

BC  =  2(1)  ==>  2

DC  =  1

In triangle ABD,

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

AB  =  2√3 

AB²  =  AD²+BD²

(2√3)²  =  AD²+(√3)²

4(3) - 3  =  AD²

AD²  =  9

AD  =  3

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, <A  ≅  <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  =  BD²+DC²

BC²  =  (4√2)²+7²

BC²  =  16(2)+49

BC²  =  81

BC  =  9

AB  =  9

Perimeter of triangle ABC  =  AB + BC + CA

=  9+9+14

=  32

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